Factoring quadratics
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Factoring quadratic expressions
Factoring Quadratic Expressions
Discussion and questions for this video
 In this video I want to do a bunch of examples of factoring
 a second degree polynomial, which is
 often called a quadratic.
 Sometimes a quadratic polynomial, or just a
 quadratic itself, or quadratic expression, but all it means
 is a second degree polynomial.
 So something that's going to have a variable raised to the
 second power.
 In this case, in all of the examples we'll do, it'll be x.
 So let's say I have the quadratic expression, x
 squared plus 10x, plus 9.
 And I want to factor it into the product of two binomials.
 How do we do that?
 Well, let's just think about what happens if we were to
 take x plus a, and multiply that by x plus b.
 If we were to multiply these two things, what happens?
 Well, we have a little bit of experience doing this.
 This will be x times x, which is x squared, plus x times b,
 which is bx, plus a times x, plus a times b plus ab.
 Or if we want to add these two in the middle right here,
 because they're both coefficients of x.
 We could right this as x squared plus I can write it
 as b plus a, or a plus b, x, plus ab.
 So in general, if we assume that this is the product of
 two binomials, we see that this middle coefficient on the
 x term, or you could say the first degree coefficient
 there, that's going to be the sum of our a and b.
 And then the constant term is going to be the product
 of our a and b.
 Notice, this would map to this, and this
 would map to this.
 And, of course, this is the same thing as this.
 So can we somehow pattern match this to that?
 Is there some a and b where a plus b is equal to 10?
 And a times b is equal to 9?
 Well, let's just think about it a little bit.
 What are the factors of 9?
 What are the things that a and b could be equal to?
 And we're assuming that everything is an integer.
 And normally when we're factoring, especially when
 we're beginning to factor, we're dealing
 with integer numbers.
 So what are the factors of 9?
 They're 1, 3, and 9.
 So this could be a 3 and a 3, or it could be a 1 and a 9.
 Now, if it's a 3 and a 3, then you'll have 3 plus 3 that
 doesn't equal 10.
 But if it's a 1 and a 9, 1 times 9 is 9.
 1 plus 9 is 10.
 So it does work.
 So a could be equal to 1, and b could be equal to 9.
 So we could factor this as being x plus 1,
 times x plus 9.
 And if you multiply these two out, using the skills we
 developed in the last few videos, you'll see that it is
 indeed x squared plus 10x, plus 9.
 So when you see something like this, when the coefficient on
 the x squared term, or the leading coefficient on this
 quadratic is a 1, you can just say, all right, what two
 numbers add up to this coefficient right here?
 And those same two numbers, when you take their product,
 have to be equal to 9.
 And of course, this has to be in standard form.
 Or if it's not in standard form, you should put it in
 that form, so that you can always say, OK, whatever's on
 the first degree coefficient, my a and b
 have to add to that.
 Whatever's my constant term, my a times b, the product has
 to be that.
 Let's do several more examples.
 I think the more examples we do the more
 sense this'll make.
 Let's say we had x squared plus 10x, plus well, I
 already did 10x, let's do a different number x squared
 plus 15x, plus 50.
 And we want to factor this.
 Well, same drill.
 We have an x squared term.
 We have a first degree term.
 This right here should be the sum of two numbers.
 And then this term, the constant term right here,
 should be the product of two numbers.
 So we need to think of two numbers that, when I multiply
 them I get 50, and when I add them, I get 15.
 And this is going to be a bit of an art that you're going to
 develop, but the more practice you do, you're going to see
 that it'll start to come naturally.
 So what could a and b be?
 Let's think about the factors of 50.
 It could be 1 times 50.
 2 times 25.
 Let's see, 4 doesn't go into 50.
 It could be 5 times 10.
 I think that's all of them.
 Let's try out these numbers, and see if any of
 these add up to 15.
 So 1 plus 50 does not add up to 15.
 2 plus 25 does not add up to 15.
 But 5 plus 10 does add up to 15.
 So this could be 5 plus 10, and this could be 5 times 10.
 So if we were to factor this, this would be equal to x plus
 5, times x plus 10.
 And multiply it out.
 I encourage you to multiply this out, and see that this is
 indeed x squared plus 15x, plus 10.
 In fact, let's do it. x times x, x squared. x
 times 10, plus 10x.
 5 times x, plus 5x.
 5 times 10, plus 50.
 Notice, the 5 times 10 gave us the 50.
 The 5x plus the 10x is giving us the 15x in between.
 So it's x squared plus 15x, plus 50.
 Let's up the stakes a little bit, introduce some negative
 signs in here.
 Let's say I had x squared minus 11x, plus 24.
 Now, it's the exact same principle.
 I need to think of two numbers, that when I add them,
 need to be equal to negative 11.
 a plus b need to be equal to negative 11.
 And a times b need to be equal to 24.
 Now, there's something for you to think about.
 When I multiply both of these numbers, I'm getting a
 positive number.
 I'm getting a 24.
 That means that both of these need to be positive, or both
 of these need to be negative.
 That's the only way I'm going to get a positive number here.
 Now, if when I add them, I get a negative number, if these
 were positive, there's no way I can add two positive numbers
 and get a negative number, so the fact that their sum is
 negative, and the fact that their product is positive,
 tells me that both a and b are negative.
 a and b have to be negative.
 Remember, one can't be negative and the other one
 can't be positive, because the product would be negative.
 And they both can't be positive, because when you add
 them it would get you a positive number.
 So let's just think about what a and b can be.
 So two negative numbers.
 So let's think about the factors of 24.
 And we'll kind of have to think of the negative factors.
 But let me see, it could be 1 times 24, 2 times 11, 3 times
 8, or 4 times 6.
 Now, which of these when I multiply these well,
 obviously when I multiply 1 times 24, I get 24.
 When I get 2 times 11 sorry, this is 2 times 12.
 I get 24.
 So we know that all these, the products are 24.
 But which two of these, which two factors, when I add them,
 should I get 11?
 And then we could say, let's take the
 negative of both of those.
 So when you look at these, 3 and 8 jump out.
 3 times 8 is equal to 24.
 3 plus 8 is equal to 11.
 But that doesn't quite work out, right?
 Because we have a negative 11 here.
 But what if we did negative 3 and negative 8?
 Negative 3 times negative 8 is equal to positive 24.
 Negative 3 plus negative 8 is equal to negative 11.
 So negative 3 and negative 8 work.
 So if we factor this, x squared minus 11x, plus 24 is
 going to be equal to x minus 3, times x minus 8.
 Let's do another one like that.
 Actually, let's mix it up a little bit.
 Let's say I had x squared plus 5x, minus 14.
 So here we have a different situation.
 The product of my two numbers is negative, right? a times b
 is equal to negative 14.
 My product is negative.
 That tells me that one of them is positive, and one of them
 is negative.
 And when I add the two, a plus b, it'd be equal to 5.
 So let's think about the factors of 14.
 And what combinations of them, when I add them, if one is
 positive and one is negative, or I'm really kind of taking
 the difference of the two, do I get 5?
 So if I take 1 and 14 I'm just going to try out things
 1 and 14, negative 1 plus 14 is negative 13.
 Negative 1 plus 14 is 13.
 So let me write all of the combinations that I could do.
 And eventually your brain will just zone in on it.
 So you've got negative 1 plus 14 is equal to 13.
 And 1 plus negative 14 is equal to negative 13.
 So those don't work.
 That doesn't equal 5.
 Now what about 2 and 7?
 If I do negative 2 let me do this in a different color if
 I do negative 2 plus 7, that is equal to 5.
 We're done!
 That worked!
 I mean, we could have tried 2 plus negative 7, but that'd be
 equal to negative 5, so that wouldn't have worked.
 But negative 2 plus 7 works.
 And negative 2 times 7 is negative 14.
 So there we have it.
 We know it's x minus 2, times x plus 7.
 That's pretty neat.
 Negative 2 times 7 is negative 14.
 Negative 2 plus 7 is positive 5.
 Let's do several more of these, just to really get well
 honed this skill.
 So let's say we have x squared minus x, minus 56.
 So the product of the two numbers have to be minus 56,
 have to be negative 56.
 And their difference, because one is going to be positive,
 and one is going to be negative, right?
 Their difference has to be negative 1.
 And the numbers that immediately jump out in my
 brain and I don't know if they jump out in your brain,
 we just learned this in the times tables
 56 is 8 times 7.
 I mean, there's other numbers.
 It's also 28 times 2.
 It's all sorts of things.
 But 8 times 7 really jumped out into my brain, because
 they're very close to each other.
 And we need numbers that are very close to each other.
 And one of these has to be positive, and one of these has
 to be negative.
 Now, the fact that when their sum is negative, tells me that
 the larger of these two should probably be negative.
 So if we take negative 8 times 7, that's
 equal to negative 56.
 And then if we take negative 8 plus 7, that is equal to
 negative 1, which is exactly the coefficient right there.
 So when I factor this, this is going to be x minus 8,
 times x plus 7.
 This is often one of the hardest concepts people learn
 in algebra, because it is a bit of an art.
 You have to look at all of the factors here, play with the
 positive and negative signs, see which of those factors
 when one is positive, one is negative, add up to the
 coefficient on the x term.
 But as you do more and more practice, you'll see that
 it'll become a bit of second nature.
 Now let's step up the stakes a little bit more.
 Let's say we had negative x squared everything we've
 done so far had a positive coefficient, a positive 1
 coefficient on the x squared term.
 But let's say we had a negative x squared
 minus 5x, plus 24.
 How do we do this?
 Well, the easiest way I can think of doing it is factor
 out a negative 1, and then it becomes just like the problems
 we've been doing before.
 So this is the same thing as negative 1 times positive x
 squared, plus 5x, minus 24.
 Right?
 I just factored a negative 1 out.
 You can multiply negative 1 times all of these, and you'll
 see it becomes this.
 Or you could factor the negative 1 out and divide all
 of these by negative 1.
 And you get that right there.
 Now, same game as before.
 I need two numbers, that when I take their product I get
 negative 24.
 So one will be positive, one will be negative.
 When I take their sum, it's going to be 5.
 So let's think about 24 is 1 and 24.
 Let's see, if this is negative 1 and 24, it'd be positive 23,
 if it was the other way around, it'd be negative 23.
 Doesn't work.
 What about 2 and 12?
 Well, if this is negative remember, one of these has to
 be negative.
 If the 2 is negative, their sum would be 10.
 If the 12 is negative, their sum would be negative 10.
 Still doesn't work.
 3 and 8.
 If the 3 is negative, their sum will be 5.
 So it works!
 So if we pick negative 3 and 8, negative 3 and 8 work.
 Because negative 3 plus 8 is 5.
 Negative 3 times 8 is negative 24.
 So this is going to be equal to can't forget that
 negative 1 out front, and then we factor the inside.
 Negative 1 times x minus 3, times x plus 8.
 And if you really wanted to, you could multiply the
 negative 1 times this, you would get 3
 minus x if you did.
 Or you don't have to.
 Let's do one more of these.
 The more practice, the better, I think.
 All right, let's say I had negative x squared
 plus 18x, minus 72.
 So once again, I like to factor out the negative 1.
 So this is equal to negative 1 times x squared,
 minus 18x, plus 72.
 Now we just have to think of two numbers, that when I
 multiply them I get positive 72.
 So they have to be the same sign.
 And that makes it easier in our head, at least in my head.
 When I multiply them, I get positive 72.
 When I add them, I get negative 18.
 So they're the same sign, and their sum is a negative
 number, they both must be negative.
 And we could go through all of the factors of 72.
 But the one that springs up, maybe you think of 8 times 9,
 but 8 times 9, or negative 8 minus 9, or negative 8 plus
 negative 9, doesn't work.
 That turns into 17.
 That was close.
 Let me show you that.
 Negative 9 plus negative 8, that is equal to negative 17.
 Close, but no cigar.
 So what other ones are there?
 We have 6 and 12.
 That actually seems pretty good.
 If we have negative 6 plus negative 12, that is equal to
 negative 18.
 Notice, it's a bit of an art.
 You have to try the different factors here.
 So this will become negative 1 don't want to forget
 that times x minus 6, times x minus 12.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

Have something that's not a question about this content? 
This discussion area is not meant for answering homework questions.
Can both a and b equal the same for example x^2+20x+100 can both a and b equal 10?
Yes the number does not have to be different.
Yes, both a and b may be 10 in that example. When both a and b are equal, it is called a perfect square trinomial. This is where the first and third terms are perfect squares, and the second term is twice the product of the square root of the other two terms. the square root of x^2 is x. the square root of 100 is ten. ten times x is 10x. Multiplied by two is 20x.
the numbers don't have to be different so yes!!
;)
;)
YA the number doesn't have to be the same as the other one
yes both them can be the same.
yes A and B can be the same.
yes that works and it has a special name: a perfect square trinomial :)
What's a coeffcient?
A coefficient is a term in mathematics which is basically the number that we used in this expression to multiply a variable. so X has a coefficient of 1, and 2X has a coefficient of 2, and so on.
watch closely in the video at 01:35, Sal (the teacher) is referring to the middle coefficient (which is of the variable " B ") as 10, meaning its 10 times B or just 10B.
further information > http://en.wikipedia.org/wiki/Coefficient
watch closely in the video at 01:35, Sal (the teacher) is referring to the middle coefficient (which is of the variable " B ") as 10, meaning its 10 times B or just 10B.
further information > http://en.wikipedia.org/wiki/Coefficient
it is the number that is multiplied by a variable
`Coefficient is just the fancy term for "the number beside the variable"`
For example.
65x → x here is the variable while 65 is the number beside it, in other words its coefficient.
For example.
65x → x here is the variable while 65 is the number beside it, in other words its coefficient.
It's a number wich is before a variable. e.g.: 3x (3 is the coefficient)
how come the constant term is the product of a and b?
Let me try (and please know that Sal already tried at 1:49 )...
Because when I you have a quadratic in intercept form `(x+a)(x+b)` like so, and you factor it (basically meaning multiply it and undo it into slandered form) you get: `x^2 + bx + ax + ab`. This of course can be combined to: `x^2 + (a+b)x + ab`. So when you write out a problem like the one he had at 5:39 `x^2 + 15x + 50`, 50, which is your "C" term (the third term) _and_ is your constant, 50 is the product of a and b (ab). This can be shown here: ```
x^2 + 15x + 50 is equal to:
(x+5)(x+10) =
x^2 + 10x + 5x + 50 =
x^2 + 15x + 50 {which is what we started with.}
```
Thank you very much for asking this question, i was wondering it for a long time and now that I know it I am glad that I am not the only one who was confused and I am glad to be of possible service. Please let me know if this helped anyone.
Because when I you have a quadratic in intercept form `(x+a)(x+b)` like so, and you factor it (basically meaning multiply it and undo it into slandered form) you get: `x^2 + bx + ax + ab`. This of course can be combined to: `x^2 + (a+b)x + ab`. So when you write out a problem like the one he had at 5:39 `x^2 + 15x + 50`, 50, which is your "C" term (the third term) _and_ is your constant, 50 is the product of a and b (ab). This can be shown here: ```
x^2 + 15x + 50 is equal to:
(x+5)(x+10) =
x^2 + 10x + 5x + 50 =
x^2 + 15x + 50 {which is what we started with.}
```
Thank you very much for asking this question, i was wondering it for a long time and now that I know it I am glad that I am not the only one who was confused and I am glad to be of possible service. Please let me know if this helped anyone.
The correct formula for what?
If you're saying it's the correct quadratic equation formula, then you are incorrect.
If you're saying it's the correct quadratic equation formula, then you are incorrect.
Why would you need to factor a quadratic in the first place?
So that we can........
1.) Find the value of a missing variable, such as x.
2.) Cancel out something that is on the numerator and the denominator to simplify rational expressions.
For instance if we have something like....
x²2x80/x²+16x+64
(x10)(x+8)/(x+8)(x+8)→As we can see we can now cancel out an (x+8) here to simplify it to.....
x10/x+8
1.) Find the value of a missing variable, such as x.
2.) Cancel out something that is on the numerator and the denominator to simplify rational expressions.
For instance if we have something like....
x²2x80/x²+16x+64
(x10)(x+8)/(x+8)(x+8)→As we can see we can now cancel out an (x+8) here to simplify it to.....
x10/x+8
say you have it quadratic expression equal to zero and you factor it, then you know that it can be zero when any one factor is zero. Thus you have found the roots or solution for x.
eg. x2 + 10x +9 = 0
(x+9)(x+1) =0
thus x+9 =0 or x+1=0.
thus x= 9 and 1 are the solutions of the equation
if you plot y = x2 +10x + 9 on graph the it will intersect the x axis (ie y=0 ) at x = 9 and x = 1
eg. x2 + 10x +9 = 0
(x+9)(x+1) =0
thus x+9 =0 or x+1=0.
thus x= 9 and 1 are the solutions of the equation
if you plot y = x2 +10x + 9 on graph the it will intersect the x axis (ie y=0 ) at x = 9 and x = 1
but remember that we are losing information as we simplify the equation. There is no way to recover the process we took to get that answer once we simplify a quadratic expression.
Factoring is one way to solve a quadratic equation. After you factor the expression, you can use the zero product property to find the solutions of x (or other variable used) and solve the equation. Factoring is mostly used only when it is quicker and easier to factor the expression rather than to solve it other ways, such as the quadratic formula.
To simplify the equation and see which factors go into the quadratic. It helps out in later math.
There are a lot of videos on factoring by grouping. We have to learn to do it by trial and error. Are there videos for that too? How would you find 8x^2 14x+3?
Even simpler (so as to avoid the fractions), if the quadratic is in the form:
`ax^2 + bx + c`
where a is not 1, you can do the following:
Multiply `a` and `c` to get a new number (say `d`). Now, you must find two numbers that add up to `b`, and multiply to `d.`
`x + y = b
x*y = d = a*c`
So in your example above, you would see that `d = a*c = 8*3 = 24.` Two numbers that add up to `b = 14` and multiply to `d = 24` are `x = 12`, and `y = 2.`
Then you can simply group and factor to get the correct answer.
`ax^2 + bx + c`
where a is not 1, you can do the following:
Multiply `a` and `c` to get a new number (say `d`). Now, you must find two numbers that add up to `b`, and multiply to `d.`
`x + y = b
x*y = d = a*c`
So in your example above, you would see that `d = a*c = 8*3 = 24.` Two numbers that add up to `b = 14` and multiply to `d = 24` are `x = 12`, and `y = 2.`
Then you can simply group and factor to get the correct answer.
Or you could do this : 8x^2 14 + 3
Knowing 12 and 2 add up to be 14 and multiply to be +24
split the 14x in the equation to 12x 2x hence :
8x^2 12x  2x + 3 = (8x^2 12x)  (2x 3) = 4x(2x  3)  (2x3)
use (2x 3) as factor then (2x3) (4x 1)
so 8x^2 14x +3 = (2x3)(4x 1)
Knowing 12 and 2 add up to be 14 and multiply to be +24
split the 14x in the equation to 12x 2x hence :
8x^2 12x  2x + 3 = (8x^2 12x)  (2x 3) = 4x(2x  3)  (2x3)
use (2x 3) as factor then (2x3) (4x 1)
so 8x^2 14x +3 = (2x3)(4x 1)
First, you have to factor out the 8:
```
8x^2  14x + 3 = 8(x^2  14/8 x + 3/8)
```
Then, you need to find the two numbers `a` and `b` such that:
```
a + b = 14/8
ab = 3/8
```
This is slightly more complex because it involves fractions instead of integer numbers, but the process is the same.
```
8x^2  14x + 3 = 8(x^2  14/8 x + 3/8)
```
Then, you need to find the two numbers `a` and `b` such that:
```
a + b = 14/8
ab = 3/8
```
This is slightly more complex because it involves fractions instead of integer numbers, but the process is the same.
how will i know what x is in the equation x^25x+24?
You cannot work out what x is in the expression x^25x+24. This is because expressions cannot be solved as they do are not equal to anything. E.g they are like statements.
If x^25x+24 was equal to a number or another expression then we would be able to work out the value of x.
For example:
x+1 (x could be equal to any number)
BUT
x+1 = 3 (x can only be equal to 2)
I hope that helps
If x^25x+24 was equal to a number or another expression then we would be able to work out the value of x.
For example:
x+1 (x could be equal to any number)
BUT
x+1 = 3 (x can only be equal to 2)
I hope that helps
You cannot solve for x, because that is only an expression. You can use various ways to find x only when it is an equation. (is equal to something)
You won't be able to know what x is until you set the equation equal to something.
You would be able to figure out what X is by using any method such as the quadratic formula, completing the square, square rooting, or even graphing the equation. The easiest and most surefire method of solving for X is the quadratic formula. This is something that will become extremely useful in Algebra 2 classes. To use this formula you take an equation such as yours, so x^25x+24 and think of the terms as A,B, and C. So _A_x^2+_B_x+_C_ where a,b, and c are 1,5,and 24 respectively. From this point the formula goes as such
http://0.tqn.com/d/create/1/0/i/9/9//Quadratic.formula.jpg
Now once you know this formula all you have to do is put 1, 5, and 24 in for A,B, and C or with whatever numbers you derive from an equation and you van solve for X.
http://0.tqn.com/d/create/1/0/i/9/9//Quadratic.formula.jpg
Now once you know this formula all you have to do is put 1, 5, and 24 in for A,B, and C or with whatever numbers you derive from an equation and you van solve for X.
x^25x+24 is only an expression, it is not an equation. Keep in mind that equations contain two expressions and an equal sign. X can only be solved if the expression is equal to another expression, so the variable remains unknown in this case.
Are there suppose to be alot of steps to this ?
i think there are around like seven steps maybe
My teacher says 8, & quadratic factoring ?
What if the numbers are prime and have no factors?
All quadratics can be factored, but not all of them can be factored with rational numbers or even real numbers. If a quadratic cannot be factored into rational factors, it is said to be irreducible. However, it is always possible to factor a quadratic, if you allow irrational or complex factors.
Here's how to factor ANY quadratic expression in the form: ax² + bx+c.
Let d = b²  4ac
(If d is not a positive perfect square, then the quadratic is "irreducible".)
The factors are:
a [x + ¹⁄₂ₐ (b + √d)] [x + ¹⁄₂ₐ (b  √d)]
If it makes for convenient numbers, you may use the distributive property to multiply a into one (but not both) of the factors.
Here's how to factor ANY quadratic expression in the form: ax² + bx+c.
Let d = b²  4ac
(If d is not a positive perfect square, then the quadratic is "irreducible".)
The factors are:
a [x + ¹⁄₂ₐ (b + √d)] [x + ¹⁄₂ₐ (b  √d)]
If it makes for convenient numbers, you may use the distributive property to multiply a into one (but not both) of the factors.
How do you factor a polynomial when the highest power is greater than ^2?
Long Division of Polynomials or Synthetic Division.
With difficulty! The more terms there are in the polynomial, the more difficult it is to find an exact factor.
Check out the video called Factoring by Grouping or use Completing the Square or the Quadratic Formula.
What would be considered a constant?
A fixed value.
In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number.
Example: in "x + 5 = 9", 5 and 9 are constants
If it is not a constant it is called a variable.
 taken from "math is fun"
In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number.
Example: in "x + 5 = 9", 5 and 9 are constants
If it is not a constant it is called a variable.
 taken from "math is fun"
`*Constants are numbers whose values don't change*`
For instance, in
15+x=y → 15 is a constant because its value doesn't change while x and y the are variables.
Note: x is called the independent or the input while y A.K.A f(x) is the dependent variable or the output.
Other examples of constants are:
integers such as 1, 2, 3, 4, 5, 6, 7, 8, 9 etc
irrational numbers such as √2, √7, e (which is roughly equal to 2.71828.....), π (which is roughly equal to 3.1415926.....) etc
For instance, in
15+x=y → 15 is a constant because its value doesn't change while x and y the are variables.
Note: x is called the independent or the input while y A.K.A f(x) is the dependent variable or the output.
Other examples of constants are:
integers such as 1, 2, 3, 4, 5, 6, 7, 8, 9 etc
irrational numbers such as √2, √7, e (which is roughly equal to 2.71828.....), π (which is roughly equal to 3.1415926.....) etc
I just came. 0__o


Do the numbers have to be whole, or could you end up with an answer like (x+2.5)(x+5.7)?
Yes, that could certainly happen. The quadratic expression x^2 + 4x + 3.75 factorises into (x + 1.5)(x + 2.5), for example. But (at least at first) you will probably find that the examples you are given factorise into integers.
At 14:49, can you do (x3)(x+8) instead of 1(x3)(x+8)?
No. The distributive property is not being properly used.
You are trying to multiply 1 by (x3). Using the distributive property, you'll get x+3 as a result. So it should be (x+3)(x+8).
You are trying to multiply 1 by (x3). Using the distributive property, you'll get x+3 as a result. So it should be (x+3)(x+8).
what if we have:
7x^231x20
How do we do that?
7x^231x20
How do we do that?
Neha,
To factor 7x^231x20
First multiply the first and last numbers. 7*20 = 140.
Factor 140 to its primes 7*5*2*2
Now find two numbers that multiply to 140 and subtract to 31
You need a number that is more than 31 so start with
7*5=35 and 2*2 = 4 And we got it first guess 354 = 31
We know the x values are
(7x ± ? )(1x ± ?)
You know the 7 need multiplies by the 5 so the second ? is 5
(7x ± ? )(1x ± 5) The other ? must be 4
(7x ± 4 )(1x ± 5)
The 20 is a 20 so one sign must be + and the other must be 
The middle factor is 31 not +31 so the 7*5 must be negative so use 5
(7x ± 4 )(1x  5) And as we already said the other sign must be +
(7x + 4 )(x  5)
7x^231x20 factors to (7x + 4)( x  5)
I hope that helps make it click for you.
To factor 7x^231x20
First multiply the first and last numbers. 7*20 = 140.
Factor 140 to its primes 7*5*2*2
Now find two numbers that multiply to 140 and subtract to 31
You need a number that is more than 31 so start with
7*5=35 and 2*2 = 4 And we got it first guess 354 = 31
We know the x values are
(7x ± ? )(1x ± ?)
You know the 7 need multiplies by the 5 so the second ? is 5
(7x ± ? )(1x ± 5) The other ? must be 4
(7x ± 4 )(1x ± 5)
The 20 is a 20 so one sign must be + and the other must be 
The middle factor is 31 not +31 so the 7*5 must be negative so use 5
(7x ± 4 )(1x  5) And as we already said the other sign must be +
(7x + 4 )(x  5)
7x^231x20 factors to (7x + 4)( x  5)
I hope that helps make it click for you.
What if you get a polynomial with 4 terms? Even 5?
Your question needs to be more clear. At first you might need to combine like terms. If this was not helpful, rephrase you question with an example.
What I mean is that polynomials with 4 or more distinct terms. Such as:
x^2 + 4x + 2xy + 10y
In which there are no like terms.
x^2 + 4x + 2xy + 10y
In which there are no like terms.
Do we use FOIl for this problem?HELP
Once you have a quadratic formula in (xa)(xb) form, you don't need to use FOIL. If you do, you just get the original equation. The answers are a and b.
so all the time the answer should be A and
I have a question on solving but I can't find a video that deals with the topic (probably because I don't know what this type of problem is called) Could someone point me in the right direction or solve this step by step for me?
Solve for x
x^413x^2+36=0
answers given on the sheet are {3, 3, 2, 2}
my first thought was to reduce to x^213x+6=0 and solve from there using the quadratic formula but my answers don't match the answers on the handout and I only got two when we are supposed to find four, how do you get four?
Thanks!
Solve for x
x^413x^2+36=0
answers given on the sheet are {3, 3, 2, 2}
my first thought was to reduce to x^213x+6=0 and solve from there using the quadratic formula but my answers don't match the answers on the handout and I only got two when we are supposed to find four, how do you get four?
Thanks!
simplify your life and from the very beginning, substitute a different variable in to stand for x^2 the whole time until you've got it solved and are ready for the final answer.
Say, g=x^2. So, just deal with this equation: g^213g+36=0.
Factor: Two numbers that multiply to be 36 and add up to be 13 are 9 and 4.
(g9)(g4) = 0. Now, set each part equal to zero.
g9=0 ; g4=0
so, g=9 ; g=4. NOW, bring back the x^2 instead of g again.
x^2 = 9 ; x^2 = 4
so, x can be 3, 3, 2, or 2.
Say, g=x^2. So, just deal with this equation: g^213g+36=0.
Factor: Two numbers that multiply to be 36 and add up to be 13 are 9 and 4.
(g9)(g4) = 0. Now, set each part equal to zero.
g9=0 ; g4=0
so, g=9 ; g=4. NOW, bring back the x^2 instead of g again.
x^2 = 9 ; x^2 = 4
so, x can be 3, 3, 2, or 2.
You had the right idea, but missed a few steps. You can't quite reduce it to x²13x+6, because that isn't the same equation and so it won't have the same roots. What you can do is define u=x², and from there you get the u²13u+6=0. Then you can figure out what u has to be, which I bet you already did. Let's say that you got that u=5 or u=8 (which isn't the actual right roots but I don't want to do your homework :) Then you substitute back, so that you need to solve x²=5 or x²=8. That is two equations that will have a total of four possible solutions, and hopefully the ones you're looking for.
Don't you get tired doing all that algebra? It would make me tired to so painstakingly explain these concepts to everyone! ; ) Thanks for your hard work.
I have no idea how to transform rational functions into graphing form
For exam le 4/4x↑2 +40x+ 10
Do I factor the bottom im so confused I'm not any good at precalculus it frustrates me because people in my class seem to get it I feel horrible and my teacher refuses to help :(
For exam le 4/4x↑2 +40x+ 10
Do I factor the bottom im so confused I'm not any good at precalculus it frustrates me because people in my class seem to get it I feel horrible and my teacher refuses to help :(
4/4x² +40x+ 10
First we simplify by cancelling out the 4's to get: 1/x²
The common denominator is x², so we multiply everything by the factor that would given them that denominator  it is just the same idea as adding fractions. In this case, one term already has that denominator, the other two need to be multiplied by x² / x² (remember what you did with adding ordinary fractions). This will give you:
1/x² + (40x)( x²/x²) + 10(x²/x²)
= (1 + 40x³ + 10x²) / x²
= (40x³ + 10x²  1) / x²
You cannot simplify this any further. If you could tell me what it is equal to, I could finish solving the problem. Of course, if it is just equal to y or f(x) then that is as simple as it gets.
First we simplify by cancelling out the 4's to get: 1/x²
The common denominator is x², so we multiply everything by the factor that would given them that denominator  it is just the same idea as adding fractions. In this case, one term already has that denominator, the other two need to be multiplied by x² / x² (remember what you did with adding ordinary fractions). This will give you:
1/x² + (40x)( x²/x²) + 10(x²/x²)
= (1 + 40x³ + 10x²) / x²
= (40x³ + 10x²  1) / x²
You cannot simplify this any further. If you could tell me what it is equal to, I could finish solving the problem. Of course, if it is just equal to y or f(x) then that is as simple as it gets.
So, what would a real life problem for using this be? How would you use it with other parts of math? Please ask me a question if you are confused about what I mean here! Thanks!
age problems, research and other things
solving word problems. especially age ones.
Do *unfactorable* polynomials exist?
For example:
```x^2+8x+36```
For example:
```x^2+8x+36```
That is a great question. The anwer is actually yes. There are unfactorable (or irreducible) polynomials and the reason why can go very deep into mathematics. It is something that people study is college and beyond. Another simple polynomial that can not be factored is x^2+1. To see why suppose it factored as (x+a)(x+b). Then we would need a+b=0 and ab=1. The first equations tells us that either both 'a' and 'b' are 0 or one and not the other is negative. From there we see that both equations can not be true at the same time (did that make sense?) The same kind of thing is true about your polynomial.
It turns out that you can always factor polynomials with degree greater than 2 into smaller polynomials but there are many degree 2 polynomials that can not be factored. If you want to know more about this you will probably have to learn about roots of equations and then maybe the imaginary numbers like the square root of 1.
It turns out that you can always factor polynomials with degree greater than 2 into smaller polynomials but there are many degree 2 polynomials that can not be factored. If you want to know more about this you will probably have to learn about roots of equations and then maybe the imaginary numbers like the square root of 1.
i do not think so
How do you know when to stop factoring a quadratic expression?
Why is it called Quadratic??
Quadatus is latin for square. Quadratic pertains to a square. If you have a term in an expression or equation that is to the second degree or squared, it is called quadratic.
What's a coeffcient?
A coefficient is a constant (like "5") that is multiplied by a variable.
Is this also referred to as " Factoring x^2+bx+c". My book goes over this and calls it so. Why is it called Factoring quadratic Expressions here on KA? I'm confused.
Is this the easiest way of factoring?
The easiest method is the one which works the best for you
No. Look up the box method. Sal dropped the ball here IMHO.
What are quadratic expressions?
Expressions of the form ax^2 + bx + c; where a is not 0.
Can you give some factoring examples that have no solution, one solution, and many solutions. And are there any examples that cannot be factored?
Well, all quadratics can be factored if you allow irrational and complex factors. But, if you mean rational factors, there are plenty that cannot be factored:
2x²+x+15 cannot be factored with real numbers.
No quadratic has "many" factors: Either there are two real, one real, or no real factors. If you have irrational or complex factors, you always have two of them (if there are no irrational or complex numbers in the quadratic expression).
As for having just one factor, it will always be the case that you really have two identical factors. For example, x²+6x+9 factors as (x+3)(x+3) also written as (x+3)².
No quadratic has more than two factors (unless you count factoring out a common factor such as:
2x² + 6x + 4 = 2(x² + 3x + 2) = 2(x+1)(x+2)
2x²+x+15 cannot be factored with real numbers.
No quadratic has "many" factors: Either there are two real, one real, or no real factors. If you have irrational or complex factors, you always have two of them (if there are no irrational or complex numbers in the quadratic expression).
As for having just one factor, it will always be the case that you really have two identical factors. For example, x²+6x+9 factors as (x+3)(x+3) also written as (x+3)².
No quadratic has more than two factors (unless you count factoring out a common factor such as:
2x² + 6x + 4 = 2(x² + 3x + 2) = 2(x+1)(x+2)
i need help with this!
5x^2=15x
please help!
ive got it to where its equal to zero but i cant get past that
5x^2=15x
please help!
ive got it to where its equal to zero but i cant get past that
5x²=15x
x²=3x
x²3x = 0
x(x3)=0
x=0 or
x3=0
x = 3
Thus, x= 3 or x=0
x²=3x
x²3x = 0
x(x3)=0
x=0 or
x3=0
x = 3
Thus, x= 3 or x=0
What Is A Degree?
A degree is a unit of measurement of angles. It is also equal to 1/360th of the circumference of a circle. º is its symbol.
I hope this helps!
I hope this helps!
What's a coeffcient?
A coeffient is the number before the variable. eg. in 3x, 3 is the coefficient. The variable is x. Variables are used to represent unknown numbers.
A coefficient is a number that comes before a variable. One time he says coefficient is at 3:24.
Will this method always work?
only if you have a trinomial
ughh i just got confused again whats 3a^2+7a+2 my techer just mixed up my brain:[ im way to new at this
Try going to math tutoring on tuesdays, wednesday, & thursdays, it helps alittle better .
i do go to tutoring :( maybe u could tutor me? if u could tutor me or help i will be VERY happy but the choice is urs
3a^2+7a+2
= 3a^2 + 6a + a + 2
= 3a(a + 2) + 1(a + 2)
=(3a+1)(a+2)
= 3a^2 + 6a + a + 2
= 3a(a + 2) + 1(a + 2)
=(3a+1)(a+2)
why do you start factoring out a 1 in the last two equations of this video?
it seems like extra work and you don't explain why you do it, other than "it makes it easier". how does it make it easier?
it seems like extra work and you don't explain why you do it, other than "it makes it easier". how does it make it easier?
What is he talking about?
factoring quadratic expressions….. JK… I really don't know.
what are polynomials?
A group of several terms.
What level of math would this be considered? Algebra 1? Algebra 2? College Algebra? Thanks!
I disagree. I learned this and far more in Algebra 1.
yea i'm learning this in Algebra 1
how would you solve questions like x squared + 19x + 17, since 17 is a prime # and the sum of it's only 2 factors add up to 18, what would you do instead?
You could complete the square to
x^2 + 19x + (19/2)^2 = (19/2)^2  17
or use the quadratic formula.
Either way, you will get two factors with expressions that include the square root of 293.
x^2 + 19x + (19/2)^2 = (19/2)^2  17
or use the quadratic formula.
Either way, you will get two factors with expressions that include the square root of 293.
What are first and second degree terms?
Examples of first degree terms: 7x, 4, 89z, 23r
Examples of second degree terms: 3x^2, 64y^2 18p^2
Examples of third degree terms: 23x^3, 34^y^2 87o^3
and so on...
Examples of second degree terms: 3x^2, 64y^2 18p^2
Examples of third degree terms: 23x^3, 34^y^2 87o^3
and so on...
3x^210x+5=0 cannot be factored, why?
Sure it can. The easiest way to do it is probably to use the quadratic formula.
https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/usingthequadraticformula
https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/usingthequadraticformula
How do you factor t^2 = 5t?
That helped a lot! Thank you so much!
Alright here's how we do it.
t^2 = 5t → This is what we were given.
t^2 − 5t = 0 → I subtracted both sides by 5t
t(t−5) = 0 → Notice that both t^2 and 5t have a common factor of t so we can just factor that out and we're done!
We now also what the value t should be so that the equation will equal 0. Which is t=5
Let's try substituting the value of t here.
t(t−5) = 0 → This is the original equation.
5(5−5) = 0 → I now substituted 5 in place of t in the original equation.
5(0)=0 →This what we'll get
0=0 → And we finally checked that our answer is correct.
Although it it might have been easier (depending to the person) to solve it by dividing both sides by t.
t^2 = 5t → This is what we were given.
(t^2)/t = 5t/t → I'm dividing both sides here by t
t=5 → and there's my answer.
t^2 = 5t → This is what we were given.
t^2 − 5t = 0 → I subtracted both sides by 5t
t(t−5) = 0 → Notice that both t^2 and 5t have a common factor of t so we can just factor that out and we're done!
We now also what the value t should be so that the equation will equal 0. Which is t=5
Let's try substituting the value of t here.
t(t−5) = 0 → This is the original equation.
5(5−5) = 0 → I now substituted 5 in place of t in the original equation.
5(0)=0 →This what we'll get
0=0 → And we finally checked that our answer is correct.
Although it it might have been easier (depending to the person) to solve it by dividing both sides by t.
t^2 = 5t → This is what we were given.
(t^2)/t = 5t/t → I'm dividing both sides here by t
t=5 → and there's my answer.
Can anyone tell me why it's called a quadratic expression?
It's from the Latin _quadratus_ which means "made square"
Why is it called quadratic if it is raised to the second power. Shouldn't it be called bidratic or something?
It actually comes from the Latin word "quadratum", which means "square". The reason lies in geometry: x^2 is the surface of a square with sides of length x.
I'm in Calculus II and I didn't know this tidbit! Neat. @ZeroFK
How can i relate factoring quadratioc equations to real life ?
Even though I watched the whole video, I still don't get it .. this is hard ! Can someone explain this to me more easier ?
trust me it gets harder can u help me on mine its like up there
Okay, my problem is 6r^2+16r32=0 im at 2(3r^2+8r16) Now what do I do?
2(3r²+8r16)
That much is correct. As for finding the rest, many people just try to guess the factors, but that takes up time and doesn't always work. Here's how to do it systematically, for when you can't easily guess the factors:
To make this easier to follow, I will just factor the part you've not factored yet (in real work, you should carry the 2 along the whole way, but we'll just ignore it for the moment.)
3r²+8r16
we will split the middle term into two terms, mr + nr such that:
m + n = 8 (this is the coefficient of the middle term)
mn = (3)(16) =  48 (this is the product of the coefficients of the a and c terms).
Looking at the pairs of factors of 48, we see which of them add to 8 (and of course, multiply to be 48). I won't go through the whole list of possible factors that you would have to eliminate, but you'd get 12 and 4.
So, we split 8x into 12x  4x
This gives us:
3r²+8r16
= 3r² + 12r  4r 16
We factor the first two and then the second two:
= 3r(r + 4)  4r 16
= 3r(r + 4)  4(r+4)
There now is a common factor, so we factor (r+4) out:
= (r + 4)(3r  4)
And, don't forget about the 2 we already factored out:
2(r + 4)(3r  4)
That much is correct. As for finding the rest, many people just try to guess the factors, but that takes up time and doesn't always work. Here's how to do it systematically, for when you can't easily guess the factors:
To make this easier to follow, I will just factor the part you've not factored yet (in real work, you should carry the 2 along the whole way, but we'll just ignore it for the moment.)
3r²+8r16
we will split the middle term into two terms, mr + nr such that:
m + n = 8 (this is the coefficient of the middle term)
mn = (3)(16) =  48 (this is the product of the coefficients of the a and c terms).
Looking at the pairs of factors of 48, we see which of them add to 8 (and of course, multiply to be 48). I won't go through the whole list of possible factors that you would have to eliminate, but you'd get 12 and 4.
So, we split 8x into 12x  4x
This gives us:
3r²+8r16
= 3r² + 12r  4r 16
We factor the first two and then the second two:
= 3r(r + 4)  4r 16
= 3r(r + 4)  4(r+4)
There now is a common factor, so we factor (r+4) out:
= (r + 4)(3r  4)
And, don't forget about the 2 we already factored out:
2(r + 4)(3r  4)
(x+a)(x+b) can be used to factor all quadratics, correct?
What's a coeffcient?
A coefficient is the number in front of a variable that is intended to be multiplied by that variable. For example:
4x has a coefficient of 4
but,
4 + x has a coefficient of 1
(if nothing is shown, the coefficient is 1)
4x has a coefficient of 4
but,
4 + x has a coefficient of 1
(if nothing is shown, the coefficient is 1)
what is a polynomial ?
what is a polynomial?
A polynomial is kAn expression of more than two algebraic terms, esp. the sum of several terms that contain different powers of the same variable(s).
hi! I have a question about this problem x^3x^2+x1 what is the answer? I am very confused
Thank you very much sir!
how do I factor the following question(x^3x^2+x1)
plz help me! I need it!
Well, you can only answer it unless you have the amount of x. Therefore, I cannot help you unless you have the value.
What is th purpose of factoring by grouping? I don't understand how, by what advatnge does it give?
Factoring is a way to find the value for x in a quadratic. (Try just solving for x the way you would do a linear equation. It doesn't work, does it :P)
There are several ways to solve for x in a quadratic (completing the square, quadratic formula, factoring, etc.) However, for simpler quadratics, factoring tends to be the quickest and easiest for most people.
Hope this helps!
There are several ways to solve for x in a quadratic (completing the square, quadratic formula, factoring, etc.) However, for simpler quadratics, factoring tends to be the quickest and easiest for most people.
Hope this helps!
At 00:54, why does Sal use the expression (x+a)(x+b) to solve the problem? Is that a formula?
No, that's not a formula. That's just your basic form for what your answer should be after factoring. The basic goal is to have a + b equal to the middle x term, and have a x b equal to the last number term.
So, by finding factors of 9: 1, 3, and 9, and then seeing which two add up to 10 ( 1 and 9) you can factor the equation to (x + 9)(x + 1).
So, by finding factors of 9: 1, 3, and 9, and then seeing which two add up to 10 ( 1 and 9) you can factor the equation to (x + 9)(x + 1).
I have a problem that's X^2 + X + 1 in my text book it just says prime.... what does that even mean?
Prime means that the expression cannot be factored any further because each term in the expression has no GCF.
Maybe it means that it cannot be factored, that it is completely factored.
x24x= 4 how do you solve? i need help.
Essentially these types of questions are asking for you to solve for x.
How you would do this:
Add 4 to both sides then you would get x^24x+4=0 which you can factor to get
(x2)(x2)=0 so then your answer would be x=2 and 2.
How you would do this:
Add 4 to both sides then you would get x^24x+4=0 which you can factor to get
(x2)(x2)=0 so then your answer would be x=2 and 2.
1(x12)(x6). should you write this like 1((x12)(x6)) or is it a rule to multiply the parentheses with each other first?
Nicholas,
You get the same answer if you do it either way.
(1(x12))(x6) = 1((x12)(x6))
This is because of the associative property of multiplication.
The associative property is (a*b)*c = a*(b*c)
You get the same answer if you do it either way.
(1(x12))(x6) = 1((x12)(x6))
This is because of the associative property of multiplication.
The associative property is (a*b)*c = a*(b*c)
At 14:49, can you simplify the problem 1(x3)(x+8)? If so, would the answer be (x+3)(x8)? Or would it be (x+3)(x+8)?
The examples on this video which starts with negative such as ( x^2 + 18x 72) , when we factor out with 1 it works just fine , but just for the intuition i try them without factoring out with 1 , i couldnt figure that out . Why i cannot make it ? is it just me or there is really no way without factoring out with 1 . Thanks for help...
In this video, Sal doesn't seem to go over problems like this: (a number)x = x^2(a number). I am having trouble finding a video on these specific kind of problems, anybody know where I can find another one of these videos on khanacademy, or elsewhere?
hey what if you have like 24 and 10x its 6 and 4 or 8 and 3 how do you tell what one to use?
In my school, they taught us a technique called "Splitting the middle term" to factor quadratics.
How it goes is, we think of two numbers that add up to the coefficient of x but also when multiplied those two numbers form the product of the coefficient of x^2 and the constant term.(We ,often, prime factorize the "product" and then go by trial and error to find the "sum").
Then we replace our x term with two other x terms which have the numbers we thought of as respective coefficients. Then we proceed to reverse engineer the polynomial to make it into the form (x+a)(x+b).
Now, can somebody explain to me why the method Sal is teaching me is so ridiculously similar?
Also, which method is easier? Splitting the Middle Term OR Sal's Factoring Quadratic expressions.
How it goes is, we think of two numbers that add up to the coefficient of x but also when multiplied those two numbers form the product of the coefficient of x^2 and the constant term.(We ,often, prime factorize the "product" and then go by trial and error to find the "sum").
Then we replace our x term with two other x terms which have the numbers we thought of as respective coefficients. Then we proceed to reverse engineer the polynomial to make it into the form (x+a)(x+b).
Now, can somebody explain to me why the method Sal is teaching me is so ridiculously similar?
Also, which method is easier? Splitting the Middle Term OR Sal's Factoring Quadratic expressions.
Okay, so my problem is x^212x+32=0... My problem is why does my answer sheet tell me that the answer is 8 and 4. I got 8, 4. which makes more since to me. I understand how 8 and 4 could be right... but checking it.. It doesn't.. Here (x+4)(x+8) >x^2+8x+4x+32= x^2+12x+32.. Not x^212x+32.. BUT, if I do 4,8 then I get (x4)(x8)>x^28x4x+32= x^212x+32. Which is the original equation.. Help?
You are confusing the roots with the factors.
x²  12x+ 32 = 0
(x4)(x8) = 0
So the factors, not the roots, are (x4) and (x8).
To get to the roots, you find those values of x for which the equation is true. In this case, (x4)(x8) = 0 is true whenever either factor equals 0.
Thus the roots (also called solutions) are:
x4 = 0 OR x8 = 0
Solving each equation gives us
x = 4 OR x = 8
The distinction between factors and roots might be easier to see in a problem like this: 6x² + x  35 = 0
This factors to: (2x+5)(3x7)=0
So, the factors are (2x+5) and (3x7).
The solutions, or roots, are:
(2x+5)=0
2x = 5
x = 5/2
OR
3x7 = 0
3x = 7
x = 7/3
So the roots, or solutions, are x=7/3 OR x= 5/2
Hopefully you can see how, though related, the factors and the roots are not the same thing.
x²  12x+ 32 = 0
(x4)(x8) = 0
So the factors, not the roots, are (x4) and (x8).
To get to the roots, you find those values of x for which the equation is true. In this case, (x4)(x8) = 0 is true whenever either factor equals 0.
Thus the roots (also called solutions) are:
x4 = 0 OR x8 = 0
Solving each equation gives us
x = 4 OR x = 8
The distinction between factors and roots might be easier to see in a problem like this: 6x² + x  35 = 0
This factors to: (2x+5)(3x7)=0
So, the factors are (2x+5) and (3x7).
The solutions, or roots, are:
(2x+5)=0
2x = 5
x = 5/2
OR
3x7 = 0
3x = 7
x = 7/3
So the roots, or solutions, are x=7/3 OR x= 5/2
Hopefully you can see how, though related, the factors and the roots are not the same thing.
Oh! That makes total since to me. Thank you very much!
So are there quadratic expressions that can't be factored? (For example if the constant term is a prime and the first degree coefficient is ≠ the constant term ± 1)
When we swap a number from a side to the other side, we make it a negative number, don't we?
I get it now. Thank you.
i really do not understand how to factor at all and i have these huge problems i can not do for example 56m^2+117m55
Well this is a tough one. If you are really struggling I would suggest working on some easier ones first since the explanation I am about to give uses a lot of problem solving skills that are easier to develop with simpler problems.
We are looking for something of the form (nx+a)(mx+b). When we multiply this out we get nmx^2+nbx+max+ab=nmx^2+(nb+ma)x+ab. We get three equalities from this found by matching the constant terms, the x terms, and the x^2 terms. First mn=56, next nb+ma=117, last ab=55. The reason this problem is difficult is that you have to choose what to try and making good choices comes with practice. I would say start with ab=55 because 55 only has two factors. So we just take a guess that a=55 and b=1. Then nb+ma=117 means that m is probably small and positive so lets let it be 2. Then we have n(1)+2(55)=117 so n=7 but unfortunately 3(7)=21 is not equal to 56.
We could try other m's but I don't think it will work so lets switch to a=11 and b=5. Well m is a factor of 56=2*2*2*7 so lets try m=8. Then 5n+11(8)=117 means n=29/5 so not right. m=14 gives n=37/5 so that doesn't work either and we went to high with m. If m=7 then n=8 which works! We are done. The answer is (8x+11)(7x5).
As you saw there was a lot of guessing and trying things. It could have taken a lot longer if I was not aware of what would be good guesses and when to stop following a track. This comes with practice.
We are looking for something of the form (nx+a)(mx+b). When we multiply this out we get nmx^2+nbx+max+ab=nmx^2+(nb+ma)x+ab. We get three equalities from this found by matching the constant terms, the x terms, and the x^2 terms. First mn=56, next nb+ma=117, last ab=55. The reason this problem is difficult is that you have to choose what to try and making good choices comes with practice. I would say start with ab=55 because 55 only has two factors. So we just take a guess that a=55 and b=1. Then nb+ma=117 means that m is probably small and positive so lets let it be 2. Then we have n(1)+2(55)=117 so n=7 but unfortunately 3(7)=21 is not equal to 56.
We could try other m's but I don't think it will work so lets switch to a=11 and b=5. Well m is a factor of 56=2*2*2*7 so lets try m=8. Then 5n+11(8)=117 means n=29/5 so not right. m=14 gives n=37/5 so that doesn't work either and we went to high with m. If m=7 then n=8 which works! We are done. The answer is (8x+11)(7x5).
As you saw there was a lot of guessing and trying things. It could have taken a lot longer if I was not aware of what would be good guesses and when to stop following a track. This comes with practice.
what if the equation was 2x^2  13x  24?
If you are trying to find the roots of this particular quadratic equation to put it onto a graph the work would look like this:
2x^2  13x  24
ac product = 48
Factors of 48:
1 times 48
2 times 24
3 times 16
Because the middle term is negative and so is the ac product, one of the two factors you replace the middle with has to be negative and the other positive. And due to how you solve this problem, the two factors that you replace the original middle term with have to have a difference of the middle term.
2x^2  16x + 3x  24
This is a big problem to factor now so you split up the problem into two separate problems to factor like so:
2x^2  16x >3x  24
2x(x  8) > 3(x 8)
Put those together and due to the commutative property you can do that and you get:
(2x + 3) (x  8)
The roots after that would be solved like this:
2x + 3 = 0 > x  8 = 0
 3 3 > +8 +8
2x = 3 > x = 8
________
2
x = 2/3
x= 2/3 or 8
2x^2  13x  24
ac product = 48
Factors of 48:
1 times 48
2 times 24
3 times 16
Because the middle term is negative and so is the ac product, one of the two factors you replace the middle with has to be negative and the other positive. And due to how you solve this problem, the two factors that you replace the original middle term with have to have a difference of the middle term.
2x^2  16x + 3x  24
This is a big problem to factor now so you split up the problem into two separate problems to factor like so:
2x^2  16x >3x  24
2x(x  8) > 3(x 8)
Put those together and due to the commutative property you can do that and you get:
(2x + 3) (x  8)
The roots after that would be solved like this:
2x + 3 = 0 > x  8 = 0
 3 3 > +8 +8
2x = 3 > x = 8
________
2
x = 2/3
x= 2/3 or 8
How would you factor 3x^2 −5x−2? The solution is (3x + 1)(x  2) but I don't understand how? Using the method, factoring 2 I cant figure anything out that satisfies how it equals to 5?
A Man,
Multiply the outside terms
3x * 2 = 6x
Multiply the inside terms
1 * x = 1x
Then add them together
6x + 1x = 5x
Does that help?
Multiply the outside terms
3x * 2 = 6x
Multiply the inside terms
1 * x = 1x
Then add them together
6x + 1x = 5x
Does that help?
Not quiet following. I understand when you say the outside terms which would be 3x and 2. Don't understand where the ^2 went though, and where the inside terms came from.
Also, the final equation how does that fit into get (3x+1)(x2)?
Also, the final equation how does that fit into get (3x+1)(x2)?
When I was solving the practice problems, I got most of them wrong because the (/+) sign was flipped. After working the equation out several times again, it still amounted to a very clear answer, yet the answer the machine accepted was the opposite sign. Why is this?
how would you factor 4x^2+13x+3? I understand how to do it when x^2 doesn't have a number in front of it, but not when it does.
You're looking for a solution on the form (ax + b)(cx + d), and you know that
1) ac = 4
2) ad + bc = 13
3) bd = 3
So I start with a guess: a = 4.
Then according to 1) c must be 1.
Solving 2) and 3) gives you b = 1 and d = 3, and that's all there is to it.
4x^2 + 13x + 3 =
(ax + b)(cx + d) =
(4x + 1)(x + 3)
1) ac = 4
2) ad + bc = 13
3) bd = 3
So I start with a guess: a = 4.
Then according to 1) c must be 1.
Solving 2) and 3) gives you b = 1 and d = 3, and that's all there is to it.
4x^2 + 13x + 3 =
(ax + b)(cx + d) =
(4x + 1)(x + 3)
use the quadratic formula. or you could get the leading coefficient equal to 1 and subtract C then complete the square. A=4, B=13, and C=3
how do u factor qudratics! i dont get this video at all plz help!
Try this: http://www.purplemath.com/modules/factquad.htm
Or this: http://www.mathsisfun.com/algebra/factoring.html
Hope these sites help you!
Or this: http://www.mathsisfun.com/algebra/factoring.html
Hope these sites help you!
I had encountered problems like 3x squared  x  72... the 3x squared makes me confused... how do i do this?
at 10:40 can't we do 7x2x? i mean that equals 5x too,so?
i'm sorry if its not a very good question.... maths takes time to get in my head
i'm sorry if its not a very good question.... maths takes time to get in my head
Where did the word Quadratic come into math and other things? Also, isnt this kinda like algebra EX : A2 + B2 = C2 ?
It is algebra for one thing, quadratics are a type of polynomial.
Polynomials are just somes of unlike terms, and a quadratic polynomial has degree 2.
Polynomials are just somes of unlike terms, and a quadratic polynomial has degree 2.
2 degree how come there isnt one when dealing with quadratic polynomial ? Also , is there anything else that will be more than a degree2 ?
Why does a+b =negative instead of positive
Carl,
a+b equals a negative number only when the coefficient on the x value is negative.
a*b = negative when the coefficient on the constant value (the one without the x) is negative.
I hope that helps
a+b equals a negative number only when the coefficient on the x value is negative.
a*b = negative when the coefficient on the constant value (the one without the x) is negative.
I hope that helps
how do i factorise this question
(3x4)^1
(3x4)^1
What do you do if, for example, you have 216 and 72. 216 has no factors that add together to get 72, so would the problem have no solution?
The problem *would* have solutions, but they wouldn't be integers, and you wouldn't be able to find those solutions by factoring. You'd have to use another method (probably the quadratic formula) instead. Later Khan Academy videos will explain those methods.
I have a question. How do you know if its (x4)(x+3) or (x3)(x+4)?
Even though a*b and b*a are the same answer, 12, look at (a+b). For your first example it is (4+3)=1. For your second example (3+4)=1. Therefore, your first example would be correct for x^2x12=0 and your second example would be correct for x^2+x12=0.
Is it always x squared, or can it be different. For example, 2x squared, 3x squared, ect.
yes it can have anything as its coefficent
How do you figure out a question that has the answer but you have to figure out x's value?
For example: x^2  3x 18 = 0
Is there an easier or faster way to figure out what x equals besides guess and check?
For example: x^2  3x 18 = 0
Is there an easier or faster way to figure out what x equals besides guess and check?
To factor the polynomial more easily look at the videos on completing the square or the quadratic formula.
what is 4+4 (4a)
it equals 4+16a
I get everything but the end. Why is the answer always the opposite on the exercise? Am I doing it wrong? When I get the numbers and its a positive, I get the numbers right but they would be negative. Then when I get a negative, it's a positive. What am I doing wrong?
Oh okay. Thank you.
In this video we factor a quadratic expression. In the exercise, we want to solve for when a quadratic expression is equal to 0. Factoring is only part of the solution.
Use the hint button on the exercise and it will help you.
Use the hint button on the exercise and it will help you.
maths sucks why?
It just does, I find it easier to just accept it and move on.
what do you mean by factoring out the negative one at 13;00 ?
The idea is that he doesn't want the x squared to be a negative, since that would make the entire equation more difficult, so the simplest way to solve it and make it like the previous equations he'd been solving for us, was to make the negative x squared into a positive. He did that by factoring out the 1, making the first term positive and making it exactly like the other ones that you had seen. Hope this helped!
what if you have a number in front of the variable squared???
What if the coeffiencent x^2 ( the first term) gets a number?
5x^2+x+10
What happens?
5x^2+x+10
What happens?
Divide the expression by 5.Sorry,I'm not sure whether its a equation or an expression. You'll get : x^2+x/5+2. Btw you do understand that the roots of the expression are irrational? So you'll have to use the discriminant method. Factorization is just hit and trial :P.
How is there A . B yet in there equation there is no mention of these letters.
Thank you for the answer>
Sal Uses The F.O.I.L Method at 1:11 right? Just Making Sure!
Yes! He is using the F.O.I.L Method!
How do I solve these when there's another number in front of the X62? Such as 2x^25x3. It has the answer as (2x+1)(x3) 1+3 isn't 5, so I know the 2 on 2x^2 is doing something, but I don't understand what.
Nix that, just found the second video.
I'm unclear with the problem at 11:29. How does it figure that the difference would have to be 1?
Ha! Nvm, I figured it out. The formation of the problem was a little bit different so it threw me off.
is there anyway they can explain this in simplier ways or can anyone help me out like coach me my thingy is rk010672@sdale.org please help if your up for a challenge
wow tht helps alot
How do you factor an equation with no GCF?
Ex. 2x^2  5x 88 = 0
Ex. 2x^2  5x 88 = 0
You could just use the quadratic formula for that one, if that is acceptable. Completing the square is also an option.
what if the second term is in form of roots. I have some sums in a book where we have to separate the number in terms of its roots and i find it very difficult
How would you do 4x^36x^228
What if the first number is not a 1x squared but say 6x squared
You would try different ways of getting for example 6x^2 which would be 3 and 2. but it could also be 6 and 1 depending on the rest of the probelm. Just makde sure you check your inner and outer to see if it works!! Hope that helps!
What about if it is 7m^454m^216
Well, 7m^454m^216 is able to be simplified.
(m^28)(7m^2+2)
If you solve this using the foil method, it is 7m^4+2m^256m^216, then simplify
(m^28)(7m^2+2)
If you solve this using the foil method, it is 7m^4+2m^256m^216, then simplify
What is the quadratic formula, and how does it exactly work?
how do i factorize x^28x+15=0
If you understand the technique in this video, then the factorization of this quadratic is very simple.
factorize lol
what about setting it up to zero?
what does coffecient mean?
A coefficient is the number in front of a variable.
5x
/\
5x
/\
why is the quadracy of factor "f" squared with the quadrint of factor "A"?
Is (a+b)x the same as x(a+b)