Quadratics and polynomials
Factoring quadratic expressions
None
More examples of factoring quadratics with a leading coefficient of 1
Can't get enough of Sal factoring simple quadratics? Here's a handful of examples just for you!
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?

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Can both a and b equal the same for example x^2+20x+100 can both a and b equal 10?
Yes, a and b can be equal. In the example you wrote, that expression is a special case. It is the square of a binomial. (x+10)*(x+10) = (x+10)^2. x+10 is a binomial that is being squared. I hope I helped.
Yes, a and b can both be equal. That is when you have a perfect square. You could write the factored form of x^2+20x+100 as (x+10)(x+10) or (x+10)^2.
Yes! In fact, if both a and b are the same, you are multiplying the same binomial by itself. For example, if a and b are 10, you would have (x + 10)(x + 10), which you can express as (x + 10)^2, if you want to simplify it even further. A neat trick that you can use to distribute this back to the quadratic expression is to use the formula (a + b)^2 = a^2 + 2ab + b^2. Try it out both ways and you can see where this comes from. I think it is a neat thing to keep in your toolkit when turning your product of binomials into quadratic expressions.
Yes, if you have the form `(x+a)(x+b)` and `a` and `b` are the same value, you have a "perfect square trinomial," which is explained in more detail here: https://www.khanacademy.org/math/algebra/multiplyingfactoringexpression/factoringspecialproducts/v/factoringperfectsquaretrinomials
For your example `(x+10)(x+10) = x^2+20x+100` if you multiply it out.
For your example `(x+10)(x+10) = x^2+20x+100` if you multiply it out.
Yes it can be the same
Yes, that is correct Jay, I would recommend watching the video by Ezequiel not all quadratics need 2 answers
Yeah just like Sunshine said it would be a perfect square trinomial so it would be written like (x^2+10)(x^2+10) or like (x^2+10)^2
yes A and B can be the same.
Yes a + b = 10 that mean equal the equation x^2 + 20x + 100
the numbers don't have to be different so yes!!
;)
;)
It's explained here: https://www.khanacademy.org/math/algebra/polynomials/multiplying_polynomials/v/specialproductsofpolynomials1
yes that works and it has a special name: a perfect square trinomial :)
yes a and b can be equal until they fit in the requirements
a+b
a.b
a+b
a.b
You can just simplify it to (x+10)^2.
So yes. The roots can be the same
So yes. The roots can be the same
Yes, it may vary in the answers.
sometimes in tests people give you a problem that they know the answer to, so there would be a answer
Yes both number could be the same and it means that it is a perfect square
i put this up about a year ago i think i was confused because the teacher said to guess if its (4x+a)(2x+b) or (8x+a)(x+b) but i understand now. Thanks!
Yes the a and b can be the same
it does not matter
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.
As long as the numbers fit the requirements (a+b=c and axb=d) then they can be any two numbers.
Yes, it doesn't matter what you use as long as you understand it!
What's a coeffcient?
it is the number that is multiplied by a variable
The number multiplying the variable, in this case x
A coefficient is a multiplying factor. For example, in the equation 2x^2 + 3x = 0, where x is a variable, the coefficient of x^2 is 2 and the coefficient of x is 3. Sometimes, the value of coefficients is not known, although they are known to stay constant as x changes, for example, ax^2 + bx = 0. In this case, a and b are constant coefficients.
See this video: https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/polynomial_tutorial/v/termscoefficientsandexponentsinapolynomial
A number or symbol multiplied with a variable or an unknown quantity in an algebraic term. For example, 4 is the coefficient in the term 4x, and x is the coefficient in x(a + b). A numerical measure of a physical or chemical property that is constant for a system under specified conditions.
A coefficient is a constant or a number that multiplies a variable in an algebraic expression. In the expression 8x, x is the variable and 8 is the coefficient of x.
It's the number that indicates how much of that variable is in the equation
EX: 5y= there are 5 y's.
EX: 5y= there are 5 y's.
A Coefficient is a number in front of a variable, and it signifies the number that variable is multiplied by (i.e. in the expression 8x, 8 is the coefficient).
it is a number that is right next to a variable.
ex: 2x
2 is the coefficent, it will be multiplied by x
ex: 2x
2 is the coefficent, it will be multiplied by x
`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.
A coefficient is a number next to a variable. For example, x has a coefficient of 1 because it is understood to have a 1 next to it in mathematics. In the example 4y, the coefficient would be 4 for y. Hope that helps clear some things out for you.
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
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
A coefficient is a word that basically describes the number in front of the variable( the amount the variable is being multiplied by). Ex. 6y+12 where in this case 6 will be the coefficient.
It's a number in front of something that indicates how many of that particular something is present. That something could be a variable, function or something else.
So in the expression 6x, the coefficient 6 indicates there are 6 x's.
So in the expression 6x, the coefficient 6 indicates there are 6 x's.
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?
We are looking for a and b for this situation:
(x+a)(x+b) = 0
When we multiply the left side out we get
x^2 + bx + ax + ab = 0
So x^2 + (a+b)x + ab = 0
Therefore ab is the constant term.
(And a + b is the coefficient of the x term.)
(x+a)(x+b) = 0
When we multiply the left side out we get
x^2 + bx + ax + ab = 0
So x^2 + (a+b)x + ab = 0
Therefore ab is the constant term.
(And a + b is the coefficient of the x term.)
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.
We are looking for a and b for this situation:
(x+a)(x+b) = 0
When we multiply the left side out we get
x^2 + bx + ax + ab = 0
So x^2 + (a+b)x + ab = 0
Therefore ab is the constant term.
(And a + b is the coefficient of the x term.)
(x+a)(x+b) = 0
When we multiply the left side out we get
x^2 + bx + ax + ab = 0
So x^2 + (a+b)x + ab = 0
Therefore ab is the constant term.
(And a + b is the coefficient of the x term.)
because a and b would be constants in a factored quadratic equation i.e. (x+2)(x+3)
You should watch all the videos in the https://www.khanacademy.org/math/algebra/multiplyingfactoringexpression/multiplyingbinomials section first. But quick answer is that for `(x+a)(x+b)` we multiply every term by every other term, to get `x*x + x*b + a*x + a*b`, which can be rewritten as `x^2 + (a+b)x + a*b`. The constants term is the last term listed: the product of `a` and `b`.
The constant term is a*b because when you do F.O.I.L. when you multiply (x+a)(x+b), you have to multiply a and b at the last stage. This will be a constant number because a and b are constant numbers. Neither is attached to a variable.
Can the 1 and the 9 switch so it's (x+9)(x+1)? Or does it have to be (x+1)(x+9)?
yes they can
remember the commutative property of multiplication
remember the commutative property of multiplication
Of course they can switch. The commutative property of multiplication states that 9*2 is the same as 2* 9.
Let's take a look at your problem:
(X+9)(x+2) = (x+2)(x+9)
Let's use the substitution property of equality and substitute x for 1.
(1+9)(1+2)=(1+2)(1+9)
10(3) = 3(10)
30 = 30
So the answer to your problem is yes
Let's take a look at your problem:
(X+9)(x+2) = (x+2)(x+9)
Let's use the substitution property of equality and substitute x for 1.
(1+9)(1+2)=(1+2)(1+9)
10(3) = 3(10)
30 = 30
So the answer to your problem is yes
Yes. If you switch them, the outcome will be exactly the same. This is because of the commutative property.
For ex., 2*3=6 and 3*2=6.
This is your problem:
Solution #1 (Beginner solution):
(x+9)(x+1). Replace (x+9) with the 2 in the example above. Replace the (x+1) with the 3. Then multiply 2 by 3. It should be 6. Now, switch (x+9) with (x+1). You should still get six.
Solution #2 (Intermediate solution):
(x+9)(x+1). Substitute x with 1 (any other number is also fine). (x+9) would equal 10. (x+1) would equal 2. 10*2 is 20. Now, switch (x+9) with (x+1). You should still get 20.
HOPE THIS HELPED!
For ex., 2*3=6 and 3*2=6.
This is your problem:
Solution #1 (Beginner solution):
(x+9)(x+1). Replace (x+9) with the 2 in the example above. Replace the (x+1) with the 3. Then multiply 2 by 3. It should be 6. Now, switch (x+9) with (x+1). You should still get six.
Solution #2 (Intermediate solution):
(x+9)(x+1). Substitute x with 1 (any other number is also fine). (x+9) would equal 10. (x+1) would equal 2. 10*2 is 20. Now, switch (x+9) with (x+1). You should still get 20.
HOPE THIS HELPED!
yes, they can switch. but when you are dealing with negative and positive signs you can not switch or it will change the problem
Yes they are able to switch and the outcome will still be the same. This is due to the commutative property
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
I think he was asking if in a realworld situation.
So you can find out the equation like if x squared + 9x + 18 = 0
You can factor and find out
You can factor and find out
You need it because on tests such as the ACT ask you to find all possible values for x
Quadratic equations are used to find the more than one possible value of a certain variable because in some instances there can be many possible values of a variable that suit the equation
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.
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.
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.
Not all quadratics are able to be factored. Some quadratic expressions don't even have roots that are real numbers.
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
I presume you mean x^25x+24=0 right because what you wrote isn't really an equation.
So you can try the straight forward way of finding the discriminant and solving it this way (instructions here http://en.wikipedia.org/wiki/Discriminant)
5*54*24=2596=71
x=(5+√71)/2
So you have two solutions for x but if you would want to gain that by factoring it then I think that it would be very hard indeed. (nigh impossible)
So you can try the straight forward way of finding the discriminant and solving it this way (instructions here http://en.wikipedia.org/wiki/Discriminant)
5*54*24=2596=71
x=(5+√71)/2
So you have two solutions for x but if you would want to gain that by factoring it then I think that it would be very hard indeed. (nigh impossible)
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)
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.
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.
You won't be able to know what x is until you set the equation equal to something.
There are no answers since 24 does not have any factors that equal 5. Since a positive times a negative is a negative there are no solutions that multiply to 24, but add to 5.
If you can't factor (which in this case you can't) use the quadratic formula.
http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula
If the equation was = to 0 then this equation is in standard form (ax^2+bx+c) so using this you need to plug in the values of a, b, and c (a=1)(b=5)(c=24) in the quadratic equations and you will get 2 values for x (roots). For this expression the roots or the values of x are not real numbers.
http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula
If the equation was = to 0 then this equation is in standard form (ax^2+bx+c) so using this you need to plug in the values of a, b, and c (a=1)(b=5)(c=24) in the quadratic equations and you will get 2 values for x (roots). For this expression the roots or the values of x are not real numbers.
Are there suppose to be alot of steps to this ?
x^23x10=0
How to factor , what is this called?
Thank you!!
How to factor , what is this called?
Thank you!!
"x" can be 5 or 2. If you wanted to factor it, you would get '(x−5)(x+2) = 0' and now it makes sense that if x is 5 in '(x5)' it turns into 0 and if x is 2 in '(x2)', x would be 0 as well.
I didn't get this could you answer me please
It depends on the technique you need
It honestly can differ on the problem you're doing.
actually once you get the hang of it you can do it in less that 3 or 4 steps
My teacher says 8, & quadratic factoring ?
How do you factor a polynomial when the highest power is greater than ^2?
Long Division of Polynomials or Synthetic Division.
You can also try to factor out x as a GCF if the others also have one more x
e.g.
x^3+5x^2+6x
x(x^2+5x+6)
Then factor what is inside it:
x((x+2)(x+3))
And there it is.
If that's not the case, than it's much more difficult.
e.g.
x^3+5x^2+6x
x(x^2+5x+6)
Then factor what is inside it:
x((x+2)(x+3))
And there it is.
If that's not the case, than it's much more difficult.
Check out the video called Factoring by Grouping or use Completing the Square or the Quadratic Formula.
With difficulty! The more terms there are in the polynomial, the more difficult it is to find an exact factor.
What would be considered a constant?
A constant is a number with no variable. It is called a constant because it's value does not change.
a number with no visible variable right next to it. it is not visisble because that variable is to the power of 0, which makes it equal 1( 1*the number is still that number)
hope this helps!
hope this helps!
`*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
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"
A constant is an element that is fixed, contrast this with variable. In the equation y = x + 5, x and y are variables, '5' is a constant
I just came. 0__o


at 14:46 we do not have to multiply negative 1 to both (x3)(x+8) ?
Remember that multiplication is associative: a(bc) = (ab)c.
So (1) ( (x3)(x+8) ) = ( (1)(x3) ) (x+8) = (x + 3)(x+8)
So (1) ( (x3)(x+8) ) = ( (1)(x3) ) (x+8) = (x + 3)(x+8)
No, because x*x would equal positive x².
The goal is to get x².
So, x*x = x²
Hope this helps!
The goal is to get x².
So, x*x = x²
Hope this helps!
what does quadratic mean
This is a really good question. It confused me for a long time, because "quad" usually means "four".
A quadratic equation is one where one of the terms is raised to the second power, and there are no terms raised to higher powers. There are probably a few more restrictions if we wanted to be precise. So a quadratic equation might look like this:
```
ax² + bx + c = d
```
A more concrete example might be:
``` 3x² + 2x + 1 = 0```
or ```2x² = 8```
I always wondered why it was called quadratic when it goes up to the second power, and not the fourth. It turns out the same word is used for square as for four. I think this is because if you want the area of a square, you multiply the length by itself. The same goes for cubic. A cubic has a term raised to the third power, and that is what you have to do to get the volume of a cube.
It stops there as far as I know. I'm not sure what something raised to the fourth power would be called. Maybe a tesseratic?
A quadratic equation is one where one of the terms is raised to the second power, and there are no terms raised to higher powers. There are probably a few more restrictions if we wanted to be precise. So a quadratic equation might look like this:
```
ax² + bx + c = d
```
A more concrete example might be:
``` 3x² + 2x + 1 = 0```
or ```2x² = 8```
I always wondered why it was called quadratic when it goes up to the second power, and not the fourth. It turns out the same word is used for square as for four. I think this is because if you want the area of a square, you multiply the length by itself. The same goes for cubic. A cubic has a term raised to the third power, and that is what you have to do to get the volume of a cube.
It stops there as far as I know. I'm not sure what something raised to the fourth power would be called. Maybe a tesseratic?
It's called that because power of 2 means square, and a square has four sides(?).
Thank You Guys For explaining everything to me
i want to learn how to factor polynomials with 4&5 degree Quartic and Quintic expressions... can someone plz help me find the video? thanks
Unlike quadratic or even cubic equations, there is no easy way to factor quartic and quintic expressions. That is an exceedingly advanced topic. With the exception of a form of quartic expression called a "biquadratic", factoring quartic and quintic expressions requires the most prodigious of mathematical skills.
where can practice on these kind of problems?
Should I work on polynomials before quadratics? I felt I needed quadratics for some parts of functions, but trooped through. I've felt sometimes I should have gone through polynomials before most everything and polynomials is misplaced in the playlist, but I'm not sure. Should I work through polynomials before quadratics first? I don't want to just know how to do the problems, I want to be able to visualize it well and have a conceptual knowledge.
Since a quadratic expression is a special form of polynomial, you are actually working on both. For most polynomial problems you will be given three terms and asked to factor it as the product of two binomials or one binomial with another polynomial. If you understand that at any time a coefficient can be replaced with another variable then it will be easier for you to understand how to factor all polynomials. You will often see the equation Ax^2 + Bx + C = 0 and that just shows the general form of all quadratic equations. When you understand how B is related to A and C then it is easier to factor these types of expressions.
There are multiple ways to visualize polynomials because they can be used to calculate real world problems. An easy visualization is that of a rectangle with a variable length and width and the area is equal to the length times the width. The two binomials define the width and length while the final polynomial represents the area.
Here is a tool to examine factoring quadratic expressions. It doesn't address polynomials with multiple variables, but is a good way to see the relationship of the a and b in the (x + a)(x + b).
http://www.khanacademy.org/cs/quadraticfactorization/1386235304
There are multiple ways to visualize polynomials because they can be used to calculate real world problems. An easy visualization is that of a rectangle with a variable length and width and the area is equal to the length times the width. The two binomials define the width and length while the final polynomial represents the area.
Here is a tool to examine factoring quadratic expressions. It doesn't address polynomials with multiple variables, but is a good way to see the relationship of the a and b in the (x + a)(x + b).
http://www.khanacademy.org/cs/quadraticfactorization/1386235304
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).
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.
`(x + 2.5)(x + 5.7)` = x(x + 5.7) + 2.5(x + 5.7)
= x^2 + 5.7x + 2.5x + 14.25
= `*x^2 + 8.2x + 14.25*`
This can be converted into integers by multiplying each term by 20
20(x + 2.5)(x + 5.7) = 20x^2 + 164x + 285
To factorise 20x^2 + 164x + 285 compute the "outer" product
i.e. 20x^2 x 285 = 5700x^2
The "middle" term is 164x so require two x terms which multiply to 5700x^2 and add to 164x (note both terms must be positive)
An exhaustive lengthy systematic search is
x 5700x, 2x 2850x, 3x 1900x, 4x 1425x, 5x 1140x, 6x 950x, 10x 570x,
12x 475x, 15x 380x, 19x 300x, 20x 285x, 25x 228x, 30x 190x, 38x 150x,
*50x 114x* (*this sums to 164x*), 57x 100x, 60x 95x, 75x 76x
20x^2 + 164x + 285 = 20x^2 + 50x + 114x + 285
= 10x(2x + 5) + 57(2x + 5)
= (10x + 57)(2x + 5)
(20x^2 + 164x + 285)/20 = (10x + 57)/10 x (2x + 5)/2
= `(x + 5.7)(x + 2.5)`
= x^2 + 5.7x + 2.5x + 14.25
= `*x^2 + 8.2x + 14.25*`
This can be converted into integers by multiplying each term by 20
20(x + 2.5)(x + 5.7) = 20x^2 + 164x + 285
To factorise 20x^2 + 164x + 285 compute the "outer" product
i.e. 20x^2 x 285 = 5700x^2
The "middle" term is 164x so require two x terms which multiply to 5700x^2 and add to 164x (note both terms must be positive)
An exhaustive lengthy systematic search is
x 5700x, 2x 2850x, 3x 1900x, 4x 1425x, 5x 1140x, 6x 950x, 10x 570x,
12x 475x, 15x 380x, 19x 300x, 20x 285x, 25x 228x, 30x 190x, 38x 150x,
*50x 114x* (*this sums to 164x*), 57x 100x, 60x 95x, 75x 76x
20x^2 + 164x + 285 = 20x^2 + 50x + 114x + 285
= 10x(2x + 5) + 57(2x + 5)
= (10x + 57)(2x + 5)
(20x^2 + 164x + 285)/20 = (10x + 57)/10 x (2x + 5)/2
= `(x + 5.7)(x + 2.5)`
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.
I would use the quadratic formula.
If the leading coefficient is 1, is it a MUST to factor out the 1?
Not necessarily a "must", but it is much easier if you do so.
So say a quadratic equation like this has numbers on a much larger scale, or with decimals. Are we supposed to just guess which set of a and b makes a+b=c and a*b=c true? Isn't there a more solid way?
In those cases you probably wouldn't factorise. The idea of factorising is that we take out factors ie whole numbers so we wouldn't do it with decimals. There are other ways of solving quadratics  both of these examples would be easiest to solve with the quadratic equation as this would work regardless of which numbers you input.
If there are no common factors in the large numbers that can be removed to make the equation more manageable, or if there is no way to convert the decimals into fractions that are easy to manipulate, then it will be faster and less error prone to use the quadratic formula to solve the quadratic in those cases.
With practice and experience, you will soon be able to look at a quadratic and know which will be the best approach to solve it.
With practice and experience, you will soon be able to look at a quadratic and know which will be the best approach to solve it.
Is it OK to switch the order of the equation? For instance, instead of x*2+10x+9, could you solve it as 9+10x+x*2? Or would that just be harder? Would it also work that way for other problems? Can you still factor it in this form?
Theoretically, that isn't incorrect, but, as mathematicians like to be neat freaks, it's better to write x² + 10x + 9. In some cases it's just easier to solve.
It is okay if you switch the order of the terms in the equation but most people prefer it in the form ax^2+bx+c (standard form) because it is easy to plug in values from standard form to the quadratic equation.
Yes, it is OK to switch the order of the equation. For example in your question, x*2 + 10x + 9 could be 9 + 10x + x*2. It could also be any other form of the equation, like 10x + 9 + x*2, and so on. However, the first form, x*2 + 10x + 9, is the most common because it is the easiest to factor out.
you can always change the expression as much as you like as long as you perform the same action to all the terms. for example 2x + 3y can be multiplied by 2 to get 2(2x) + 2(3y) = 4x + 6y. you can still factor these problems and often messing with them a little makes them easier to factor. if the difficulty increases then you are probably not supposed to do that. remember math is your friend even through its maze of complexity. sorry my comma key is messed up and I cant be bothered to capitalize. this is the internet. hope this helps.
u is right ma homie dawg dont try an play me dawg!
What if the leading coefficient is greater than 1?
There are other techniques you will learn soon that deal with that situation.
What if you get a polynomial with 4 terms? Even 5?
The process of factoring 3rd degree or 4th degree polynomial is quite long for one answer.
Factoring 3rd degree Polynomials:
http://m.wikihow.com/FactoraCubicPolynomial
Factoring 4th degree Polynomials:
http://mathcentral.uregina.ca/QQ/database/QQ.09.05/kyle2.html
I hope this helps!
Factoring 3rd degree Polynomials:
http://m.wikihow.com/FactoraCubicPolynomial
Factoring 4th degree Polynomials:
http://mathcentral.uregina.ca/QQ/database/QQ.09.05/kyle2.html
I hope this helps!
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.
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.
Isn't there a box method you could use to find the same answer?
The box method is the same thing as what was done in this video except that it has more visualization so it takes less time to find the answer.
Im not sure about a box but you can use a diamond.Have it so there are 4 smaller diamonds inside ( Like North, South, East, West; one on top, one on the bottom, and two to the sides). You want to leave the side diamonds open because youre trying to find 2 numbers that multiply to be a nuimber (which goes in the top box) and the same two numbers whose sum is the bottom number. ex: (x)^2+12x+7 top #: 12 Bottom #: 7 Left #: 4 or 3 Right #: 3 or 4. Since 4x3=12 and 4+3=7, it works
At 14:47 and 16:27, did Sal mean 1 ( ( x + a ) + ( x + b ) )
Given:
``` x^2  5•x + 24```
When attempting to factorize, it's preferable to have your x^2 positive.
So factor out a `1`:
``` 1•(x^2 + 5•x  24)```
For factoring `(x^2 + 5•x  24)`, we need to find number pairs that add to the middle coefficient, and whose product equals the last term:
```n1 n2 sum product
1 24 23 24
2 12 10 24
3 8 5 24 << found terms```
Group with new terms:
``` (x  3)•(x + 8)```
From above, we still need to account for the originally factored `1`:
``` [1•(x  3)•(x + 8)]```
``` x^2  5•x + 24```
When attempting to factorize, it's preferable to have your x^2 positive.
So factor out a `1`:
``` 1•(x^2 + 5•x  24)```
For factoring `(x^2 + 5•x  24)`, we need to find number pairs that add to the middle coefficient, and whose product equals the last term:
```n1 n2 sum product
1 24 23 24
2 12 10 24
3 8 5 24 << found terms```
Group with new terms:
``` (x  3)•(x + 8)```
From above, we still need to account for the originally factored `1`:
``` [1•(x  3)•(x + 8)]```
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.
Hi! I just wanted to ask about: What is the meaning of "quadratic"?
A polynomial or equation in general is considered a "quadratic" when it has a degree of 2 (meaning the highest exponent a variable in the equation is being raised to is 2).
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.
This will only work if you mean (x  a)(x  b) = 0
If you meant that it equals 0, then one factor has to be equal to zero.
So, for (x  a) to be equal to 0, x would have to be a, because a  a = 0
Also, for (x  b) to be equal to 0, x would have to be b, because b  b = 0.
Overall, the two answers for this equation would be x = a and b
However, be careful, because this only works if the expression is equal to 0
If you meant that it equals 0, then one factor has to be equal to zero.
So, for (x  a) to be equal to 0, x would have to be a, because a  a = 0
Also, for (x  b) to be equal to 0, x would have to be b, because b  b = 0.
Overall, the two answers for this equation would be x = a and b
However, be careful, because this only works if the expression is equal to 0
thank that help
so all the time the answer should be A and
Why is the last term called a constant term?
The last term is called a constant because when x changes, the last term doesn't change since it doesn't have x in it. It's value stays *constant*.
Is a quadratic equation also a trinomial equation, or are exponents not supposed to be in polynomial equations?
Yes, a quadratic equation is a trinomial. A trinomial is simply any algebraic expression with 3 terms. In fact, the 3 terms can even all have different variables. For instance, "2x + 5y  7z" is also a trinomial.
Great question.
Great question.
This is honestly VERY confusing. I'm in high school, failing my Algebra I class, and i'm using Khan Academy to help...But how do variables like a and b and x and y add together? They're different variables! Please help!
Different variables (ex. x and y) can't be combined because they aren't *like terms*
_Example_
7x + 2x  3y
= 9x  3y *Can't be simplified further*
Let me know if you need this explained any further!
_Example_
7x + 2x  3y
= 9x  3y *Can't be simplified further*
Let me know if you need this explained any further!
Post a specific problem.
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.
A quadratic equation is second degree, meaning it's squared. In Latin, "quadratus" means square.
ok, so i get the concept, but in the problems, it says X = _____ and X = _____
how am i supposed to give my answer in that form?
how am i supposed to give my answer in that form?
Suppose you have x^2  11x + 24.
In this case a=8 and b=3.
So the factoring results in (x  8)(x  3).
Now, what values of x make the expression equal to zero?
When x = 8, we have (8  8)(8  3) = (0)(5) = 0
When x = 3, we have (3  8)(3  3) = (5)(0) = 0
Therefore the solution is: x = 8 and x = 3.
In this case a=8 and b=3.
So the factoring results in (x  8)(x  3).
Now, what values of x make the expression equal to zero?
When x = 8, we have (8  8)(8  3) = (0)(5) = 0
When x = 3, we have (3  8)(3  3) = (5)(0) = 0
Therefore the solution is: x = 8 and x = 3.
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
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
I'd bet that this is on the SAT. Many jobs require math skills, especially computer related ones (and there are LOTS). Best to learn as much as you can, right?
Quadratic equations are used in planetary motion, missile trajectories, pendulum periods, quantum theory, falling objects, sundials, airplane lift, wireless communications, microelectronics, hydrodynamics and acoustics just to name a few. Any motion or change in state that resembles a parabola can be described with a quadratic equation. The factoring part is important when converting these equations to other forms and its necessary to solve for the second degree variable (x^2, k^2, etc.)
solving word problems. especially age ones.
Is there any kind of official, mathematical way for finding the 'a' and 'b' in a quadratic? Sal used a guessandcheck method, but I that takes me a long time.
Yes, there is a method for directly factoring any quadratic without any trial and error. It always works but it is more difficult than the simpler methods you start out with. It makes use of the quadratic formula. Here is how to do it:
For a quadratic in the form of ax² + bx + c
Let q = √(b²  4ac)
x₁ = ½ (q  b) / a
x₂ = −½(q+b)/a
The factors will be: a(xx₁)(xx₂)
This will often involve fractions. If you do have them, be *sure* to reduce the fractions to the simplest form  *this is important*.
To get rid of the fractions, you multiply and divide by the denominators and then use the distributive property for the multiplication part. This doesn't change any values because if you multiply and divide by the same number (other than 0) then it is the same as multiplying by 1. For example if your factors were:
6(x½)(x+⅔)
Then you would get rid of the fractions like this:
6∙(½)(2)(x½)∙(⅓)(3)(x+⅔)
6∙(½)(2x1)∙(⅓)(3x+2)
6∙(½)(⅓)∙(2x1)∙(3x+2)
(2x1)(3x+2)
Now let us apply this to an example:
10x² − 7x − 12
a=10, b= −7, c= −12
q = √(b²  4ac)
q = √[−7)²− 4(10)(−12)]
q= √[49− (−480]
q = √529
q = 23
x₁ = ½ (q  b) / a
x₁ = ½ [23 −(−7)] /10
x₁ = ½ (30) /10
x₁ = ³⁄₂
x₂ = −½(q+b)/a
x₂ = −½(237)/10
x₂ = −½(16)/10
x₂ = −⅘
So the factored form is:
a(xx₁)(xx₂)
10(x³⁄₂)(x+⅘)
Now multiply and divide by the denominators to get rid of the fractions:
10(½)(2)(x³⁄₂)∙(⅕)(5)(x+⅘)
=10(½)(2x3)∙(⅕)(5x+4)
=10(½)(⅕)∙(2x3)(5x+4)
=(¹⁰⁄₁₀)(2x3)(5x+4)
=(2x3)(5x+4)
For a quadratic in the form of ax² + bx + c
Let q = √(b²  4ac)
x₁ = ½ (q  b) / a
x₂ = −½(q+b)/a
The factors will be: a(xx₁)(xx₂)
This will often involve fractions. If you do have them, be *sure* to reduce the fractions to the simplest form  *this is important*.
To get rid of the fractions, you multiply and divide by the denominators and then use the distributive property for the multiplication part. This doesn't change any values because if you multiply and divide by the same number (other than 0) then it is the same as multiplying by 1. For example if your factors were:
6(x½)(x+⅔)
Then you would get rid of the fractions like this:
6∙(½)(2)(x½)∙(⅓)(3)(x+⅔)
6∙(½)(2x1)∙(⅓)(3x+2)
6∙(½)(⅓)∙(2x1)∙(3x+2)
(2x1)(3x+2)
Now let us apply this to an example:
10x² − 7x − 12
a=10, b= −7, c= −12
q = √(b²  4ac)
q = √[−7)²− 4(10)(−12)]
q= √[49− (−480]
q = √529
q = 23
x₁ = ½ (q  b) / a
x₁ = ½ [23 −(−7)] /10
x₁ = ½ (30) /10
x₁ = ³⁄₂
x₂ = −½(q+b)/a
x₂ = −½(237)/10
x₂ = −½(16)/10
x₂ = −⅘
So the factored form is:
a(xx₁)(xx₂)
10(x³⁄₂)(x+⅘)
Now multiply and divide by the denominators to get rid of the fractions:
10(½)(2)(x³⁄₂)∙(⅕)(5)(x+⅘)
=10(½)(2x3)∙(⅕)(5x+4)
=10(½)(⅕)∙(2x3)(5x+4)
=(¹⁰⁄₁₀)(2x3)(5x+4)
=(2x3)(5x+4)
Isnt the formula for quadratic ax^2+bx+c=0?
That, I believe, is the standard form of a quadratic equation.
The quadratic formula is used to solve equations like these and is x = [b plus or minus sqrt(b^2  4 * a * c)] / (2 * a)
The quadratic formula is used to solve equations like these and is x = [b plus or minus sqrt(b^2  4 * a * c)] / (2 * a)
That is the standard formula for all quadratic equations.
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