Quadratics and polynomials
Factoring quadratic expressions
None
Introduction to grouping
Sal factors t^2+8t+15 as (t+3)(t+5), first using the "sumproduct" method, then using the grouping method.
Discussion and questions for this video
 Factor t squared plus 8t plus 15.
 So let's just think about what happens
 if we multiply two binomials t plus a times t plus b.
 And I'm using t here because t is
 the variable in the polynomial that we need to factor.
 So if you multiply this out, applying
 the distributive property twice or using FOIL,
 you get t times t, which is t squared.
 Plus t times b, which is bt.
 Plus a times t, which is at.
 Plus a times b, which is ab.
 We essentially multiplied every term here
 by every term over there.
 And then we have two t terms, I guess you could call them,
 this bt plus at.
 So we could combine those.
 And we get t squared plus a plus bt
 I could have written this as b plus at as well plus ab.
 Now, if we compare this to this right over here,
 we see that we have a similar pattern.
 Our coefficient on the second degree term is 1.
 Our coefficient on the second degree term here is 1.
 We didn't have to write it.
 Then a plus b is the coefficient on t.
 So 8 right over here, this 8 could be a plus b.
 And then finally, our constant term, ab, that could be 15.
 So if we wanted to factor this out,
 we just have to find and a and a b where their product is 15
 and their sum is 8 in general.
 In general, if you ever see and I'll write it
 in the more traditional with an x variable
 if you see anything of the form x squared plus bx plus c,
 the coefficient here is 1.
 Then you just have to find two numbers
 whose sum is equal to this thing right here
 and whose product is equal to that thing right there.
 Whose sum is equal to 8 and whose product is equal to 15.
 So what are two numbers that add up to 8 whose products are 15?
 So if you just factor 15, it could be 1 and 15.
 Those don't add up to 8 in any way.
 3 and 5, those do add up to 8.
 So a and b could be 3 and 5.
 So a and b, this could be 3 times 5,
 and then 8 is 3 plus 5.
 Now we could just go straight and factor this and say,
 hey, this is t plus 3 times t plus 5,
 since we already figured out a and b are.
 But what I want to do is kind of factor this by grouping.
 So I'm essentially going to go in reverse steps
 from what I just showed you.
 So this polynomial right here, I'm
 going to write it as t squared plus instead of writing
 8t I'm going to write it as the sum of at plus bt,
 or as the sum of 3t plus 5t.
 So plus 3t plus 5t.
 So essentially I've started here.
 And then I'm going to this step where I break up
 that middle term into the coefficients that
 add up to the 8.
 And then finally plus 15.
 And now I can factor it by grouping.
 These two guys right over here have a common factor t.
 And then these two guys over here have a common factor 5.
 So let's factor out the t in this first expression
 right over here, or this part of the expression.
 So it's t times t plus 3.
 Plus and then over here, if you factor out a 5
 you get a 5 times t plus 3.
 5t divided by 5 is t.
 15 divided by 5 is 3.
 And now you can factor out a t plus 3.
 You have a t plus 3 being multiplied
 times both of these terms.
 So let's factor that out.
 So it becomes t plus 3 times t plus 5.
 Let me write that plus a little bit neater.
 Times t plus 5.
 And we are done.
 But we actually didn't even have to do this grouping step,
 although hopefully you see that it does work.
 We could have just said, look, from this pattern
 over here I have two numbers that add up to 8
 and their product is 15.
 So this must be t plus 3 times t plus 5.
 Or t plus 5 times t plus 3, either way.
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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This discussion area is not meant for answering homework questions.
What happens at 3:35???? I got really confused at that part.
He factors (t+3) out. You can treat t*(t + 3) + 5*(t+3) as though it has parenthesis around it (t*(t + 3) + 5*(t+3)). If you do that you can see that you have two things (t and 5) multiplied by (t+3) so you can factor out a (t+3). This leaves you with a (t+3) multiplied by what is left inside the parenthesis which is (t+5). So the expression ends up being (t+3)(t+5)
In other videos he has used the phrase "undistributing".
Looking back at the distributive property, if you start with
x(a+b) you can "distribute" x among the terms inside the parentheses like this:
`x(a+b) = xa + xb`
Or, going the other way, if you have xa+xb, you can factor out ("undistribute") the x term:
`xa + xb = x(a+b)`
Now, in the video example, instead of a simple "x" we have "(t+3)". Use "(t+3)" the same way we used "x" in my example above. If you start with (t+3)(t+5) you can distribute the (t+3) term like this:
`(t+3)(t+5) = (t+3)t + (t+3)5 `
Or, to make it a little more readable:
`(t+3)(t+5) = t(t+3) + 5(t+3)`
Now in his example in the video, he just "undistributed" (t+3) in the expression above:
`t(x) + 5(x) = (t+5)(x)`
`t(t+3) + 5(t+3) = (t+5)(t+3)`
` = (t+3)(t+5)`
It is a little overcomplicated for this example. But it's a skill you will need in order to tackle more complex problems.
Looking back at the distributive property, if you start with
x(a+b) you can "distribute" x among the terms inside the parentheses like this:
`x(a+b) = xa + xb`
Or, going the other way, if you have xa+xb, you can factor out ("undistribute") the x term:
`xa + xb = x(a+b)`
Now, in the video example, instead of a simple "x" we have "(t+3)". Use "(t+3)" the same way we used "x" in my example above. If you start with (t+3)(t+5) you can distribute the (t+3) term like this:
`(t+3)(t+5) = (t+3)t + (t+3)5 `
Or, to make it a little more readable:
`(t+3)(t+5) = t(t+3) + 5(t+3)`
Now in his example in the video, he just "undistributed" (t+3) in the expression above:
`t(x) + 5(x) = (t+5)(x)`
`t(t+3) + 5(t+3) = (t+5)(t+3)`
` = (t+3)(t+5)`
It is a little overcomplicated for this example. But it's a skill you will need in order to tackle more complex problems.
It is kind of confusing, i know, but he over complicated it in my opinion.
where is the other t+3 gone can anyone explain please
You didn't specify at which time but I guess that you meant somewhere at 3:30 where we have:
(t²+3t) + (5t+15)
t(t+3) + 5(t+3) → We factored out t from (t²+3t) and 5 from (5t+15)
(t+3) (t+5) → now this is part that you were asking for, you're wondering where did the other (t+3) go, actually if you look at t(t+3) + 5(t+3) carefully you might realize that they both have (t+3) so we can just factor that out and add their coefficients in another parenthesis as another variable.
And that is why (t+3) (t+5) is our final answer!
(t²+3t) + (5t+15)
t(t+3) + 5(t+3) → We factored out t from (t²+3t) and 5 from (5t+15)
(t+3) (t+5) → now this is part that you were asking for, you're wondering where did the other (t+3) go, actually if you look at t(t+3) + 5(t+3) carefully you might realize that they both have (t+3) so we can just factor that out and add their coefficients in another parenthesis as another variable.
And that is why (t+3) (t+5) is our final answer!
I have a question?
HOW DO YOU SOLVE THIS!
Example:
3/2x+4=9
Thanks, Please help!
HOW DO YOU SOLVE THIS!
Example:
3/2x+4=9
Thanks, Please help!
3/2x + 4 = 9. To solve for x, we can do a few things. I would suggest one method though, let me show you...
3/2x + 4 = 9
subtract 4 from both sides.
3/2x = 13
Now, multiply both sides by 2/3.
(2/3)*3/2x = (2/3)*(13/1)
6/6x = 26/3
6/6 = 1, so
x = 26/3
and we are done. Tell me if that helped or not.
3/2x + 4 = 9
subtract 4 from both sides.
3/2x = 13
Now, multiply both sides by 2/3.
(2/3)*3/2x = (2/3)*(13/1)
6/6x = 26/3
6/6 = 1, so
x = 26/3
and we are done. Tell me if that helped or not.
Why do the (t+3)'s you factored out of the trinomial cancel each other, and if they did wouldn't they leave a 1 instead of a a single (t+3) remaining? sorry if this was a badly worded question btw
When doing the KhanAcademy lessons for Factoring Polynomials 1 (https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_factoring/e/factoring_polynomials_1) I was given this problem to factor.
x2−10x+24 (x to the second power minus by ten x plus twenty four)
I wrote down this answer and it solves the right way,
(x + 2)(x  12)
But KhanAcademy said that I got it wrong and that the answer was actually,
(x−6)(x−4)
Can anybody tell me where I went wrong? I just don't get it. They both solve the same way.
x2−10x+24 (x to the second power minus by ten x plus twenty four)
I wrote down this answer and it solves the right way,
(x + 2)(x  12)
But KhanAcademy said that I got it wrong and that the answer was actually,
(x−6)(x−4)
Can anybody tell me where I went wrong? I just don't get it. They both solve the same way.
So the equation they want is x^210x+24 Correct?
If you multiplied yours out, it would be x^210x24.
(x+2)(x12) = x*x12x+2x24 = x^210x24 because 2 times 12 is 24.
While the answer provided
(x6)(x4) = x*x6x4x+24 = x^210x+24. Negative 6 times negative 4 is positive 24.Which is what they wanted.
You got really close, just gotta watch that minus sign on the last term.
If you multiplied yours out, it would be x^210x24.
(x+2)(x12) = x*x12x+2x24 = x^210x24 because 2 times 12 is 24.
While the answer provided
(x6)(x4) = x*x6x4x+24 = x^210x+24. Negative 6 times negative 4 is positive 24.Which is what they wanted.
You got really close, just gotta watch that minus sign on the last term.
You went wrong with the last term. The last term has to be +24, while your answer gives 24.
what if the problem is x^22x8. the factors of 8 dont add up to 2? like in the video 5 and3 add up to 8
sifuentesj,
4*2 = 8
and 4+2 = 2
So x²2x8 can be factored.
There are other methods such as "completing the square" and the quadratic formula that can be used when you can't easily factor the expression.
Here is a video on completing the square
https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/solvingquadraticequationsbycompletingthesquare
And on the quadratic formula https://www.khanacademy.org/math/algebra/quadratics/quadratic_formula/v/usingthequadraticformula
I hope that helps make it click for you.
4*2 = 8
and 4+2 = 2
So x²2x8 can be factored.
There are other methods such as "completing the square" and the quadratic formula that can be used when you can't easily factor the expression.
Here is a video on completing the square
https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/solvingquadraticequationsbycompletingthesquare
And on the quadratic formula https://www.khanacademy.org/math/algebra/quadratics/quadratic_formula/v/usingthequadraticformula
I hope that helps make it click for you.
Because it is subtracting 8 and not addition for the last term, one of the factors of 8 is going to be negative. let's say 2 and 4.
2 + (4) can also be written as 2  4
2  4 = 2.
you have to pay attention to the negatives.
2 + (4) can also be written as 2  4
2  4 = 2.
you have to pay attention to the negatives.
This question is a bit off topic but can anyone explain how I would factor an expression such as this one
x^2+(3y+4)x+(2y+3)(y+1) I have been stuck on this question for a while and any help would be useful, Thank you in advance.
x^2+(3y+4)x+(2y+3)(y+1) I have been stuck on this question for a while and any help would be useful, Thank you in advance.
The sum of two numbers is 3y + 4 and the product of two numbers is (2y + 3)(y + 1)
So this equation can be factored as (x + 2y + 3)(x + y + 1)
This problem is of the form (x + a)(x + b) where a = 2y + 3 and b = y + 1
So this equation can be factored as (x + 2y + 3)(x + y + 1)
This problem is of the form (x + a)(x + b) where a = 2y + 3 and b = y + 1
How would you factor an equation ax^2+bx+c
To factor the expression ax² + bx + c, first solve the equation ax² + bx + c = 0 to find the roots of the polynomial (this can always be accomplished using the quadratic formula). Suppose ß and µ are roots of said equation. Then both (x  ß) and (x  µ) must be factors of the expression. We may then write ax² + bx + c = a(x  ß)(x  µ).
Note: ß and µ need not be distinct, and they may even be complex conjugates of each other.
Note: ß and µ need not be distinct, and they may even be complex conjugates of each other.
"A quadratic polynomial is always factorable, though not always with rational roots."
( Help was provided by Simon )
( Help was provided by Simon )
I want to see an example of factoring a trinomial in the type of ax(squared) + bx + c where a is greater than 1 and a, b, c do not have a common factor greater than 1.
Julie  Chuck's answer leads you to the quadratic formula, which is a great onesizefitsall solution to polynomials like those in this video. But it doesn't answer your question: if you're looking for a way to factor that's analogous to the way above, where "a" is not 1, check out this video:
https://www.khanacademy.org/math/algebra/multiplyingfactoringexpression/factoringquadraticexpressions/v/factoringtrinomialswithacommonfactor
The quadratic formula becomes more important when you encounter quadratic polynomials that you can't factor easily  often roots and irrational values result.
https://www.khanacademy.org/math/algebra/multiplyingfactoringexpression/factoringquadraticexpressions/v/factoringtrinomialswithacommonfactor
The quadratic formula becomes more important when you encounter quadratic polynomials that you can't factor easily  often roots and irrational values result.
Julie,
This is close to what you wanted: http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/usingthequadraticformula
And the practice session here :http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/e/quadratic_equation
has some problems that fit your request. If you use the "i'd like a hint" button, it will show you how to work each problem step by step.
This is close to what you wanted: http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/usingthequadraticformula
And the practice session here :http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/e/quadratic_equation
has some problems that fit your request. If you use the "i'd like a hint" button, it will show you how to work each problem step by step.
at 2:27 he mentions grouping, what is grouping
Taking out separate thing and putting them into groups
What if at 3:40 you had t(t+3) + 5(t3). How do you deal with one of the values being negative?
At the end, I seem to have lost my step. What did you do? I was lost.
2xto the forth +2xto the thirdxsquaredx
ok, now what if the question is x^2 x 8 = 0 ?
I'm just having trouble coming up with the two numbers for the factoring...
I'm just having trouble coming up with the two numbers for the factoring...
Sal is an awesome teacher explaining it in short times and no questions needed for his methods
Is there an easier way of doing this?
You find factors for the last term and then add them together and the ones that add together to equal the middle coefficient are the ones that you put after x.
Eg. t^2+8t+15
The factors of 15 are 5 and 3, and 1 and 15. 5 and 3 add together to make 8, which is the middle coefficient, so the factored equation is (x+3)(x+5)
Eg. t^2+8t+15
The factors of 15 are 5 and 3, and 1 and 15. 5 and 3 add together to make 8, which is the middle coefficient, so the factored equation is (x+3)(x+5)
(x  1)(x + 3) = 4 what is the next step
Expand the polynomial, to get x^2 + 2x  3 = 4
Set the equation equal to zero, and then factor.
Set the equation equal to zero, and then factor.
I know this is not a quadratic equation, but how would I factor a polynomial such as this one: 2p^3 + 5p^2 + 6p + 15 ?
There are a variety of methods, the one that you try first is factoring by grouping. But note it won't always work. Here's how to do it with your example:
2p³ + 5p² + 6p + 15
Step 1: Group
(2p³ + 5p²) + (6p + 15)
Step 2: Factor each group's GCF
p²(2p + 5) + 3(2p + 5)
Step 3: If there is a common factor, factor that out of the sum:
(2p + 5)(p² + 3)
Step 4: If the quadratic factor can be factored, do so:
(p²+3) has no real factors
So the fully factored form is:
(2p + 5)(p² + 3)
2p³ + 5p² + 6p + 15
Step 1: Group
(2p³ + 5p²) + (6p + 15)
Step 2: Factor each group's GCF
p²(2p + 5) + 3(2p + 5)
Step 3: If there is a common factor, factor that out of the sum:
(2p + 5)(p² + 3)
Step 4: If the quadratic factor can be factored, do so:
(p²+3) has no real factors
So the fully factored form is:
(2p + 5)(p² + 3)
if a polynomial cannot be factored completely and you use the quadratic formula to solve for the variable should you simplify and factor the beginning polynomial as far as you can before you use the quadratic formula?
Thanks!
Thanks!
so the final answer, the thing you would write down and box is "(t+3)(t+5)" right?
What if the leading coefficient is 1?
It does not make a difference. Just do the same thing that you normally do. Ex: x^2+5x6
=> (x+2)(x3).
=> (x+2)(x3).
Please Help! How would I factor the quadratic 6x^245x24=0, thanks.
First lets notice that all of the coefficients are divisible by 3. We can factor that out to get 3(2x^215x8).
So now we have ax^2 + bx + c and we're looking for something with the form of (mx + p)(nx + q) where a = mn, b = pn + mq, and c = pq.
a = 2 so m and n have to be 1 and 2. We now have 3(2x+p)(x+q).
b = p + 2q and c = pq. c also equals 8 so p and q must be { 1, 1, 2, 2, 4, 4, 8, 8}. If you test out all of the option you'll see that 8 * 1 = 8 (c) and 1 + 2 * 8 = 15 (b).
Our answer is 3(2x  1)(x + 8)
So now we have ax^2 + bx + c and we're looking for something with the form of (mx + p)(nx + q) where a = mn, b = pn + mq, and c = pq.
a = 2 so m and n have to be 1 and 2. We now have 3(2x+p)(x+q).
b = p + 2q and c = pq. c also equals 8 so p and q must be { 1, 1, 2, 2, 4, 4, 8, 8}. If you test out all of the option you'll see that 8 * 1 = 8 (c) and 1 + 2 * 8 = 15 (b).
Our answer is 3(2x  1)(x + 8)
Thank you! Thank you! Thank you!
I understand everything except at 3:28, when t(t+3) + 5(t+3) equals (t+3)(t+5), I do not know how Sal got this, plz someone explain to me....
By the way Sal is awesome!!! LOL
By the way Sal is awesome!!! LOL
Watch 4:28
your welcome
Thanks Jules :)
You factor even further. In t(t+3) + 5(t+3) you can see the two terms have (t+3) in common. You then factor that out of t(t+3) + 5(t+3) and you get (t+3) times what is left over, or t+5 which you write as (t+3)(t+5) :)
You distribute
How does he get from t(t + 3) + 5(t + 3) to (t + 3)(t + 5)
I'm interested why and how the factoring out of t + 3 works, exactly which places are factored out.
I'm interested why and how the factoring out of t + 3 works, exactly which places are factored out.
Answering my own question with the help of my GF, important is the following math rule:
a * b + a * c = a (b + c)
*Why is this?*
(a * b) + (a * c)
____________ * a = (b + c) * a
a
So to get from t(t + 3) + 5( + 3) to (t + 3)(t + 5) we can do the following:
t(t+3) + 5(t+3)
____________ * (t+3) = (t+5) * (t+3)
(t+3)
a * b + a * c = a (b + c)
*Why is this?*
(a * b) + (a * c)
____________ * a = (b + c) * a
a
So to get from t(t + 3) + 5( + 3) to (t + 3)(t + 5) we can do the following:
t(t+3) + 5(t+3)
____________ * (t+3) = (t+5) * (t+3)
(t+3)
Am I performing a correct arithmetic operation if I factor like following:
(x+4)(4+4)=8x+32
I haven't seen anything like it, so I got a bit confused.
(x+4)(4+4)=8x+32
I haven't seen anything like it, so I got a bit confused.
Is 6x^213x+5 = (3x5)(2x+1)?
No. Use FOIL First Inside Outside Last to do it backwards
3x(2x) 10x +3x  5 =
6x^2 7x 5
3x(2x) 10x +3x  5 =
6x^2 7x 5
Why is B there in (t+b)?
A trinomial follows the form a^2+2ab+b^2, where a and b do not equal 0. When factored, the trinomial equals (x+a)(x+b). A and b are the variables commonly used to write out the formula for factoring a trinomial, but really any two symbols can be used.
I need some homework help.
The problem is 20t^2  12t  32
I know the answer is 4(5t8)(t+1) but I don't understand how my teacher got to that answer.
The problem is 20t^2  12t  32
I know the answer is 4(5t8)(t+1) but I don't understand how my teacher got to that answer.
i need help i don't understand how my instructor got the answer
x^2+xz/3; use x=3, and z=5
x^2+xz/3; use x=3, and z=5
x^2+xz/3: use x=3, and z=5
thank u.u have helped me very much.can you do simultaneous equations please?
It is kind of a homework question, practice question. I have worked on it for two days!! x(4x5) =114. I cannot find anything that factors out. 4x5x114=0 What adds to 5 and when multiplies gives you 114??
Hi Patty,
Nikola is correct in the factoring but I wanted to point out where I think you were having trouble. It sounds like you were forgetting about the 4 in front of the x^2 when you were looking for factors. Remember that for Ax^2 + Bx + C = 0, when we are looking to break down the expression, we multiply A*C to get a value. In this case, A*C = 456. We then break down the 456 into two factors that, when added together, give us B, in this case, 24 and 19 are the two factors we want. We then rewrite the original equation, using our two factors to replace B:
4x^2  24x + 19x  114 = 0
We can then group and factor:
(4x^2  24x) + (19x  114) = 0
From the first group we can factor out a 4x and from the second group we can factor out a 19:
4x(x  6) + 19(x  6) = 0
At this point, since we have like terms (x6), we can combine the 4x and the 19:
(4x + 19)(x  6) = 0
From there, if you are solving for x, you would set each factor equal to zero:
x  6 = 0
x = 6
or
4x + 19 = 0
4x = 19
x = 19/4
So your roots would be x = 19/4 or x = 6
Hope that helps.
Nikola is correct in the factoring but I wanted to point out where I think you were having trouble. It sounds like you were forgetting about the 4 in front of the x^2 when you were looking for factors. Remember that for Ax^2 + Bx + C = 0, when we are looking to break down the expression, we multiply A*C to get a value. In this case, A*C = 456. We then break down the 456 into two factors that, when added together, give us B, in this case, 24 and 19 are the two factors we want. We then rewrite the original equation, using our two factors to replace B:
4x^2  24x + 19x  114 = 0
We can then group and factor:
(4x^2  24x) + (19x  114) = 0
From the first group we can factor out a 4x and from the second group we can factor out a 19:
4x(x  6) + 19(x  6) = 0
At this point, since we have like terms (x6), we can combine the 4x and the 19:
(4x + 19)(x  6) = 0
From there, if you are solving for x, you would set each factor equal to zero:
x  6 = 0
x = 6
or
4x + 19 = 0
4x = 19
x = 19/4
So your roots would be x = 19/4 or x = 6
Hope that helps.
It factors to (x6)(4x+19)
Is there any way this system of factoring can be incorrect?
what does a stand for
Again, How do you factor a quadratic expression that looks like 6x^223+15
See video titled: Factor by grouping and factoring completely
What's coefficient?
The number in front of the variable. An example is 9x where 9 is the coefficient and x is the variable.
how do i get the coefficient, and the numbers,variables, to have a similar factor?
need a more clearer question, what the topic? You can multiply an entire equation by some factor to match up other factors, say in elimination.
x^2+4x+5 can be multiplied by say 2 to get 2[x^2+4x+5] = 2x^2+8x+10 or by whatever you want to change the coeffients to.
x^2+4x+5 can be multiplied by say 2 to get 2[x^2+4x+5] = 2x^2+8x+10 or by whatever you want to change the coeffients to.
How would you solve this
6k^210kl+4l^2
6k^210kl+4l^2
He mentions foil at 0:20, what is foil?
The FOIL Method is a process used in algebra to multiply two binomials. The lesson on the Distributive Property, explained how to multiply a monomial or a single term such as 7 by a binomial such as (4 + 9x).
What does the "leading 1 coefficient" mean?
The leading term, first term, `x^2` has no explicit coefficient since, by convention, we do not write the one, but it is there.
At 3:45, how does Mr Khan go from (tsq.+3t)+(5t+15) to (t+3)*(t+5)?
(t^2+3t)+(5t+15)
t(t+3)+5(t+3)
(t+3)*(t+5)
t(t+3)+5(t+3)
(t+3)*(t+5)
What is a coefficient?
A coefficient is the constant number in front of a variable that is multiplying by it. So if you have "5x", 5 is the coefficient.
i was looking for an example where i could find how to solve quadratics when a is more than one. Anybody know where I could find that?
www.mathway.com has a worksheet creating tool where it gives you questions for free and you can choose the topic and how many questions you need
www.mathway.com has a worksheet creating tool where you can choose how many questions you need and what topic to do
How do you factor a problem such as m^2+m20 ?
m^2 5m + 4m 20 =0
m(m5)+4(m5)=0
m=4,5
m(m5)+4(m5)=0
m=4,5
I'm only confused about problems where "a" (in ax^2 + bx + c) does not have a value of 1. How do you take this into account when factoring?
Without using the quadratic formula.
Without using the quadratic formula.
If you have ax^2 + bx + c = 0 and you want to factor, the first thing I would do is divide both sides by a. Here is an example: 4x^2 + 12x + 9 = 0. Dividing both sides by 4 yields: x^2 + 3x + (9/4) = 0. You might be able to recognize this as a perfect square: The square root of 9/4 is 3/2 and (3/2)x + (3/2)x = (6/2)x = 3x. So we have x^2+3x+(9/4) = (x + 3/2)^2
ok, what if there is a variable in the middle like X to the second,minus x, minus 20?
You can still factor x2x20. the x would be the same as 1x. I prefer plugging these things into the quadratic equation for then you get exact answers without the guess and check method.
How do you do this is the question is t squared + 8t MINUS 15?
equals what? you have to complete the square or use the quadratic formula if you wanted to find t
Well, to be fair to teachers, they do have questions asked all the time. . . . but as for the 'taught it better' part, that depends on the teacher. A bit of a redundant statement, even by my standards. And the awesome thing is that we can ask our questions in the comments and know that we won't be shunned by the world for asking a redundant question, or a simple question that many people know but a few just can't grasp.
what if you had something like 2x(x+2)3(x2). How would you further simplify it?
You would have to expand it out using the distributive property. Then you could simplify it into a trinomial. That particular problem cannot be factored using real numbers, so a trinomial would be its simplest form.
Hi everyone, for factoring quadratic expressions how do you know what numbers to put in the parenthesis. I understand that you find numbers that multiply to whatever x and add up to the other number but does anyone have tips for finding out these numbers??
You can factor the last number (15 in the example) to see which of its factors add up to the middle number (8 in this case). The factors of 15 are {15,1} and {5,3}. Since 5 +3 = 8, you can try them to see if they work.
(t + 5)(t+3).= (t^2 + 8t +15)
Also, since the middle number was positive, we didn't need to consider the negative factors of 15, namely {15,1} and {5,3}. If the middle number was negative (8) then {5,3} would have been the answer since
(t  5)(t  3).= (t^2  8t +15)
(t + 5)(t+3).= (t^2 + 8t +15)
Also, since the middle number was positive, we didn't need to consider the negative factors of 15, namely {15,1} and {5,3}. If the middle number was negative (8) then {5,3} would have been the answer since
(t  5)(t  3).= (t^2  8t +15)
At 4:00, is that the answer for this problem? I still don't get it how does that when mutiplied do we get t squared? Is it the the one in purple with green or the one you were talking about t +3 times t+5. I still don't understand.
what does the t mean ?
t is the variable. It has a number that could be anything (variable: varyable)
Why does the 1 go out in front?
The 1 doesn't have to be put since there is only one "t", but if there were 3 t's, then you would need to put 3t, because 1t=t, you do not need to worry about the leading coefficient.
How do I factor this:
2x^25x12
2x^25x12
it is fully factored
where do u get the a and b from?
Sal is just showing us the same problem in "color Coded" form. he is showing us that if you replaced a and b with the real numbers in this problem you will get the same "form" of answer.
what if you had a question like: 3x squared + x  4?
How would you factor an expression with numbers that are more than squared? Something along the lines of X to the 3rd + bX squared + cX. How would that work? Btw just as a side note...how do you get a computer to type in exponents?
Ok, I run into an issue when trying to factor my own equation: X^2+4x3.
I cannot seem to get both the product of 3 and the sum of 4. I am running into many problems like this. does the method in this video not apply here??
I cannot seem to get both the product of 3 and the sum of 4. I am running into many problems like this. does the method in this video not apply here??
You are correct in that this method does not work for all quadratics. Your expression's roots are not rational numbers so you would need to solve for the roots with the quadratic formula.
What do I do if the exponent with the first monomial is a 3?
Ex: 6x^313x^228x
Ex: 6x^313x^228x
Thank you! That just never even clicked in my mind! Glad I found out before EOC's though!
For that problem in particular, you can factor out an x from all the terms so that you have x(6x^213x28), and doing the factoring we've been learning in the last few videos, we get x(3x+4)(2x7).
Is there a simpler way to do this?
what if you have a problem such as 9x^2+72x+142=0 where none of the numbers you mutiply together to get 142 and have a sum of 72
Ok, so I have a question. If i had a problem like this: x^2+x90, what would i do? because the middle term only has a variable x. Can someone help please?
Ok! I got it! Thanks!
You can view the x variable in the quadratic as having an implied 1 in front of it. In other words you could rewrite it as
x^2+1x90
so now all that is left is finding factors of 90 whose sum is 1. 10 and 9 work.
The factored version of the quadratic would be
(x+10)(x9)
or if you factor it through grouping the steps would be
x^2+10x9x90
x(x+10)9(x+10)
(x+10)(x9)
x^2+1x90
so now all that is left is finding factors of 90 whose sum is 1. 10 and 9 work.
The factored version of the quadratic would be
(x+10)(x9)
or if you factor it through grouping the steps would be
x^2+10x9x90
x(x+10)9(x+10)
(x+10)(x9)
ok, so here goes nothing! so we know that we have to find a value for a and b. a times b should be equal to 15 and their sum should be equal to 8. so a and b are 5 and 3. this makes the equation: t(t+3)+5(t+3). so since there are two (t+3)'s then we can factor it out to this: (t+3)(t+5) and that would be the answer! does that help? hope ie does!
I'm working on Factoring Quadratics 2 and I got a question that looks like this: 2x^318x^2+20x
I'm confused about the 18 and the +20. I tried watching a video to know how to factor that part, but all of the videos either show two negatives or two positives. Can anyone help? Thanks!
I'm confused about the 18 and the +20. I tried watching a video to know how to factor that part, but all of the videos either show two negatives or two positives. Can anyone help? Thanks!
```2•x^3  18•x^2 + 20•x
2•x•(x^2 + 9•x  10)
[2•x•(x + 10)•(x  1)]```
2•x•(x^2 + 9•x  10)
[2•x•(x + 10)•(x  1)]```
For that equation, first you need to lower the whole equation, so that the leading coefficient is 1.
To do that, simply divide the whole thing by the current leading coefficient, putting the rest into brackets, and multiplying the whole thing by the leading coefficient.
2x(x9x+10)
Then you factor whatever is inside the brackets, just like normal.
Hope this helped.
To do that, simply divide the whole thing by the current leading coefficient, putting the rest into brackets, and multiplying the whole thing by the leading coefficient.
2x(x9x+10)
Then you factor whatever is inside the brackets, just like normal.
Hope this helped.
why don't just do the magic x method?
There's no specific time in the video but I was wondering why, for integers a and b and the form Ax^2 + Bx + C, ab = AC and not ab = C. Isn't it just C. Sorry if this is confusing
What do you do when it's 2x to the 2nd + 8t + 15?
This may seem like a dumb question, but what is the mathematical definition of factoring?
I'm starting to get the concept of the factorizing idea. But what if I had to imply on harder polynomials, for example this one in my book:
81p^418p^2q^2+q^4
81p^418p^2q^2+q^4
The answer is (9p^2q^2)^2
or
(9p^2q^2)(9p^2q^2)
Note that you can treat the p^4 as (p^2)^2 and q^4 as (q^2)^2.
It may be simpler to look at it this this way.
Make the definition of
p^2 = a
and the definition of
q^2 = b.
Then the original expression can be written as 81a^2  18ab + b^2.
This factors into (9a  b)(9a  b).
Then make the substitution a = p^2 and b = q^2. This will give you
(9p^2  q^2)(9p^2  q^2)
or
(9p^2  q^2)^2
or
(9p^2q^2)(9p^2q^2)
Note that you can treat the p^4 as (p^2)^2 and q^4 as (q^2)^2.
It may be simpler to look at it this this way.
Make the definition of
p^2 = a
and the definition of
q^2 = b.
Then the original expression can be written as 81a^2  18ab + b^2.
This factors into (9a  b)(9a  b).
Then make the substitution a = p^2 and b = q^2. This will give you
(9p^2  q^2)(9p^2  q^2)
or
(9p^2  q^2)^2
What's FOIL in 0:20?
First
Outside
Inside
Last
It's used to factor equations.
Outside
Inside
Last
It's used to factor equations.
Show all 2 answers to CreeperJocky204((The #26 AntiButton Proponent))'s question
•
Answer this question
FOIL is a method for multiplying binomials. It goes like this: you multiply the first terms, then the outside terms, then the inside terms, and then the last ones. It's an acronym to remember the process. That's how you multiply a binomial using FOIL. It's not the ideal way to do it, though, because it requires memorisation instead of logic. For more info, check out this video: https://www.khanacademy.org/math/algebra/multiplyingfactoringexpression/multiplyingbinomials/v/multiplyingpolynomials1
I've been swearing over this for a couple of days now. :) Is it really not possible to solve a second degree polynomial (equal to 0) without grouping? I either seem to go in loops, or end up with x = +/ 1, which is false in most cases. Can't seem to find any videos about it either, it just goes unsaid. I'm guessing it's due to the raised var being added to the var, or that there's two solutions, but it really really bugs me. Any clarifying thoughts or ideas about what I'm missing?
Here's the last go I had without factoring much or grouping:
x^26x+5=0
x^26x=5
x(x6)=5
x6=5/x
x5/x=6
x/15/x=6
x^2/x5/x=6
x^25=6x
Bear in mind, I did learn all about factoring, I just want to understand why I can't do this.
Here's the last go I had without factoring much or grouping:
x^26x+5=0
x^26x=5
x(x6)=5
x6=5/x
x5/x=6
x/15/x=6
x^2/x5/x=6
x^25=6x
Bear in mind, I did learn all about factoring, I just want to understand why I can't do this.
There are usually 2 solutions to the Quadratic Equation where it is equal to zero. It is much easier to find solutions by grouping. While your method results in rearranging the expression several different ways, it is very difficult to come up with solutions. It is very helpful to factor it out into groups. (This is further helpful when you need to multiply, divide, or otherwise further process the polynomial, perhaps using another polynomial (i.e., you will more easily be able to cancel certain groups out).
So, to solve where x^2 6x +5 = 0, the factors (x5)(x1) will tell you your answers: what values of x result in the expression to equal zero? If x = 5, the first factor turns to zero, making the whole thing zero; and where x =1 it is also zero. So the solutions are x = 1 and x = 5.
I see an error in one of your steps which makes all that follows incorrect:
x6=5/x does not equal x5/x=6. It should be x+5/x =6
Either way, you still have either a fraction of an x or a factor of an x to deal with at some point. You could do trial and error with plugging in numbers to find where x + 5/x =6 and come up with the 2 solutions, but there are cases where it is more difficult. I think that (x5)(x1) is nicer to deal with in the long run.
So, to solve where x^2 6x +5 = 0, the factors (x5)(x1) will tell you your answers: what values of x result in the expression to equal zero? If x = 5, the first factor turns to zero, making the whole thing zero; and where x =1 it is also zero. So the solutions are x = 1 and x = 5.
I see an error in one of your steps which makes all that follows incorrect:
x6=5/x does not equal x5/x=6. It should be x+5/x =6
Either way, you still have either a fraction of an x or a factor of an x to deal with at some point. You could do trial and error with plugging in numbers to find where x + 5/x =6 and come up with the 2 solutions, but there are cases where it is more difficult. I think that (x5)(x1) is nicer to deal with in the long run.
at 0:33, where does he get a and b from? why not x and y?
Mathematical convention. a and b are coefficients. The accepted general form of a quadratic equation is `ax^2 + bx + c = 0`
Using x and y would be confusing because the expression already has avariable named x.
Using x and y would be confusing because the expression already has avariable named x.
I am working on Factoring Trinomials using FOIL. However, I am having a problem finding a video to explain that. Can you point me in the right direction? Thanks!
I am sorry, but Sal doesn't have a video on this topic. I found someone else who does a really good job on this. http://www.youtube.com/watch?v=uFenBVfPxio
Show steps to answering a2+11a+18
If you meant "a^2 + 11a + 18", AKA "a² + 11a + 18", then here are the steps:
first, find two numbers that multiply to 18, but add to 11. In this case, the two numbers are "9" and "2", because "9 * 2 = 18", and "9 + 2 = 11". So now, you can factor:
(a + 9) * (a + 2)
If you want to doublecheck that, you can just FOIL it out to see that it works.
Next, if we want to find when the whole thing equals zero (aka a solution), set that equal to zero:
(a + 9) * (a + 2) = 0
This will only equal zero when "a + 9" equals zero, or when "a + 2" equals zero. So the solutions are "a = 9" and "a = 2".
I hope this helps.
first, find two numbers that multiply to 18, but add to 11. In this case, the two numbers are "9" and "2", because "9 * 2 = 18", and "9 + 2 = 11". So now, you can factor:
(a + 9) * (a + 2)
If you want to doublecheck that, you can just FOIL it out to see that it works.
Next, if we want to find when the whole thing equals zero (aka a solution), set that equal to zero:
(a + 9) * (a + 2) = 0
This will only equal zero when "a + 9" equals zero, or when "a + 2" equals zero. So the solutions are "a = 9" and "a = 2".
I hope this helps.
I'm assuming you mean 2a when you say a2. (We usually put the coefficient before the variable.)
We have two a terms, which sum to 13a. That's all we can combine, so our final result is 13a + 18.
We have two a terms, which sum to 13a. That's all we can combine, so our final result is 13a + 18.
4x (2x1)+(2x1)^2
The answer to that question is 4x+2x^2+1.
Here are the steps I did:
4x (2x1) + (2x1)^2 (Original Problem)
4x (2x1) + 2x^2 + 1 ( I applied the two squared to the second pair of parentheses.)
8x4x + 2x^2 + 1 ( I distributed the 4x into the parentheses.)
4x + 2x^2 + 1 ( I combined like terms.)
Hope that helps!
Here are the steps I did:
4x (2x1) + (2x1)^2 (Original Problem)
4x (2x1) + 2x^2 + 1 ( I applied the two squared to the second pair of parentheses.)
8x4x + 2x^2 + 1 ( I distributed the 4x into the parentheses.)
4x + 2x^2 + 1 ( I combined like terms.)
Hope that helps!
what if the question was T^2  8T + 15
Find factors of 15 (2 numbers that multiply to create 15) and those same 2 numbers need to add to 8
3 and 5 work
This let's you factor the polynomial into: (T  3) (T  5)
3 and 5 work
This let's you factor the polynomial into: (T  3) (T  5)
Sorry Sal, you really dropped the ball at 3:35.
I would like to know what video would inform me on how to do a problem like y=(x+9)². It is a quadratic equation.
So what happens when ax is not equal to 1?
these vids seem a little backwards in their order
I don't get from 3:35 to the end. how did he factor it out to where from t(t+3) + 5(t+3) to the answer being (t+3)(t+5) ? Please Help!
okay this follows the distributive property, he distributed the t+3 out because
(t+3)(t+5) can be rewritten as t(t+3) + 5(t+3) if you distribute the t+3...did that make sense? its like when you have 15+30 and you can make it (5)(3)+(5)(6) = (5)(3+6)
(t+3)(t+5) can be rewritten as t(t+3) + 5(t+3) if you distribute the t+3...did that make sense? its like when you have 15+30 and you can make it (5)(3)+(5)(6) = (5)(3+6)
I understand how to factor these equations but I don't understand when the leading coefficient is greater than 1. For example, a problem I received in class was 4x^2+12x7. How would this be solved? Someone please help me. Thanks.
y2+y20
(x+5)(x4)
what is the factor trinomial for y2+y20
(x+5)(x4)
Okay, so i understand this, but what if the second number in the trinomial was posotive and the third one was negative? How would you solve the problem x squared +14x49???
Ajborys: You don't know basic grammar. Just saying.
The example you gave isn't factorable with this method. However in general if the second term is positive but the third is negative, then that means the factored expressions will be (t+a)(tb), ie. In the brackets one will have a positive in it while one will have a negative.
The expression you gave is not factorable. Since the roots of that expression are not integers, you cannot easily factor it using only integers.
you spelled positive wrong. just saying
This is a Great video but the entire thing was confusing to me can you help me Khan
Lots of people here would love to help you, but please ask a more specific question. What was the first thing in the video that confused you?
is the a and b step really necessary i feel that there is a easier way to factoring if so please tell me
Once you become better at factoring quadratics, you'll realize that the "a" and "b" step is practically useless, as you asked. Basically, for simple quadratics without a coefficient in front of the squared variable, you just have to find two numbers that multiply to equal the constant, or number without the variable. Also, the two numbers must be added to equal a sum that is the coefficient in front of the second term. Hopefully, that will make sense.
The video is showing more detail of how factoring works.
When you are solving problems with a first coefficient of 1 (on the x^2), just look for factors that have:
sum = coefficient of x term
product = constant term
Example:
`Factor: x^2 + 6x + 8`
you need 2 numbers with a sum of 6 and a product of 8
The numbers will be 4 and 2
`(x+4)(x+2) will be the answer.`
Harder example:
`Factor: x^2 + 5x  6`
You need two numbers with a sum of 5 and product of 6
The numbers are 6 and 1
`(x+6)(x1) is the answer`
When you are solving problems with a first coefficient of 1 (on the x^2), just look for factors that have:
sum = coefficient of x term
product = constant term
Example:
`Factor: x^2 + 6x + 8`
you need 2 numbers with a sum of 6 and a product of 8
The numbers will be 4 and 2
`(x+4)(x+2) will be the answer.`
Harder example:
`Factor: x^2 + 5x  6`
You need two numbers with a sum of 5 and product of 6
The numbers are 6 and 1
`(x+6)(x1) is the answer`
how would you factorise 2x squared +3x 7 = 0
Perhaps that quadratic equation cannot be factored. Try using the Quadratic Formula instead and see what kind of solutions you get for x. If they involve irrational numbers, then the equation is "prime" and cannot be factored.
i can't use what sal is doing in the problem in my homework..
2a^25a4^2
2a^25a4^2
There are a couple of ways to approach this, but factoring always starts by:
1) Look for a GCF (greatest common factor)
2a^25a16 There are no common factors.
2) Look for a common pattern (a^2b^2)=(a+b)(ab); a^2+2ab+b^2=(a+b)^2; a^22ab+b^2=(ab)^2
These are three patterns you just have to brute force memorize called difference of squares and perfect square patterns.
None of these patterns work.
3) Use a generic factoring method (ac method, slide and divide, grouping, educated guess and check, etc.)
ac method: 2•16=32, so look for factor pairs that multiply to equal 32 and add to equal 5
1+32=31
2+16=14
4+8=4
There are no other factor pairs of 32 so this will not factor evenly (aka it is 'prime')
4) For more advanced factoring of quadratics you can use the quadratic formula to get irrational and imaginary factors. a*x^2+b*x+c will factor as a(xr1)(xr2) where r1=(b+√(b^24ac))/(2a) and r2=(b√(b^24ac))/(2a)
[(5)±√((5)^24(2)(16))]/[2(2)]
[5±√(25+128)]/4
[5±√153]/4
[5±√(3•3•17)]/4
[5±3√17]/4
2(a+[5+3√17]/4)(a[53√17]/4)
1) Look for a GCF (greatest common factor)
2a^25a16 There are no common factors.
2) Look for a common pattern (a^2b^2)=(a+b)(ab); a^2+2ab+b^2=(a+b)^2; a^22ab+b^2=(ab)^2
These are three patterns you just have to brute force memorize called difference of squares and perfect square patterns.
None of these patterns work.
3) Use a generic factoring method (ac method, slide and divide, grouping, educated guess and check, etc.)
ac method: 2•16=32, so look for factor pairs that multiply to equal 32 and add to equal 5
1+32=31
2+16=14
4+8=4
There are no other factor pairs of 32 so this will not factor evenly (aka it is 'prime')
4) For more advanced factoring of quadratics you can use the quadratic formula to get irrational and imaginary factors. a*x^2+b*x+c will factor as a(xr1)(xr2) where r1=(b+√(b^24ac))/(2a) and r2=(b√(b^24ac))/(2a)
[(5)±√((5)^24(2)(16))]/[2(2)]
[5±√(25+128)]/4
[5±√153]/4
[5±√(3•3•17)]/4
[5±3√17]/4
2(a+[5+3√17]/4)(a[53√17]/4)
how do you solve 4x2  6x  30 ???
I've tried factoring it but can't seenm to find an answer
I've tried factoring it but can't seenm to find an answer
There are bunch of ways to solve quadratic equations. Depending on if you have a binomial (expression with two terms) or a trinomial (expression with three terms).
Binomial:
Difference of two squares
a^2  b^2 =(a+b)(ab)
ex. 16x^2  9x^2 =
(4x)^2  (3x)^2=
(4x+3x)(4x3x)
Difference of two cubes:
a^3  b^3 = (ab)(a^2+ab+b^2)
ex. 27x^3 + 1 = (3x)^3 + 1^3
= (3x + 1)((3x)^2 – (3x)(1) + 1^2)
= (3x + 1)(9x^2 – 3x + 1)
Sum of two cubes:
a^3 + b^3 = (a+b)(a^2ab+b^2)
Works the same way as Difference of two cubes
Trinomial:
FOIL
You already know this method
Quadratic Formula
(b±√(b^24ac))/2a
Completing the square:
This is a bit hard to explain without a diagram so here is a website for help. http://www.purplemath.com/modules/sqrquad.htm
Binomial:
Difference of two squares
a^2  b^2 =(a+b)(ab)
ex. 16x^2  9x^2 =
(4x)^2  (3x)^2=
(4x+3x)(4x3x)
Difference of two cubes:
a^3  b^3 = (ab)(a^2+ab+b^2)
ex. 27x^3 + 1 = (3x)^3 + 1^3
= (3x + 1)((3x)^2 – (3x)(1) + 1^2)
= (3x + 1)(9x^2 – 3x + 1)
Sum of two cubes:
a^3 + b^3 = (a+b)(a^2ab+b^2)
Works the same way as Difference of two cubes
Trinomial:
FOIL
You already know this method
Quadratic Formula
(b±√(b^24ac))/2a
Completing the square:
This is a bit hard to explain without a diagram so here is a website for help. http://www.purplemath.com/modules/sqrquad.htm
Since the coefficient of x^2 is 4. You have to solve this:
a + b = 6
a.b = 120 [the coefficient multiplied by 30]
a + b = 6
a.b = 120 [the coefficient multiplied by 30]
I think it is
2(2x^2  3x  15) ?
2(2x^2  3x  15) ?
Do you have videos that show you how to factor this type of polynomial > x38 or 10x325
Does the method shown in the Factoring quadratic expression video only work if the term with an exponent is squared? Could it work if it was an equation with an exponent of 4?
And what about this? Can you factor this out by grouping?
4x^2+4x+13
Cause i always see examples that i can solve very quickly, but i haven't see an example what i can't factor by grouping. Is that possible or i just can't see the factors in my example?
4x^2+4x+13
Cause i always see examples that i can solve very quickly, but i haven't see an example what i can't factor by grouping. Is that possible or i just can't see the factors in my example?