Factoring quadratic expressions
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Example 1: Factoring quadratic expressions
Factoring trinomials with a leading 1 coefficient
Discussion and questions for this video
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 Factor t squared plus 8t plus 15.
 So let's just think about what happen if we multiply 2 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 de distributive property twice or using F.O.I.L,
 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 essentialy multiplied every term here by
 every term over there. And then we have 2 terms, 2 t terms
 I guess we could call them. This bt plus at.
 So we can combine those. So we get t squared
 plus a plus b t, I can write this is b plus a t as well.
 plus ab. If we compare this to this right over here.
 we see that we have a similar pattern.
 Our coeffient, Our coeffient on the second degree term is one
 Our coeffient on the second degree term here is one.
 We didn't have to write it. Then, a plus b is a 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.
 That could be 15. So if we want to factor this out.
 We just have to find an a and a b where their product is 15
 and their sum is 8. In general, in general if you ever see
 I'll write it in kind of the more traditional with the 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
 2 numbers whose sum is equal to this thing right here
 and his whose product is equal that thing right there.
 Whose sum is equal to 8 and whose product is equal to 15
 So what are 2 numbers that add up to 8 and
 whose product are 15. So if we just factor 15,
 We have 1 and 15. Those don't add up to 8 in anyway.
 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 can just go straight and factor this and say, hey,
 This is t plus 3 time t plus 5. Since we already figured out what 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 step.
 From what I just showed you. So this first, this polynomial right here,
 I'm going to write it as t squared plus, instead of
 8t i'm going to write it as a sum of at plus bt.
 Or as a sum of 3 t plus 5 t. So plus 3t plus 5t.
 So, I'm essentially, I starded here and 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 by grouping.
 These 2 guys right over here have a common factor t.
 And these 2 guys over here have a common factor 5.
 So let's factor out the t in this first expression 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 multplying times both of these terms
 So, let's factor, let's factor that out.
 So it becomes t plus 3 times, times, t plus 5
 times, I'll write that plus a little bit neater,
 times t plus 5. And we are done. We didn't actually
 have to do this grouping step altough 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.
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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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)
It is kind of confusing, i know, but he over complicated it in my opinion.
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.
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!
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.
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
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.
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,
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.
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.
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.
at 2:27 he mentions grouping, what is grouping
Taking out separate thing and putting them into groups
2xto the forth +2xto the thirdxsquaredx
Sal is an awesome teacher explaining it in short times and no questions needed for his methods
What if at 3:40 you had t(t+3) + 5(t3). How do you deal with one of the values being negative?
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)
After watching this video, I still cant figure out how to solve my problem. t squared plus 7t+12.
Sometimes you may not be able to group them easily and that is why there are other methods. BUT this one is easy and it can be solved by grouping method.
t^2 + 7t + 12 =0
=> t^2 + 4t + 3t + 12 =0
=>t(t+4)+3(t+4) = 0
=>(t+4)(t + 3) = 0
=> t+4 = 0 OR t+3 = 0
=> t =  4 or t =  3
Do watch the other videos by sal about solving quadratic equations.
I hope this helps !
t^2 + 7t + 12 =0
=> t^2 + 4t + 3t + 12 =0
=>t(t+4)+3(t+4) = 0
=>(t+4)(t + 3) = 0
=> t+4 = 0 OR t+3 = 0
=> t =  4 or t =  3
Do watch the other videos by sal about solving quadratic equations.
I hope this helps !
so the final answer, the thing you would write down and box is "(t+3)(t+5)" right?
Yes; all you're doing when you're factoring a polynomial is breaking it down into its respective binomials, or the terms you multiplied together to create the polynomial.
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!
(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.
Do you have videos that show you how to factor this type of polynomial > x38 or 10x325
Is there a simpler way to do this?
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 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
What if the leading coefficient is 1?
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
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.
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.
My teacher in the past two years dont help and i been watching you for one hour and i alraedy get it? I need u as my teacher
Thanks... This was really helpful.
say the sum is 4 x^3 y^3  12 x^4 y^5 + 16 x^6 y^6
for the multiplication and addition part im unable to get the right numbers
please help me out
for the multiplication and addition part im unable to get the right numbers
please help me out
what's the advantage of factoring. If you simplify you go the other way right? Remove parentheses. Just wondering.
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
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!
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?
My question for my my friend, is where does your name come from? Ive never heard it before and it is interesting...
thank u.u have helped me very much.can you do simultaneous equations please?
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
What's coefficient?
The number in front of the variable. An example is 9x where 9 is the coefficient and x is the variable.
why not just use the xbox method?
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.
x^2+xz/3: use x=3, and z=5
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
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.
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.
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
What do you do when it's 2x to the 2nd + 8t + 15?
What would I do if I had a c that couldn't be divided by my b?
Like in b^2+*b=7?
Like in b^2+*b=7?
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.
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.
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)
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
how about x2+11x+35
What do i do with the 11x
i don't think that's factorable.
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.
these vids seem a little backwards in their order
oops, sorry only just realised they were different :/
belongs in commetns but i agree entirely :)
Is there any way this system of factoring can be incorrect?
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)
is the a and b step really necessary i feel that there is a easier way to factoring if so please tell me
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`
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.
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)
why don't just do the magic x method?
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)
I need help determining when to factor and when to apply the quadratic equation. For instance, in the problem:
Solve for x:
2x^2+6x−140=0
My first inclination was to apply the quadratic formula, which gave me x= (3+or 2(70)^1/2)/2
And that's OK, but not pretty.
But when I entered it, it was wrong. So I tried again and got the same answer. So I asked for a hint and they started in factoring. And I ran with the whole factoring thing and got the nice, pretty answer x=10 or x=7.
Why did the two methods get me different answers?
And how can I tell when looking at a problem which one to use?
Solve for x:
2x^2+6x−140=0
My first inclination was to apply the quadratic formula, which gave me x= (3+or 2(70)^1/2)/2
And that's OK, but not pretty.
But when I entered it, it was wrong. So I tried again and got the same answer. So I asked for a hint and they started in factoring. And I ran with the whole factoring thing and got the nice, pretty answer x=10 or x=7.
Why did the two methods get me different answers?
And how can I tell when looking at a problem which one to use?
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
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) :)
Thanks Jules :)
You distribute
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 do I do if the exponent with the first monomial is a 3?
Ex: 6x^313x^228x
Ex: 6x^313x^228x
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).
Thank you! That just never even clicked in my mind! Glad I found out before EOC's though!
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)
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
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]
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
I think it is
2(2x^2  3x  15) ?
2(2x^2  3x  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.
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???
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.
Ajborys: You don't know basic grammar. Just saying.
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
what is the factor trinomial for y2+y20
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)
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?
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.
What is a polynomial? (at 0:13)
It is an expression that has two or more terms, in this case there are 3 terms.
Thank you, Ashley.
you are 9 and know some stuff about polnomials? i'm in tenth grade and I just learned about them last year. Thats really cool, I wish I knew stuff like this when I was 9 I'd b a genius rite now :). This website is really great no teacher sets ur pace, the skies the limit!
At 1:58, Sal says you can use this method if the coefficient is 1.
What if the coefficient of x term raised to the second power, is greater than one?
Like: 2x^2 + 2x + 1
Can you use this same factoring method?
What if the coefficient of x term raised to the second power, is greater than one?
Like: 2x^2 + 2x + 1
Can you use this same factoring method?
You can do something similar.
2X^2 +10X + 8.
Multiply A and C. 2*8 = 16.
Find factors of AC (16) that add to equal B (10). These are 8 and 2.
Divide each factor by A. 8/2 and 2/2.
Have each coefficient of X as 1
(X+ )(X+ )
and put your factors in
(X+8/2)(X+2/2)
Multiply this by A
2(X+8/2)(X+2/2). This reduces to
2(x+4)(x+1).
The same thing that happens when A is 1, but multiplying and dividing by 1 don't change your values.
2X^2 +10X + 8.
Multiply A and C. 2*8 = 16.
Find factors of AC (16) that add to equal B (10). These are 8 and 2.
Divide each factor by A. 8/2 and 2/2.
Have each coefficient of X as 1
(X+ )(X+ )
and put your factors in
(X+8/2)(X+2/2)
Multiply this by A
2(X+8/2)(X+2/2). This reduces to
2(x+4)(x+1).
The same thing that happens when A is 1, but multiplying and dividing by 1 don't change your values.
Can x^2+24+143 be factored, having trouble with the operation.
You can save some time by noticing that you only have to check for divisibility up to the square root of the number. For example, when checking factors of 143, we only have to go up to 11 because sqrt(143)~=11.9. All of the factors of 143 that are above 11.9 will be flushed out when you test the lower numbers.
In other words, you can stop checking factors when the result you get is smaller then the number you divided by.
In other words, you can stop checking factors when the result you get is smaller then the number you divided by.
I'm assuming that you meant to say 24x in which case you could just use the quadratic formula of ( b + Squareroot( b^2  4ac) / 2a where a=1 b=24 and c=143. so (24 + squareroot(24^2  4*1*143)) /2 (24 + squareroot( 576  572)) / 2 (24 + squareroot( 4)) /2 (24 + 2) / 2 (24 + 2) /2 22/2 = 11 (242) / 2 26/2 = 13
So the factors would be (x + 11) (x +13)
There you go. sorry if lengthy. If 24 was 24 not 24x I'm sorry. Hope this helped.
So the factors would be (x + 11) (x +13)
There you go. sorry if lengthy. If 24 was 24 not 24x I'm sorry. Hope this helped.
Whats going on? Where did Sal get a and b from? can someone explain this to me?
The a and b are what he uses to represent the numbers and can be used to represent the the coefficients of terms in any quadratic expression
How do you do the problem a^39a+3a^227? How do you factor it. I'm really confused
Continuing from Kate Ballard's explanation, the first quantity can be factored more using the difference of two squares. The final answer should be (a3)(a+3)(a+3) and this can be changed to (a3)(a+3)^2.
It helps if you start by putting the variables in order from biggest exponent to smallest. So, a^3+3a^29a27. Then you split the middle and have a^3+3a^2 9a27. Then factor them separately and you have a^2(a+3) 9(a+3). They both have a+3 so you would end up with (a^29)(a+3)
I was factoring polynomials 1 and the problem was x2(exponent)+3x54 and I got +9 and 6 and it said it was wrong what did I do wrong?
Andy,
You probably factored the polynomial correctly into (x+9)(x6). But remember, if you were trying to solve for x, the equation is x^2+3x54=0. Don't forget about the "equals 0" part.
So what does that mean? Well we get everything factored into:
(x+9)(x6)=0 That means that we have 2 quantities multiplied together that equal 0, so either the first one is 0, or the second one is. So we have two possibilities, which is why we get two answers.
First possibility: x+9=0 so x=9
Second possibility x6=0, so x=6.
Does that sound about right?
You probably factored the polynomial correctly into (x+9)(x6). But remember, if you were trying to solve for x, the equation is x^2+3x54=0. Don't forget about the "equals 0" part.
So what does that mean? Well we get everything factored into:
(x+9)(x6)=0 That means that we have 2 quantities multiplied together that equal 0, so either the first one is 0, or the second one is. So we have two possibilities, which is why we get two answers.
First possibility: x+9=0 so x=9
Second possibility x6=0, so x=6.
Does that sound about right?
x^2+3x54=0
x^2+(9x6x)54=0
(x^2+9x)(6x+54)=0
x(x+9)6(x+9)=0
(x+9)(x6)=0
=>x+9=0 and x6=0
=>x=9 and x=6
x^2+(9x6x)54=0
(x^2+9x)(6x+54)=0
x(x+9)6(x+9)=0
(x+9)(x6)=0
=>x+9=0 and x6=0
=>x=9 and x=6
ifigured out the answer I forgot to but in x+ terms
A good shortcut would be finding the factors of the last monomial or variable which you can add or subtract and the answer would be the numeral coefficient in the middle.
That sounded more complicated
That sounded more complicated
I think he was just showing the basis of the concept.
Yeah, the way you explained was how I was taught too.
Yeah, the way you explained was how I was taught too.
the terminology you used is wrong. The last thing in that expression, the +15, isn't a monomial or a variable, it's a "term". A monomial is an expression with only two terms. This expression has three terms, so it's a trinomial.
Sal is factoring whole numbers. What about fractions? Ex: x^2  4/3x + 4/9
The factors of 4/9 are: 1, 4/9 or 2/3,2/3.
because in the original expression the middle term is negative the 4/9 is positive we are looking for two negative factors that add up to 4/3.
(x2/3)(x2/3)
because in the original expression the middle term is negative the 4/9 is positive we are looking for two negative factors that add up to 4/3.
(x2/3)(x2/3)
how do I factor by grouping polynomials to the third power? ie.
12n to the third 28n squared 15n + 35?
12n to the third 28n squared 15n + 35?
You factor by grouping.
12n^328n^215n+35
You essentially create two binomials through grouping, so you have:
(12n^328n^2) and (15n+35)
Then, you reduce each binomial by taking out a term (i.e. dividing by 5 or 5n or something) so the binomials left int the parentheses are equivalent:
4n^2(3n7) 5(3n7)
So in the equation above, 4n^2 was taken out of the first binomial by dividing, and 5 was taken out of the second binomial through division as well. Then, you can reduce this even further by taking out, or factoring out the 3n7 and leaving 4n^25:
(3n7)(4n^25)
The above is your answer.
12n^328n^215n+35
You essentially create two binomials through grouping, so you have:
(12n^328n^2) and (15n+35)
Then, you reduce each binomial by taking out a term (i.e. dividing by 5 or 5n or something) so the binomials left int the parentheses are equivalent:
4n^2(3n7) 5(3n7)
So in the equation above, 4n^2 was taken out of the first binomial by dividing, and 5 was taken out of the second binomial through division as well. Then, you can reduce this even further by taking out, or factoring out the 3n7 and leaving 4n^25:
(3n7)(4n^25)
The above is your answer.
sigh... I'm so terrible at grasping concepts... I will give you a problem i dont understand..
If i factor by grouping: 7w squared+14w+wb+2b... how does it turn out to be (w+2)(7w+b)????
If i factor by grouping: 7w squared+14w+wb+2b... how does it turn out to be (w+2)(7w+b)????
(w+2)(7w+b)= w times 7w which gives you 7w squared plus w times b which gives you wb plus 2 timea 7w plus 2 times b.this gives you 7w squared+wb+14w+2b.when u put them in order from greatest to least you get 7w squared+14w+wb+2b.Hope this is helpful!
Report a mistake in the video
Example:
At 2:33, Sal said "single bonds" but meant "covalent bonds."
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