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### Course: Algebra (all content)>Unit 10

Lesson 16: Factoring polynomials with quadratic forms

Sal factors x^2+4xy-5y^2 as (x-y)(x+5y). Created by Sal Khan.

## Want to join the conversation?

• Can you not just use the difference/sum of a square (as in x^2 +- 2ax + b^2) to factor x^2+4xy-5y^2? It would be much shorter and more convenient. Also, what if you have an equation like 6x^2 + 12x + y + 13?
• The middle term isn't a square so you can't do a difference of two squares. This equation should be in the form (x - cy)(x + dy). The factors of 5 are 1 & 5 so to make +4xy, c=1 and d=5.
6x^2 + 12x + y + 13 is not in the form of a quadratic as y has no other terms in this expression.
If you rearrange it to y = -6x^2 - 12x - 13 and used the quadratic formula you could find the terms solutions for x.
• Why does Sal switch the x and y of the 4?
(1 vote)
• Because he wanted to let all the variables end in x to make it easier to factor.
• I just got one more extra idea.....I'm not sure if its correct...Hope someone helps....
So Y disguised as a co-factor by appearing with 4x and -5. It could even disguise itself more by appearing with the x^2 term like this

(Y)x^2+(4Y^2)x-5Y^3=0 right?

here we need to find factors of (-5Y^4)x^2......which when added up should give (+4Y^2)x

So factors will be (-Y^2)x and (5Y^2)x
• The original Equation should be factorized as following:
(Yx + 5Y^2)(x - Y) = 0
Then, it is solved according to that.
I will post the easy way to factorize a trinominal equation in the near future I had developed.
• How would I factorize y=x^2+7x-5?
• Either we solve the quadratic equation 𝑥² + 7𝑥 − 5 = 0 ⇒ 𝑥 = (−7 ± √(7² − 4 ∙ (−5)))∕2 =
= (−7 ± √69)∕2, which gives us that the quadratic expression 𝑥² + 7𝑥 − 5 can be factorized as
(𝑥 + (7 − √69)∕2)(𝑥 + (7 + √69)∕2)

Or, we complete the square:
𝑥² + 7𝑥 − 5 = (𝑥 + 7∕2)² − (7∕2)² − 5 = (𝑥 + 7∕2)² − 49∕4 − 20∕4 =
= (𝑥 + 7∕2)² − (√69∕2)² = (𝑥 + (7 − √69)∕2)(𝑥 + (7 + √69)∕2)
• I have been trying to study and learn this section on factoring polynomials: quadratic methods(advanced). Is there another video or additional tips or coaching that can be offered?
• So if I already was given a problem that was already factored at example:

x^4-y^4 and then I have to turn it in a quadratic function

How do I do that and can I get a video to go along with it if that is too much to ask
• What you have is a difference of two squares. We know that (a²-b²)=(a+b)(a-b) (multiply it out to check for yourself).

If we apply this formula to your problem, letting x²=a and y²=b, we get
x⁴-y⁴=(x²+y²)(x²-y²)

Then notice we have a difference of squares again: x²-y². This factors as (x+y)(x-y). So the final factored form is
(x²+y²)(x+y)(x-y)
• What should I do if the first variable is raised to the fourth or third power such as in: x^4-9x^2+20xy as well as having variables in the other two numbers. How do you compute this?
• The only thing you can do to factor this is to factor out a GCF = X
The rest would not be factorable.
• I think that I am on the wrong topic, but I really don't understand what Sal was doing at -. Can someone please explain what he just did?
(1 vote)
• (x–1)(x+5) is the factorization of x²+4x–5.

He got those numbers by assuming that x²+4x–5 = 0 has integer solutions, in which case there are two integers (s and t) for which s+t = 4 and s∙t = -5 (because (x+s)(x+t) = x² + (s+t)x + s∙t).
Now, -5 is a negative number, which means that either s is positive and t is negative, or s is negative and t is positive.
Also, 5 is a prime which means that there are only four ways to choose s and t:
s = -5, t = 1
s = -1, t = 5
s = 1, t = -5
s = 5, t = -1
Then he plugs each of these number pairs into s+t = 4 to see if any of them satisfies this equation.
Luckily enough, s = 5, t = -1 is indeed a solution, and the factorization of x²+4x–5 is (x+5)(x–1).
• Sir, I have asked for
(2x^2)y-(3x^2)+4y