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Factoring two-variable quadratics

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

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  • female robot amelia style avatar for user anoonan
    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?
    (5 votes)
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    • duskpin ultimate style avatar for user Helen White
      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.
      (5 votes)
  • male robot donald style avatar for user Chris Brackamonte
    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
    (4 votes)
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  • female robot ada style avatar for user Samantha
    Why does Sal switch the x and y of the 4?
    (1 vote)
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  • duskpin ultimate style avatar for user Sadat Ahmmed
    How would I factorize y=x^2+7x-5?
    (3 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      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)
      (3 votes)
  • blobby green style avatar for user erichutchins64
    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?
    (4 votes)
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  • duskpin sapling style avatar for user mluna51
    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
    (2 votes)
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    • leaf green style avatar for user kubleeka
      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)
      (4 votes)
  • piceratops ultimate style avatar for user Leonardo Padro
    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?
    (2 votes)
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  • marcimus purple style avatar for user Alexander Hermann
    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)
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    • cacteye blue style avatar for user Jerry Nilsson
      (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).
      (2 votes)
  • mr pink orange style avatar for user sreenu.maddisetty
    Sir, I have asked for
    (2x^2)y-(3x^2)+4y
    (2 votes)
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  • starky ultimate style avatar for user Zéca
    at khan swapped the x by the y (4xy =>4yx). Why did he do this? Could not it be 4xy or is it necessary to change to 4yx?
    (1 vote)
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Video transcript

We already have the tools in our tool kit to factor something like x squared plus 4x minus 5. And the way that we've already thought about it is we've said, hey, let's think of two numbers that if we were to take their product, we'd get negative 5. And if we were to add the two numbers, we'd get positive 4. And the fact that their product is negative tells you one of them is going to be positive and one of them is going to be negative. And so there's a couple of ways you could think about it. Well, you could say, well, maybe one of the numbers is negative 1 and then the other one is positive 5. Actually, this one seems to work. Negative 1 times 5 is negative 5. Negative 1 plus 5 is positive 4. So this one actually seems to work. The other option would have been-- since we're just going to deal with the factors of 5, and 5's a prime number, the other option would have been something like 1 and negative 5. There's only two factors for 5. So 1 and negative 5-- their product would have been negative 5. But if you were to add these two numbers, you would have gotten a negative 4 right over here. So we're going to go with this right over here. And so this tells us that if we want to factor this using the tools that we already know about, we will get-- And let me write these numbers in a different color so we can keep track of them. So negative 1 and 5. We know that this would factor out to be x minus 1 times x plus 5. And you can verify this for yourself that if you were to multiply this out, you will get x squared plus 4x minus 5. You can even see this here. x times x is x squared. Negative x plus 5x is going to be 4x. And then negative 1 times 5 is negative 5. Fair enough. This is all review for us at this point. Now I want to tackle something a little bit more interesting. Let's say we wanted to factor x squared plus 4xy minus 5y squared. And at first, this looks really daunting. All of a sudden, I've introduced a y and a y squared here. I have two variables. How would I tackle it? But the important thing is to just take a deep breath and realize that we're not fundamentally doing something different. Now, the one little tricky thing I've done when I've written it this way-- and I encourage you to pause this and try this on your own before I explain any further. But the one tricky thing I did right over here is I wrote the x before the y. And that tends to be the convention. You just write them kind of in alphabetical order. But if we wanted it in a form that's a little bit closer to this and something that would fit this mold a little bit more is if we swapped these two. Because then we could write it as x squared plus 4yx minus 5y squared. And now it becomes pretty clear that this 4y term right over here-- this right over here is the coefficient on the x term, the same way that 4 was the coefficient on x right here. And this negative 5y squared corresponds to the negative 5 right over here. So we can do the exact same thought process. Let's think of two-- now not just numbers. They're going to have variables in them. Let's think of two terms or two expressions that if I multiply them, I get negative 5y squared. And then when I take the sum, I get 4y. So let's think about how we could do this. So one option would be positive. Let's say positive y and negative 5y. So where would this take us? Positive y times negative 5y would indeed be equal to negative 5y squared. But then if I add y to negative 5y, I'm going to get negative 4y. So this doesn't work. But let's see if we swap the two signs. So what about negative y and positive 5y? Well, here, if I take the product of negative y and 5y, it will be negative 5y squared. And if I take the sum negative y and 5y, it will be positive 4y. So we know how to factor it now. So once again, let me put this in the same color. So this I'm going to put in this mauve color, this light purple. And this I'll put in a darker purple. So now we know how to factor this. And this is the same exact mold that we did up here, same exact idea. This is going to be x. Instead of just a minus 1 here, now we've factored-- here we factored into a negative 1 and a 5. Here we factored into a negative y and 5y. So instead of a negative 1, it's going to be a negative y, x minus y times x plus 5y. And we can verify that when you multiply this out, it indeed does equal x squared plus 4xy minus 5y. Let me do that here just so we know for sure. So x times x is going to be x squared. Let me do everything in a different color. x times 5y is going to be plus 5xy. Then negative y times x is negative yx. And then finally-- and I'm running out of colors-- if we take negative y times 5y, that's negative 5y squared. And now we just have to simplify. We have to combine these middle two terms right over here. And at first, it looks a little bit-- this is xy. This is yx. It's not so obvious. But we just have to rewrite it. This is the same thing as 5yx minus yx. And so here, you're saying, look, I have 5yx's. And I'm going to subtract yx's. So I'm going to have 4yx's. So this is just going to be 4yx's. I have 5yx's. Take out another yx. I'm going to have 4yx's. So this is going to be x squared plus 4yx minus 5y squared. And it all works out.