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### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 1

Lesson 2: Topic B: Lessons 12-13: Factorization- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping
- Factor polynomials: quadratic methods
- Factoring two-variable quadratics
- Factoring two-variable quadratics: rearranging
- Factoring two-variable quadratics: grouping
- Factor polynomials: quadratic methods (challenge)
- Factoring using the perfect square pattern
- Factoring using the difference of squares pattern
- Factoring difference of squares: two variables (example 2)
- Factor polynomials using structure
- Factoring higher-degree polynomials
- Factoring sum of cubes
- Factoring difference of cubes
- Finding zeros of polynomials (1 of 2)
- Finding zeros of polynomials (2 of 2)
- Finding zeros of polynomials (example 2)
- Zeros of polynomials (with factoring)

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# Factoring two-variable quadratics

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?(5 votes)
- 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)

- 3:02Why 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.(2 votes)

- 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)- 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.(6 votes)

- How would I factorize y=x^2+7x-5?(3 votes)
- 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)

- 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)
- 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)- 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)

- 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)
- The only thing you can do to factor this is to factor out a GCF = X

The rest would not be factorable.(2 votes)

- I think that I am on the wrong topic, but I really don't understand what Sal was doing at1:16-1:28. 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).(2 votes)

- Sir, I have asked for

(2x^2)y-(3x^2)+4y(2 votes)- You cannot really factor this because the greatest common factor is 1 for the three terms(1 vote)

- at2:41khan 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)
- The order of the variables does not matter.

The commutative property of multiplication tells us the 4xy and 4yx are the same.(3 votes)

## 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.