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## Algebra II (2018 edition)

### Unit 4: Lesson 3

Factoring polynomials - Quadratic forms- Factoring quadratics as (x+a)(x+b) (example 2)
- More examples of factoring quadratics as (x+a)(x+b)
- Factoring quadratics intro
- Intro to grouping
- Factoring quadratics by grouping
- Factor quadratics by grouping
- 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)

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# Factoring two-variable quadratics: rearranging

CCSS.Math: , ,

Sal rearranges 30x^2+11xy+y^2 as y^2+11xy+30x^2 and then factors it as (y+5x)(y+6x). Created by Sal Khan.

## Want to join the conversation?

- What if you have a non 1 coefficient and you need to solve a problem like 18x^2 + 3xy - 10y^2? I'm not asking for how to solve that specific problem, I'm asking how to solve problems of its nature.(20 votes)
- You can do it with factoring by grouping. Starting with for example
`18x^2 + 3yx - 10y^2`

, you pretend the`y`

terms are the numerical portions of the grouping. (I rewrote 3xy as 3yx to make this more obvious.) So you need 2 terms that multiply together to make`-18*10y^2`

, and add up to`3y`

. Well, looking at the factors of 180, -12 and 15 work, so`-12y`

and`15y`

are the factors. That yields`18x^2 - 12yx + 15yx - 10y^2`

, which undistributes to`6x(3x-2y) + 5y(3x-2y)`

, and then factoring out the 3x-2y gives`(3x-2y)(6x+5y)`

. And we are done!™

See the https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-by-grouping series for the basics of that technique.

Also see http://tinyurl.com/nvzstcv where I did another example from another user's question.(32 votes)

- I dont understand the way Sal's doing these. I paused at0:06and factored it by grouping, because it seemed like the obvious way to go. Like this:

30x^2 + 11xy + y^2

30x^2 + 5xy + 6xy + y^2

5x(6x+y) + y(6x+y)

=(5x+y)(6x+y)

Multiplying it out, i get back to the beginning.

I did the problem in the previous video the same way. Is this correct? Or am i doing it wrong and getting the right answers just by accident?(17 votes)- I think you are correct. Sal's reference to not knowing another way to do this is probably because this video was originally in a different order. If they move the videos around, the learning sequence doesn't follow exactly. I think the important thing is that you learned that skill, and you are able to apply it. I think it is good to see the different ways to do these problems, because it helps you to remember to keep analyzing these problems. Sal teaches us to really look at the problem so we don't get in the habit of always just plugging in formulas... rather we learn to analyze problems - Really look at them and understand them so we can choose the best strategy for the given situation.(10 votes)

- how would you factor an expression like this: x^3-x^2*y-x*y^2+y^3? please reply!!(4 votes)
- 1) Always look to see if there is a common factor contained in all terms. In this case there isn't. If there had been, factor out the common factor as your 1st step.

2) Next - as there are 4 terms, factor by using grouping. Look for common factors in the 1st 2 terms, then look for common factors in last 2 terms. Here are the steps for grouping:

x^3-x^2*y-x*y^2+y^3 =

x^2 (x - y) -y^2 (x - y) =

(x - y) (x^2 - y^2)

3) The new expression is not completely factored as the 2nd binomial is a difference of 2 squares, so it can be factored further. This gives you:

(x - y) (x - y) (x + y)(9 votes)

- Is there a way to check if an expression is not factorizable? Like in arithmetic we have a check for divisibility by numbers e.g. if the sums of the digits of a number don't add up to a multiple of 3, that number is not divisible by 3.(4 votes)
- Everything is factorable, the question is: Is it pretty, and is it a real number?

The fast way is look at the rules for synthetic division and Descartes's Rule of Signs.(6 votes)

- What do you do with that left over Y?(5 votes)
- If you mean the y in the 11xy, it is not left over.

Sal is ignoring it a while because it is easiest and best to ignore it when you are factoring. The y part of the quadratic is the variable that is easier to factor.

(The only way you can get y² in one of these factoring and multiplication problems is to multiply y times y.)

After you have figured out the more difficult variable (30x²) and written down the

(y + 5x)(y + 6x),

then you can and**should**check by multiplying the two binomials. You can either FOIL or distribute. First the y

(`y`

+ 5x)(`y + 6x`

) → y² + 6xy

Next the 5x

( y +`5x`

)(`y + 6x`

) → y² + 6xy +`5xy + 30x²`

Combine the like terms to complete your check

y² + 11xy + 30x²

Ta da! That means your (y + 5x)(y + 6x) binomial factors were the correct factors for this polynomial in two variables. And notice that there were no extra y's(2 votes)

- how do we know in 11xy in the example whether 11x or 11y is the coefficient?(1 vote)
- A coefficient is always the number in front of the variable.

In this case, 11 is with the variable xy.

So 11 is coefficient(5 votes)

- is there a way to factor this by grouping?(2 votes)
- Yes, you can. You find factors of 30 that add to 11. There are 6 and 5

Expand the quadratic: 30x^2 + 6xy + 5xy + y^2

Then use grouping: 6x (5x + y) + y (5x + y) = (5x + y)(6x + y)

Hope this helps.(2 votes)

- How would you factor two-variable quadratics when both the x^2 and y^2 terms don't have a coefficient of one?(1 vote)
- You can use factoring by grouping. When you expand the middle term into 2 terms, make sure both terms carry "xy" variables.(3 votes)

- What if you have two variables like y^2+5x+4 / y^2-3x-4 ?

I'm not sure if I can answer this. It's part of my assignment in my math book. I'm asking how to solve this problem...(1 vote)- Your polynomials are not factorable. So, your fraction is fully reduced in its current form.(2 votes)

- How do you factor an expression like 2x^2-8y^2+16y-8? I can figure out the answer through trial & error, but if there's a logical way to get to it, I'm not seeing it.(1 vote)
- This is how I would factorize 2x^2 - 8y^2 + 16y - 8

2(x^2 - 4y^2 + 8y - 4)

2[x^2 - 4(y^2 - 2y + 1)]

2[x^2 - 4(y - 1)^2]

2[x^2 - {2(y - 1)}^2]

2[x^2 - (2y - 2)^2]

2[x - (2y - 2)][x + (2y - 2)]

2(x - 2y + 2)(x + 2y - 2)**TRIAL AND ERROR**is a valid method in mathematics. This method:

1. gives us insight into the problem

2. it is used when there are only a few possible answers

3. it is used when you can systematically check all possibilities

4. it is used when there is no other way to solve the problem

Proof of the four color map theorem is a type of trial and error proof. It is also called brute force search.

I think this is a bit off-topic but I thought you'd like to know. If there is a logical method to approach your question I don't know.(2 votes)

## Video transcript

Let's see if we can use our
existing factoring skills to factor 30x squared
plus 11xy plus y squared. And I encourage you
to pause the video and see if you can
handle it yourself. Now, the first hint
I will give you-- and this might open up
what's going on here-- is to maybe rearrange
this a little bit. We could rewrite
this as y squared plus 11xy plus 30x squared. And my whole
motivation for doing that-- there are ways to
factor a quadratic where your first coefficient, your
coefficient on this first term, is something other than 1. But we haven't seen that yet. And so rearranging it this way,
this got us a little bit more into our comfort zone. Now our coefficient is a
1 on the y squared term. So now we can start to think
of this in the same form that we've looked at some of
the other factoring problems. Can we think of two numbers
whose product is 30x squared and whose sum is 11x? Notice, 11x is the
coefficient on y. We have y squared,
some coefficient on y. And then in terms
of y, this isn't in any way dependent on y. So one way to think about this,
if you knew what x was, then this would be a
quadratic in terms of y. And that's how we're really
thinking about it here. So can we find two
numbers whose product is 30x squared and two
numbers whose sum is the coefficient on this y term
right here, whose sum is 11x? So let's just think about all
of the different possibilities. If we were just thinking
about two numbers whose product was 30
and whose sum was 11, we would be thinking of 5 and 6. 5 times 6 is 30. 5 plus 6 is 11. It's some trial and error. You could have tried 3 and 10. Well, that would have been--
13 would be their sum. You could have tried 2 and 15. That wouldn't have worked. But 5 and 6 does work
here, so we've already seen that multiple times. So 5 and 6 would work for
30, but we have 30x squared. So what if we have 5x and 6x? Well, 5x times 6x is 30x
squared, and 5x plus 6x is 11x. So this actually works. So then our factoring
or our factorization of this expression
is just going to be y plus 5x times y plus 6x. And I'll leave it up to you to
verify that this does indeed, when you multiply it
out, equal this up here.