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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: grouping

Sal factors 5rs+25r-3s-15 as (s+5)(5r-3). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Can you use FOIL to do this problem?(29 votes)
- So...FOIL "undoes" factoring and Factoring "undoes" foiling.(10 votes)

- I dont get this extra credit question my teacher gave me, and i am not sure if sal have videos about this type of question:

a^2 x^2 + 4ab +b^2

thank you very much!(6 votes)- you just need to apply the power and the add the exponents(1 vote)

- Doesn't FOIL stand for: First, Outside, Inner, and Last?(5 votes)
- Yes it does. Basically what that means is that when multiply to polynomials. In the case on (x+1)(y+2), you multiply the first (x and y), the outer (x and 2), the inner (1 and y), and the last (1 and 2). In the end you get x * y + x * 2 + 1 * y + 1 * 2. This can be simplified to xy+2x+y+2.(7 votes)

- how can you do the with variables which have at least variable in common. for ex; 25x-5x-15xy-3y

how can i factor 15 and 3? I know they're both divisible by 3 but, i don't know where to put the x.(5 votes)- This is slightly tricky since you have a term with an xy. So here's what I'd do. I'd factor out the 5x's and leave the y's.

5x(5 - 1 - 3y) - 3y

5x(4 - 3y) - 3y

If it was just -15xy-3y alone, I would just leave the x since the second term doesn't have an x in it.

-3y(5x - 1)

Like that.(5 votes)

- Isn't the title of this video potentially misleading? The title is "Factoring two-variable quadratics: grouping", but the polynomial in the video 5rs+25r-3s-15 has no second degree exponent and is therefore a linear polynomial rather than a quadratic.(5 votes)
- It must have been factoring two variable polynomials: quadratics or grouping method.(1 vote)

- I'm curious about how useful this type of technique is. It seems that the problems it can solve are very artificial - they conveniently happen to have a common factor (in this case, s + 5) that can easily be factored out. How often, when solving real-world problems that weren't designed specifically to illustrate this technique, are you going to be fortunate enough to have something where this technique works out so nicely? I feel like you'd almost always just use the quadratic formula. Can't think of any situation outside of homework where this video's technique would be useful, but if anyone can, please enlighten me.(2 votes)
- As far as I know, this would work on any quadratic equation. Yes, the quadratic formula would be quicker than this method, but in my opinion, this is much easier, and much more fun.(1 vote)

- how to solve this? x^2-y^2/xy - xy-y^2/xy-x^2(1 vote)
- You can't solve it, as it doesn't equal anything. You can simplify it to -2y/x -xy.(2 votes)

- How does this apply to the real world?(1 vote)
- Any kind of engineering, computer software, and a few other random jobs require you to know this. Not this specifically, but factoring polynomials in general.(2 votes)

- what does foil mean?(0 votes)
- first

outside

inside

last(4 votes)

- what is the answer with the different factors ? b^2-2b+1-y^2(1 vote)
- This equation can't be factored that easily, because you would have to add some components first. By adding components I mean taking something like
`+ a - a`

to the equation. As`a - a = 0`

, this doesn't change the value of the equation. In this case I add`- b*y + b*y - y + y`

, so I can factorize the equation:`b^2 - 2*b + 1 - y^2`

= b^2 - b - b + 1 - y^2 - b*y + b*y - y + y

= b(b - y - 1) + y(b - y - 1) - 1 (b - y - 1)

= (b + y - 1)(b - y - 1)

You can check this is the solution by multiplying the factors again.(2 votes)

## Video transcript

We're asked to factor this
expression by grouping. Now, they mention grouping,
we're going to see what grouping is, but we're going
to see very quickly that we have to do this thing called
grouping because you can't just factor this expression. If you look at these, each of
the terms, all but one of them is divisible by 5. So you can't just
factor out a 5. Not all of them are divisible
by either r or s. This is only divisible by r,
that's only divisible by s, that's divisible by neither. So there is no common factor
across all four of these terms. That's why we have to
group them into groups where there are common factors, and
then see if that simplifies the whole thing. And there is a little bit of an
art to recognizing when you can factor by grouping, but
they've set this problem up nicely for us. So if you look at these first
two terms right here. You have a 5rs and a 25r. These two guys clearly have
some common factors. They're both divisible by 5,
they're both divisible by r. So if I just wanted to factor
this one out, or if I wanted to rewrite it as a product
of two expressions, how could I write it? Well, I could write it as a
product of 5r times-- what's 5rs divided by 5r? Well, you still have
an s left over, you just have an s there. Plus-- what's 25r
divided by 5r? Well, 25 divided by 5 is 5, and
r divided by r is just 1. So 25r over 5r is 5. So these first two terms can
be factored into these two expressions. And then let's look at these
second two terms. Well, they definitely have a common factor,
you have a negative 3 or positive 3 common
to both of these. Let's just go with
the negative 3. And our goal is really to factor
it into a negative 3 times, hopefully, something
very similar to s plus 5. And you might already be seeing
that it's going to factor into s plus it. 5. So let's factor out
that negative 3. So these two terms you can
rewrite as negative 3 times-- what's negative 3s divided
by negative 3? Well, you're just going to
have an s left over. And then what's negative 15
divided by negative 3? Well, that's just positive 5. And just like that, we've
grouped them and we're able to factor each of those groups, and
then something interesting might pop out at you. And one, you can always verify
that you factored this properly by distributing each
of these expressions. Distributing the 5r times s
plus 5, and the negative 3 times s plus 5, you'll
get exactly this. But something maybe jumped
at out you just now. You have 5r times s plus 5. Then you have negative
3 times s plus 5. So now this expression, we
have two terms instead of four, right, this is one term,
this is another term. And they both have s plus 5 as a
common factor, so we can now factor out s plus 5. So this whole thing can
be rewritten as s plus 5 times 5r. Right? If you take 5r times s plus 5,
and you factor out the s plus 5, you're just left
with the 5r. And then similarly, if you take
negative 3 times s plus 5, and you factor out the s plus
5, or divide by s plus 5, you just have a negative
3, just like that. And then we're done! We've factored this expression
by grouping. It's s plus 5 times
5r minus 3. And you can verify it by
multiplying it out. If you distribute the s plus
5 onto each of these terms, you'll get this expression
up here, and then if you distribute the 5r over there
you're going to get that expression. If you distribute the negative
3, you're going to get that expression. So this does simplify to that,
so we have factored