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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 difference of squares: two variables (example 2)

Sal factors 49x^2-49y^2 as (7x+7y)(7x-7y) or as 49(x+y)(x-y). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- How does i solve something like 32 x^3 - 2 w^4 x^3(8 votes)
- First, look for a common factor between the two terms. Both are divisible by 2 and by x^3. Pull that out of both terms, and write what is left.

2x^3(16 - w^4)

Now, factor the part in parentheses as a difference of squares:

2x^3(16-w^4)

= 2x^3(4^2 - (w^2)^2)

= 2x^3(4 - w^2)(4 + w^2)

Now you can see that the (4-w^2) is also a difference of squares, so factor that part again:

2x^3(4 - w^2)(4 + w^2)

2x^3(2 - w)(2 + w)(4 + w^2)(11 votes)

- how do i factor this expression?

4x^2 - 25?

please help(8 votes)- Well, we know that 4x^2 = (2x)(2x) or (2x)^2, right? We know 25 = 5^2, right?

If we have a^2-b^2, we can figure that into (a+b) (a-b). in our case, a=2x, b=5, so you factor that out as (2x-5)(2x+5 )(9 votes)

- how to solve x^2+6x+9-y^2(8 votes)
- We can't solve this because we have no equation. However, we can factor this expression. Notice that x^2 + 6x + 9 is equal to (x+3)^2. This can be shown by multiplication of the two binomials x+3. Now notice that we can write the new expression as (x+3)^2 - y^2. This is a difference of squares. Let a equal (x+3), and b equal y. Then a^2 - b^2 = (a+b) (a-b). Now replace a and b in the new expression of two multiplied terms. this leaves us with (x+3+y)(x+3-y). Now we are done factoring this expression.(8 votes)

- This example shows 6x^2-x-12...Can you explain to me step by step how i can solve this problem(7 votes)
- That would not be a "difference of squares" problem...

Try guessing factor pairs of the form (ax + b)*(cx + d) where a*c = 6 and b*d = -12. (You can see why that is necessary if you FOIL out.) Then check the "cross terms", ad+bc to see which gives you -1 to match the coefficient of the middle term.

(6x+1)(x-12)

(2x+ 3)(3x- 4)

That's actually it! Any other possibility would result in a quadratic with a common factor. For example the first factor in (6x+2)(x-6) has a common factor of 2. Because 2 is not a common factor of the original polynomial, I don't need to consider this one.

So FOIL them out:

(6x+1)(x-12) = 6x^2 -71x -12

(2x+ 3)(3x- 4) = 6x^2 + x - 12

The first doesn't work. The second ALMOST works, but you need the middle term to be negative. This is easily fixed -- switch the + and - signs. So your factorization is (2x - 3)(3x + 4).(3 votes)

- Is there a video that shows how to factor a trinomial in the form of ax^2 + bx + c

ex. 3p^2 - 2p - 5(4 votes)- There are a variety of ways to solve that. There is a complex method for finding the factors, but we usually reserve that for difficult cases. Here is the method for simple cases:

Given: ax² + bx + c, factor into two binomials.

Step 1: Find two numbers that multiply to give a×c, and add to give b. Let us call those two numbers d and g.

Step 2: Rewrite the expression, replacing "bx" with the sum of those two numbers times x. This would give you:

ax² + bx + c = ax² + dx + gx + c

Step 3: Factor the first two and last two terms individually:

Find GCF of a and d and factor that out of ax² + dx. Let us call the GCF m.

mx(ax/m + dx/m)

Now find the GCF of d and g and factor that of gx + c. Let us call this GCF j.

j(gx/j + c/j)

Putting your quadratic expressing back together we get:

mx(ax/m + dx/m) + j(gx/j + c/j)

If you have done the math correctly, you will always find that

(ax/m + dx/m) = (gx/j + c/j)

So let us replace (gx/j + c/j) with (ax/m + dx/m)

mx(ax/m + dx/m) + j(ax/m + dx/m)

We can factor out the (ax/m + dx/m) to get:

(ax/m + dx/m)(mx + j)

I know this looks hard, but if you do it with actual numbers you'll find it isn't too difficult.(5 votes)

- Sorry for being off topic, but how do I do 3(x + 2)(x – 4) ?(2 votes)
- How would I determine if a Trinomial like 4x^2+12x+9 was a perfect square or not?(3 votes)
- First check if the first and last terms are perfect squares. They are in our case:

(2x)² = 4x²

3² = 9

Now check if the square roots of the first and last terms when multiplied by each other and then doubled, yields the middle term. In other words:

2x • 3 • 2 = 12x

Since this matches the second term, we know that are equation fits the form:

a² + 2ab + b² = (a + b)²

Where:

a = 2x

b = 3

Comment if you have questions.(3 votes)

- well... i came to this 49x^2-49y^2 =49x^2+0+(-49y^2)

49x^2+7xy-7xy+(49y^2)

7xy(7x/y)+7xy(-7y/x)

7xy(7x/y - 7y/x) .....

it seems easier for me this way, what do you think?(3 votes)- While this is an equivalent expression, all you have done here is factor out a 7xy from the original difference of squares. This is neither simpler than the original statement nor particularly helpful for solving an equation in most cases.(1 vote)

- How would i factor 3x-3y-a(x-y)(3 votes)
- What does difference of squares mean? It is all kind of confusing to me...(3 votes)
- It's difference in value between squares of two numbers. You can visualise it as a difference in surface of two squares of given surfaces.

The formula used for factoring difference of squares is:

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

For example:

64 - 25 = (8-5) * (8+5)

Or:

x^2 - 25 = (x - 5) * (x + 5)

Hope it helped.(1 vote)

## Video transcript

We need to factor 49x squared
minus 49y squared. Now here there's a pattern
that you might already be familiar with. But just to make sure you are,
let's think about what happens if we multiply a plus b-- where
these are just two terms in a binomial-- times
a minus b. If you multiply this out, you
have a times a, which is a squared, plus a times negative
b, which is negative ab-- that's a times negative b-- plus
b times a, which is the same thing as ab. And then you have b times
negative b, which is negative b squared. So when you do that, you have
a negative ab and a positive ab, they cancel out. And you're just going
to be left with an a squared minus a b squared. Now, this thing that we have
here is exactly that pattern. 49x squared is a
perfect square. 49y squared is a
perfect square. We can rewrite it like that. We could rewrite this over here
as 7x squared minus-- and I'll do it in blue--
minus 7y squared. And so you see it's a pattern. It's a squared minus
b squared. So if you wanted to factor
this-- if you would just use this pattern that we just
derived-- you would say that this is the same thing as a,
7x plus b plus 7y times 7x minus b, minus 7y. And you'd be done. Now there's one alternate way
that you could factor this and it'd be completely legitimate. You could start from
the beginning and say, you know what? 49 is a common factor here, so
let me just factor that out. So you could say it's equivalent
to 49 times x squared minus y squared. And you say, oh, this fits the
pattern of-- this is a squared minus b squared. So this will be x plus
y times x minus y. So the whole thing would
be 49 times x plus y times x minus y. And to see that this, right
here, is the exact same thing as this right over here, you
could just factor 7 out of both of these. You'd factor out a 7 out
of that term, a factor 7 out of that term. And when you multiply them,
you'd get the 49. So these are-- this or this--
these are both ways to factor this expression.