Lesson: Factoring - Difference of Squares

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Hi Brent, thanks for all these great resources!
@3.38, you showed x^2+81 and you factored it as 1(x^2+81),but isn't this a "Square of a Sum" and it can be written as (x+9)^2 = x^2+18x+81?
Would you please explain. Many thanks.
gmat-admin's picture

You are correct in that (x +9)² = x² + 18x + 81
So, for example, if we want to factor the expression x² + 18x + 81, we can write: x² + 18x + 81 = (x+9)(x+9) = (x+9)²

However, in the video, we are trying to find a way to factor x² + 81. That is we want to write x² + 81 as the PRODUCT of two expressions. In other words, we want to write x² + 81 = (something)(something)

The ONLY way to do this is as follows x² + 81 = (1)(x² + 81)

Hi Brent, this question is not essentially tied in with this video, but I have been scanning these vids trying to find a solution!
So I was on one of the gmat forums, and a member factored 6^4 + 3^4 as 3^4(2^4 +1). Would that not yield 6^8 + 3^4 since you add the exponents when multiplying through the brackets? Sorry if this sounds like a novice question, haven't attempted maths in a long time!
gmat-admin's picture

What you need here is the Combining Bases Law. It can be found at 3:20 of https://www.gmatprepnow.com/module/gmat-powers-and-roots/video/1029

This law says that, if we have the same EXPONENTS in a product, we can combine the bases.
That is, (a^n)(b^n) = (ab)^n

So, (3^4)(2^4) = (3 x 2)^4 = 6^4

The rule you're referring to is correct (when we multiply two powers with the same BASE, then we add the exponents). However, in the case of (3^4)(2^4), we do not have the same bases, so we can't use that rule.

Hi Brent,

Grateful to let me have your answer for the following question please:
X^8 - Y^8 =

B.(X^4 + Y^4)(X^2 + Y^2)(X + Y)(X − Y)
C.(X^6 + Y^2)(X^2 − Y^6)
D.(X^4 − Y^4)(X^2 − Y^2)(X − Y)(X + Y)
E.(X^2 − Y^2)^4

gmat-admin's picture

First, we need to recognize that X^8 - Y^8 is a difference of squares, because X^8 = (X^4)(X^4) = (X^4)^2 and Y^8 = (Y^4)(Y^4) = (Y^4)^2

So, we can write: X^8 - Y^8 = (X^4 + Y^4)(X^4 - Y^4)

Then we must recognize that (X^4 - Y^4) is a difference of squares, which can be factored as (X^2 + Y^2)(X^2 - Y^2)

So, we get: X^8 - Y^8 = (X^4 + Y^4)(X^4 - Y^4)
= (X^4 + Y^4)(X^2 + Y^2)(X^2 - Y^2)

Finally, (X^2 - Y^2) is a difference of squares, which can be factored as (X + Y)(X − Y)

So, we get: X^8 - Y^8 = (X^4 + Y^4)(X^4 - Y^4)
= (X^4 + Y^4)(X^2 + Y^2)(X^2 - Y^2)
= (X^4 + Y^4)(X^2 + Y^2)(X + Y)(X − Y)

Answer: B

Does that help?


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