# Quadratic factoring w/ 3 brackets: solve (x^2 - 4) - 2(x -2)(x - 1) = 0 w/o expanding

#### Oxygenthief

##### New member
Hey guys. I came across a question the other day that has me completely stumped.

The question is:

Solve (x^2 - 4) - 2(x -2)(x - 1) = 0 by factoring, do not expand first.

For the life of me I can't figure it out. Can anyone get me started?

Thanks!!

#### Subhotosh Khan

##### Super Moderator
Staff member
Hey guys. I came across a question the other day that has me completely stumped.

The question is:

Solve (x^2 - 4) - 2(x -2)(x - 1) = 0 by factoring, do not expand first.

For the life of me I can't figure it out. Can anyone get me started?

Thanks!!
Hint:

(x^2 - 4 ) = (x^2 - 22 ) = (x - 2)(x + 2)

Now factor out the common factor.....

#### Denis

##### Senior Member
Solve (x^2 - 4) - 2(x -2)(x - 1) = 0 by factoring
Subhotosh's hint:
(x^2 - 4 ) = (x^2 - 2^2 ) = (x - 2)(x + 2)

May be easier if you now let u = x - 2:
u(x + 2) - 2u(x - 1) = 0

Go to town with that, substitute back in to wrap up...

Last edited by a moderator:

#### Oxygenthief

##### New member
Thanks so much guys, that really helped. Very appreciated. It's been 15 years since I've done any algebra so it's coming back very slowly.

#### mmm4444bot

##### Super Moderator
Staff member
(a^2 - b^2) = (a + b)(a - b)

This pattern is called "a difference of squares". Here's a short video showing its use, in a different situation.