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

Oxygenthief

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Jul 7, 2018
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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

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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

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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...
 
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Oxygenthief

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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

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(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.

https://www.youtube.com/watch?v=Wdb0V2hfYSU&feature=youtu.be

A difference of cubes, a sum of cubes, and perfect-square trinomials each have factoring patterns. Here's another video.

https://www.youtube.com/watch?v=1_kxLXFtqHg

These videos came from the following page (this site covers many beginning algebra topics, in addition to other subjects).

http://www.mathispower4u.com/algebra.php

You can also google keywords factoring patterns, to find other sites. :cool:
 
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