# Thread: Factoring Expressions raised to powers

1. ## Factoring Expressions raised to powers

Question 1 solved

Question 2 I have a question for the expression

when the author factors out they leave the term the same resulting in the answer
...doesn't this term need to be "broken up" like all of the other terms once its factored?

Many Thanks

2. Hi

Didn't you understand the previous clarification ?
http://www.mathhelpforum.com/math-he...ed-powers.html

3. Originally Posted by running-gag
Hi

Didn't you understand the previous clarification ?
http://www.mathhelpforum.com/math-he...ed-powers.html
Not completely regarding the question in this topic, thus the reason for making a new thread and not bumping the previous.

4. Originally Posted by allyourbass2212

In the expression

$2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2$
$=[2(x^2-6)^7+(x^2-6)^2+4x^2-24{\color{red} +1}](x^2-6)^2$

I.
a
.Where is the +1 coming from?
b.I am assuming they are using the coefficient from $+ (x^2-6)^2$
c.If this is correct then why does the +1 get grouped within $x^2-24$ resulting in $-23$, rather staying on the outside of the brackets? e.g.
$=[2(x^2-6)^7+(x^2-6)^2+4x^2-24]{\color{red} +1}(x^2-6)^2$

II.
And if thats the case why are they not using a +1 in the second term as well?
$=[2(x^2-6)^7{\color{red} +1}(x^2-6)^2+4x^2-24](x^2-6)^2$

Many Thanks for clarification!
$2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2$

let $u = x^2-6$

$2u^9+u^4+4u^3+u^2$

factor out $u^2$ ...

$u^2(2u^7 + u^2 + 4u + 1)$ ... see the "1" ?

back substitute ...

$(x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4(x^2-6) + 1]
$

distribute the $4(x^2-6)$ term ...

$(x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4x^2 - 24 + 1]$

$(x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4x^2 - 23]$

5. I have a question for the expression

when the author factors out they leave the term the same resulting in the answer
...doesn't this term need to be "broken up" like all of the other terms once its factored?

Many Thanks

7. Originally Posted by allyourbass2212
I have a question for the expression

when the author factors out they leave the term the same resulting in the answer
...doesn't this term need to be "broken up" like all of the other terms once its factored?

Many Thanks
$(15xy-1)(2x-1)^3 - 8(2x-1)^2$

let $u = (2x-1)$

$(15xy-1)u^3 - 8u^2$

factor out $u^2$ ...

$u^2[(15xy-1)u - 8]$

back substitute $(2x-1)$ for $u$ ...

$(2x-1)^2[(15xy-1)(2x-1) - 8]$

note that it is permissible to stop here, or one can expand the binomial product in the brackets and combine the resulting constant terms ...

$(2x-1)^2[(30x^2y - 15xy - 2x + 1) - 8]
$

$(2x-1)^2(30x^2y - 15xy - 2x - 7)$

really no big advantage in doing this, however.