Factoring Expressions raised to powers

**allyourbass2212** · Jun 30th 2009, 10:41 AM

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

**running-gag** · Jun 30th 2009, 10:47 AM

Hi

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

**allyourbass2212** · Jun 30th 2009, 10:52 AM

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.

**skeeter** · Jun 30th 2009, 10:53 AM

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 ...

$(x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4(x^2-6) + 1]<br />$

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

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

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

**allyourbass2212** · Jul 1st 2009, 07:20 AM

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

**allyourbass2212** · Jul 1st 2009, 03:51 PM

Bump, would someone please help?

**skeeter** · Jul 1st 2009, 05:06 PM

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 ...

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

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

really no big advantage in doing this, however.

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