Factoring expressions raised to powers

**allyourbass2212** · June 19th 2009, 11:14 AM

The book I am using gave factoring expressions raised to powers a very brief run through and a few examples, but the authors examples dont make any sense.

Could some one please provide detailed, clear instructions to walk me through this problem and explain whats going on?

expression
$2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2$

the book gives the answer as $[2(x^2-6)^7+(x^2-6)^2+4x^2-23](x^2-6)^2$

**Twig** · June 19th 2009, 11:28 AM

Hi

$<br /> 2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2<br />$

Note that the common factor $(x^{2}-6)$ appears in all the terms. This means you can factor this whole parenthesis out, which is what is done in the answer.

You know that $(x^{2}-6)^{9} = (x^{2}-6)^{7} \cdot (x^{2}-6)^{2}$

Now you get: $(x^{2}-6)^{2}(1 + 4(x^{2}-6) + ...)$

$1 + 4(x^{2}-6) = 4x^{2}-23$

So, the answer just leavs the other terms as they are and we get the same answer as you wrote.

If you want me to explain better let me know!

**allyourbass2212** · June 19th 2009, 11:32 AM

I am still confused unfortunately

.

**Twig** · June 19th 2009, 11:40 AM

hi

Ok. What part is most confusing?

You do know that $x^{a} \cdot x^{b} = x^{a+b}$ ?

Example)

$a + a^2 + a^3 + a^4 = a(1+a+a^2+a^3)$

Now think of our whole paraenthesis $(x^{2}-6)$ as the 'a' from the example above.

They do the exact same thing in your question.

$\boxed{(x^{2}-6)^{2}}[1+4\boxed{(x^{2}-6)} + \boxed{(x^{2}-6)^{2}} + 2\boxed{(x^{2}-6)^{7}}]$

Do you see it? The boxes act as our 'a' from the above example.
You´ll have to focus to read it here, the parenthesis show kinda bad.

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