Factoring completely.

[perryc]

I've gotten really stuck on the following problem.

Factor the expression completely:
5 * (x^2+4)^4 * (2x) * (x-2)^4 + (x^2+4)^5 * 4 * (x-2)^3

The answer in the back of the book is:
2 * (x^2+4)^4 * (x-2)^3 * (7x^2-10x+8)

I'm aware of special factoring forms. I can do all the math for (x+y)^4 etc, etc, but it gets really long and just feels like the wrong way to approach this problem. After scouring my book for any clues with no luck I feel really lost! Any help would be greatly appreciated!

 

[pka]

Factor the expression completely:
5 * (x^2+4)^4 * (2x) * (x-2)^4 + (x^2+4)^5 * 4 * (x-2)^3
The answer in the back of the book is:
2 * (x^2+4)^4 * (x-2)^3 * (7x^2-10x+8)

Let \(\displaystyle y=(x^2+4)~\&~z=(x-2)\) now we have:
\(\displaystyle 5y^4(2x)z^4+y^5(4)z^3\).

That is easy to factor. Then substitute.

 

[soroban]

Hello, perryc!

\(\displaystyle \text{Factor completely: }\:5\cdot (x^2+4)^4 \cdot (2x) \cdot (x-2)^4 + (x^2+4)^5\cdot 4\cdot (x-2)^3\)

\(\displaystyle \text{Book's answer: }\:2 \cdot (x^2+4)^4 \cdot (x-2)^3 \cdot (7x^2-10x+8)\:\)

I wondered why the problem was given in such a strange form . . . then I got it!


\(\displaystyle \text{You were told to differentiate: }f(x) \:=\x^2+4)^5(x-2)^4 \quad\hdots\text{ right?}\)


\(\displaystyle \text{Product rule: }\:f'(x) \:=\:5(x^2+4)2x\cdot(x-2)^4 + (x^2+4)^5\cdot 4(x-2)^3\)

. . . . . . . . . . . . . . .\(\displaystyle =\;10x(x^2+4)^4(x-2)^4 + 4(x^2+4)^5(x-2)^3\)


\(\displaystyle \text{We have two terms, separated by the "plus": }\;\underbrace{10x(x^2+4)^4(x-2)^4}\;+\;\underbrace{4(x^2+4)^5(x-2)^3}\)


\(\displaystyle \text{The two terms have some common factors.}\)

. . \(\displaystyle 10x(x^2+4)^4(x-2)^4 + 4(x^2+4)^5(x-2)^3\)
. . .\(\displaystyle \uparrow\, _{\text{ - - common factor 2 - - }} \, \uparrow\)

. . \(\displaystyle 10x\underbrace{(x^2+4)^4}(x-2)^4 + 4\underbrace{(x^2+4)^5}(x-2)^3\)
. . . . . . . .\(\displaystyle \uparrow\;_{\text{ - common }(x^2+4)^4\text{ - }}\;\uparrow\)

. . \(\displaystyle 10x(x^2+4)^4\underbrace{(x-2)^4} + 4(x^2+4)^5\underbrace{(x-2)^3}\)
. . . . . . . . . . . . . .\(\displaystyle \uparrow\;_{\text{ - common }(x-2)^3\text{ - }} \;\uparrow\)


\(\displaystyle \text{Factor out the common factors: }\:2(x^2+4)^4(x-2)^3\,\bigg[5x(x-2) + 2(x^2+4)\bigg]\)

\(\displaystyle \text{and we have: }\:2(x^2+4)^4(x-2)^3\,\bigg[5x^2 - 10x + 2x^2 + 8\bigg]\)

. . . . . . . . \(\displaystyle =\;2(x^2+4)^4(x-2)^3(7x^2 - 10x + 8)\,\)

 

[mmm4444bot]

Sam Spade, PI

Soroban, MI

 

[perryc]

Thank you soroban! Your explanation was exactly what I needed to understand this problem. It's no fun to be stuck!

-Perry (Math Student)

 

[tkhunny]

Now, let's see you do one.