Question

Can someone please teach me how to do this? I have to factor it but I don't know how to do these types of problems. Thanks
5(x^6+1)^4(6x^5)(3x+2)^3+3(3x+2)^2(3)(x^6+1)^5

Answer 1

\[
5(x^6+1)^4 * 6x^5 * (3x+2)^3 + 3 * (3x+2)^2 * 3(x^6+1)^5
\]Is this what needs to be factored?

Answer 2

Yes

Answer 3

\[5(x^6+1)^4 * 6x^5 * (3x+2)^3 + 3 * (3x+2)^2 * 3(x^6+1)^5 = \\
30 * (x^6+1)^4 * x^5 * (3x+2)^3 + 9 * (3x+2)^2 * (x^6+1)^5
\]What is the GCF of 30 and 9?
GCF of \((x^6+1)^4\) and \((x^6+1)^5\) ?
GCF of \((3x+2)^3\) and \((3x+2)^2\) ?

Answer 4

GCF of 30 and 9 is 3
GCF of \[(x^6+1)^4 and (x^6+1)^5 is (x^6+1)^4]
GCF GCF of (3x+2)^3 and (3x+2)^2 is (3x+2)^2 right?

Answer 5

The attached solution shows what Mathematica 9 is capable of running on a modern desk top computer.

Not to be submitted as orginal student work using pencil and paper.

Answer 6

Thanks

Answer 7

Your welcome.

Answer 8

GCF of 30 and 9 is 3
GCF of (x^6+1)^4 and (x^6+1)^5 is (x^6+1)^4
GCF of (3x+2)^3 and (3x+2)^2 is (3x+2)^2 right?

Correct. So factor out the GCF:\[ \large
30 * (x^6+1)^4 * x^5 * (3x+2)^3 + 9 * (3x+2)^2 * (x^6+1)^5 = \\ \large
3(x^6+1)^4(3x+2)^2\{10x^5(3x+2) + 3(x^6+1)\} = \\ \large
3(x^6+1)^4(3x+2)^2\{30x^6 +20x^5 + 3x^6+3\} = \\ \large
3(x^6+1)^4(3x+2)^2(33x^6 +20x^5 +3) = \\ \large
3\{(x^2)^3+1^3\}^4(3x+2)^2(33x^6 +20x^5 +3) = \\ \large
3\{(x^2+1)(x^4-x^2+1\}^4(3x+2)^2(33x^6 +20x^5 +3) = \\ \large
3(x^2+1)^4(x^4-x^2+1)^4(3x+2)^2(33x^6 +20x^5 +3) = \\ \large
\]Note: In the line before the last, we made use of the algebraic identity:\[\large a^3 + b^3 = (a+b)(a^2-ab+b^2)\]Therefore, \[\large (x^6 + 1) = (x^2)^3 + 1^3 = (x^2+1)(x^4 - x^2 + 1)\]