Factoring completely Factor each polynomial completely. a. [latex] 2 w^{4}-32[/latex] b. [latex] -6 x^{7}+6 x[/latex]

a. [latex] 2 w^{4}-32=2\left(w^{4}-16\right)[/latex] Factor out the greatest common factor. [latex] =2\left(w^{2}-4\right)\left(w^{2}+4\right)[/latex] Difference of two squares [latex] =2(w-2)(w+2)\left(w^{2}+4\right)[/latex] Difference of two squares The polynomial is now factored completely because [latex] w^{2}+4[/latex] is prime. b. [latex] -6 x^{7}+6 x=-6 x\left(x^{6}=1\right)[/latex] Greatest common factor [latex] =-6 x\left(x^{3}-1\right)\left(x^{3}+1\right)[/latex] Difference of two squares [latex] =-6 x(x-1)\left(x^{2}+x+1\right)(x+1)\left(x^{2}-x+1\right)[/latex] Difference of two cubes; sum of two cubes The polynomial is factored completely because all of the factors are prime.

Question B.4.7: Factoring completely Factor each polynomial completely. a. 2...

Precalculus Functions and Graphs

Factoring completely

Factor each polynomial completely.

a. $2 w^{4}-32$ b. $-6 x^{7}+6 x$

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a. $2 w^{4}-32=2\left(w^{4}-16\right)$ Factor out the greatest common factor.

$=2\left(w^{2}-4\right)\left(w^{2}+4\right)$ Difference of two squares

$=2(w-2)(w+2)\left(w^{2}+4\right)$ Difference of two squares

The polynomial is now factored completely because $w^{2}+4$ is prime.

b. $-6 x^{7}+6 x=-6 x\left(x^{6}=1\right)$ Greatest common factor

$=-6 x\left(x^{3}-1\right)\left(x^{3}+1\right)$ Difference of two squares

$=-6 x(x-1)\left(x^{2}+x+1\right)(x+1)\left(x^{2}-x+1\right)$ Difference of two cubes; sum of two cubes

The polynomial is factored completely because all of the factors are prime.