Chapter A.3, Problem 113AYU

To determine

To find: The factor completely each polynomial. If the polynomial cannot be factored, say it is prime.

Answer to Problem 113AYU

${\left(3x-2\right)}^3-27=\left(3x-5\right)\left(9x^2-3x+7\right)$

Explanation of Solution

Given:

${\left(3x-2\right)}^3-27$

Calculation:

${\left(3x-2\right)}^3-27={\left(3x-2\right)}^3-3^3=\left(3x-2-3\right)\left({\left(3x-2\right)}^2+\left(3x-2\right)\left(3\right)+3^2\right)$, since $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$ where $a=3x-2$ and $b=3$

$=\left(3x-5\right)\left(9x^2-12x+4+9x-6+9\right)$

$=\left(3x-5\right)\left(9x^2-3x+7\right)$