Chapter 8.7, Problem 18CYU

To determine

To factor: The given polynomial, if possible. If the polynomial cannot be factored, write prime.

Answer to Problem 18CYU

The factored polynomial is $w^3-3w^2-9w+27=\left(w-3\right)\left(w-3\right)\left(w+3\right)$ .

Explanation of Solution

Given information:

The given polynomial is $w^3-3w^2-9w+27$ .

Concept used:

The polynomial of the form $a{c}^3+b{x}^2+cx+d$ are factored by grouping the terms having one or more common factor(s).

Calculations:

The given polynomial E (say) is $E=w^3-3w^2-9w+27$ .

  $E=\left(w^3-3w^2\right)-\left(9w-27\right)$

  $\Rightarrow E=w^2\left(w-3\right)-9\left(w-3\right)$       Factor $\left(2c+3\right)$ is common.

  $\Rightarrow E=\left(w-3\right)\left(w^2-9\right)$

  $\Rightarrow E=\left(w-3\right)\left({(w)}^2-{(3)}^2\right)$

  $\Rightarrow E=\left(w-3\right)\left(w-3\right)\left(w+3\right)$       Difference of two squares

Thus, the factored polynomial is $w^3-3w^2-9w+27=\left(w-3\right)\left(w-3\right)\left(w+3\right)$ .

Conclusion:

The factored polynomial is $w^3-3w^2-9w+27=\left(w-3\right)\left(w-3\right)\left(w+3\right)$ .