Chapter 8.7, Problem 17CYU

To determine

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

Answer to Problem 17CYU

The factored polynomial is $3t^3+2t^2-48t-32=\left(3t+2\right)\left(t-4\right)\left(t+4\right)$ .

Explanation of Solution

Given information:

The given polynomial is $3t^3+2t^2-48t-32$ .

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 $E=3t^3+2t^2-48t-32$ .

  $E=\left(3t^3+2t^2\right)-\left(48t+32\right)$

  $\Rightarrow E=t^2\left(3t+2\right)-16\left(3t+2\right)$       Factor $\left(2c+3\right)$ is common.

  $\Rightarrow E=\left(3t+2\right)\left(t^2-16\right)$

  $\Rightarrow E=\left(3t+2\right)\left({(t)}^2-{(4)}^2\right)$

  $\Rightarrow E=\left(3t+2\right)\left(t-4\right)\left(t+4\right)$       Difference of two squares

Thus, the factored polynomial is $3t^3+2t^2-48t-32=\left(3t+2\right)\left(t-4\right)\left(t+4\right)$ .

Conclusion:

The factored polynomial is $3t^3+2t^2-48t-32=\left(3t+2\right)\left(t-4\right)\left(t+4\right)$ .