Chapter 6.7, Problem 34E

To determine

To factor : the expression $4x^3-12x^2-37x-15$.

Answer to Problem 34E

The complete factor t is $4x^3-12x^2-37x-15=\left(2x+1\right)(2x-15)(x-1)$.

Explanation of Solution

Given information:

  $(2x+1){\text{is a factor of 4x}}^3-12x^2-37x-15$.

Formula used:

  $\text{Dividend=Quotient×Divisor+Remainder}$

Calculation:

We have ,

  $(2x+1){\text{ factor of ( 4x}}^3-12x^2-37x-15)$

This implies that ${\text{(4x}}^3-12x^2-37x-15)$ can be divided by $\text{(2x+1})$.

Now, using polynomial division

Step 1.

  $2x+1\begin{array}{l}\begin{array}{c}\hfill 2x^2\\ \hfill \overline{)\begin{array}{l}4x^3-12x^2-37x-15\\ 4x^3+2x^2\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $-14x^2-37x-15$

Step 2.

  $2x+1\begin{array}{l}\begin{array}{c}\hfill 2x^2-7x\\ \hfill \overline{)\begin{array}{l}-14x^2-37x-15\\ -14x^2-7x\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $-30x-15$

Step 3.

  $2x+1\begin{array}{l}\begin{array}{c}\hfill 2x^2-7x-15\\ \hfill \overline{)\begin{array}{l}-30x-15\\ -30x-15\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $0$

  $\begin{array}{l}=4x^3-12x^2-37x-15=(2x+1)(2x^2-7x-15)+0\\ =4x^3-12x^2-37x-15=\left(2x+1\right)\left(2x^2-7x-15\right)\end{array}$

Further,

  $=2x^2-7x-15=2x^2-2x-15x-15$

  $\begin{array}{l}=2x(x-1)-15(x-1)\\ =(2x-15)(x-1)\\ =(2x-15)(x-1)\end{array}$

Hence, the complete factor of ${\text{(4x}}^3-12x^2-37x-15)$ :-

  $4x^3-12x^2-37x-15=\left(2x+1\right)(2x-15)(x-1)$