Chapter 2.3, Problem 56E

To determine

the

To show: $x=-6$ is a solution of the polynomial $x^3-52x-96=0$ and using synthetic

division factor the polynomial completely.

Answer to Problem 56E

  $\left(x-8\right)\left(x+2\right)\left(x+6\right)$

Solutions: $8,-2,-6$

Explanation of Solution

Calculation:

First, find the remainder obtained on dividing $p\left(x\right)=x^3-52x-96$ by $x+6$ using synthetic division and if the remainder is 0, then $x=-6$ is a solution of the given polynomial.

Now, the polynomial can be written as:

  $p\left(x\right)=x^3+0x^2-52x-96$

Perform the synthetic division,

  $\begin{array}{l}\text{ 1 0 }-52-96\\ -6\text{ 36 }-96\\ -6\overline{)\text{1 }-6-\text{16 0}}=R\end{array}$

Since, the remainder obtained is 0.

So, $x=-6$ is a solution of the given polynomial.

Factor theorem:

"For a polynomial $P\left(x\right)$ r is a zero of a polynomial if and only if $x-r$ is a factor of $P\left(x\right)$ ”.

Since -6 is a zero of $P\left(x\right)$ , so by factor theorem $x+6$ is a factor of $P\left(x\right)$ .

Also, from the above division, the quotient obtained on division is $x^2-6x-16$ .

So, the polynomial can be factored as shown below:

  $\begin{array}{c}{x}^3-52x-96=\left(x^2-6x-16\right)\left(x+6\right)\\ =\left(x^2-8x+2x-16\right)\left(x+6\right)\\ =\left(x\left(x-8\right)+2\left(x-8\right)\right)\left(x+6\right)\\ =\left(x-8\right)\left(x+2\right)\left(x+6\right)\end{array}$

Now, find the zeros of the polynomial equation $x^3-52x-96=0$

Using above factorization, $x^3-52x-96=0$ can be written as:

  $\begin{array}{c}{x}^3-52x-96=0\\ \left(x-8\right)\left(x+2\right)\left(x+6\right)=0\end{array}$

It is known that if $ab=0$ then either $a=0$ or $b=0$ (or both $a=0$ and $b=0$ are zero).

This gives that,

  $\begin{array}{c}x-8=0\\ x=8\end{array}$ Or $\begin{array}{c}x+2=0\\ x=-2\end{array}$ Or $\begin{array}{c}x+6=0\\ x=-6\end{array}$

Therefore, the zeros of given polynomial equation are: $-1,-2,-3$