Chapter 3, Problem 1PR

For Exercises 1-8, refer to $p(x)=x^3+3x^2-6x-8$ and $q(x)=x^3-2x^2-5x+6$ .

Find the zeros of p(x).

To determine

To find: The zeros of $p\left(x\right)$ .

Answer to Problem 1PR

The zeros of the function are $x=-1,-4,2$ .

Explanation of Solution

Given information:

The function is $p\left(x\right)=x^3+3x^2-6x-8$ .

Concept used:

Let $a{x}^3+b{x}^2+cx+d=0$ be a polynomial. The factor of the polynomial can be found by following such steps:

Step.1: First find the possible multiple of $d$ for all real numbers.

Step.2: Check whether the multiple of the $d$ satisfies the polynomial or not.

Step.3: Let $z$ be the multiple of $d$ satisfy the polynomial, then equate the other zeros of the polynomial.

Consider the given function.

  $p\left(x\right)=x^3+3x^2-6x-8$ .

The possible multiple of $8$ is $\pm 1,\pm 2,\pm 4$ .

Now for $x=-1$ ,

  $\begin{array}{c}p\left(-1\right)={\left(-1\right)}^3+3{\left(-1\right)}^2-6\left(-1\right)-8\\ =-1+3+6-8\\ =-9+9\\ =0\end{array}$

Since, $x=-1$ satisfy the polynomial, so $x+1$ is a factor of given function.

The other root is,

  $\begin{array}{c}p\left(x\right)=x^3+3x^2-6x-8\\ =\left(x+1\right)\left(x^2+2x-8\right)\\ =\left(x+1\right)\left(x^2+4x-2x-8\right)\\ =\left(x+1\right)\left(x\left(x+4\right)-2\left(x+4\right)\right)\end{array}$

Further, solve the above expression.

  $p\left(x\right)=\left(x+1\right)\left(x+4\right)\left(x-2\right)$ .

Since, $p\left(x\right)=0$ , then $\left(x+1\right)\left(x+4\right)\left(x-2\right)=0$ .

  If $x+1=0$ , then $x=-1$ .

  If $x+4=0$ , then $x=-4$ .

  If $x-2=0$ , then $x=2$ .

Therefore, the zeros of the function are $x=-1,-4,2$ .