Chapter 4.5, Problem 35E

To determine

To find: all real zeros of the given function.

Answer to Problem 35E

The real zeros are $-2,-3$ , and $-5$ .

Explanation of Solution

Given information:

Given function,

  $h\left(x\right)=x^3+10x^2+31x+30$

Calculation:

Consider the function.

  $h\left(x\right)=x^3+10x^2+31x+30$

The above function is of degree 3, so it has 3 zeros.

Here, the leading coefficient is $1$ and constant term is $30$ .

The possible rational zeros are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\text{and}\pm 30$ .

Test possible zeros using synthetic division until a solution is found.

For $x=-2$

  $\begin{array}{ccccc}-2& 1& 10& 31& 30\\ & & -2& -16& -30\\ & 1& 8& 15& 0\end{array}$

Therefore, $x+2$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}h\left(x\right)=x^3+10x^2+31x+30\\ =\left(x+2\right)\left(x^2+8x+15\right)\\ =\left(x+2\right)\left(x+3\right)\left(x+5\right)\end{array}$

Solve further for $h\left(x\right)=0$ .

  $\begin{array}{c}\left(x+2\right)\left(x+3\right)\left(x+5\right)=0\\ x=-2,-3,-5\end{array}$

Therefore, all real zeros are $-2,-3$ , and $-5$ .