Chapter 2, Problem 33REAE

Solution

To find: All zeros of the polynomial function.

The zeros of the function would be $-3,2,5$ .

Given information:

A polynomial function: $f\left(x\right)=x^3-4x^2-11x+30$ .

Formula used:

The Rational Zero Theorem: If $f\left(x\right)=a_n{x}^n+\mathrm{...}+a_{1}x+a_0$ has integer coefficients, then every rational zero of $f$ has the following form:

  $\frac{p}{q}=\frac{\text{Factor of constant term }{a}_0}{\text{Factor of leading coefficient term }{a}_n}$

Calculation:

The leading coefficient is 1 and the constant term is 30.

Factors of 1 are $\pm 1$ and factors of 30 are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$ .

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{5}{1},\pm \frac{6}{1},\pm \frac{10}{1},\pm \frac{15}{1},\pm \frac{30}{1}\\ =\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30\end{array}$

By checking possible zeros in the given function, it can be seen that $x=2$ is a zero of the function. Now, use synthetic division as shown below:

  $\begin{array}{l}2\overline{)\begin{array}{l}1-4-1130\\ \downarrow 2-4-30\end{array}}\\ 1-2-150\end{array}$

Because $x=2$ is a zero of the function, so the function can be written $f\left(x\right)=\left(x-2\right)\left(x^2-2x-15\right)$ .

Solve for the quadratic equation as:

  $\begin{array}{c}\left(x^2-2x-15\right)=0\\ x^2-5x+3x-15=0\\ x\left(x-5\right)+3\left(x-5\right)=0\\ \left(x-5\right)\left(x+3\right)=0\\ x=-3,5\end{array}$

Therefore, the zeros of the function would be $-3,2,5$ .