Chapter 2.8, Problem 2Q

Solution

To find: All zeros of the polynomial function.

The zeros of the function would be $-4,\pm \frac{3}{\sqrt{2}},5$ .

Given information:

A polynomial function: $f\left(x\right)=2x^4-2x^3-49x^2+9x+180$ .

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 2 and the constant term is 180.

Factors of 2 are $\pm 1,\pm 2$ and factors of 180 are,

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 9,\pm 10,\pm 12,\pm 15,\pm 18,\pm 20,\pm 30,\pm 36,\pm 45,\pm 60,\pm 90,\pm 180$ .

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{1}{2},\pm \frac{2}{1},\pm \frac{2}{2},\pm \frac{3}{1},\pm \frac{3}{2},\pm \frac{4}{1},\pm \frac{4}{2},\pm \frac{5}{1},\pm \frac{5}{2},\pm \frac{6}{1},\pm \frac{6}{2},\\ \pm \frac{9}{1},\pm \frac{9}{2},\pm \frac{10}{1},\pm \frac{10}{2},\pm \frac{12}{1},\pm \frac{12}{2},\pm \frac{15}{1},\pm \frac{15}{2},\pm \frac{30}{1},\mathrm{...},\pm \frac{180}{2}\\ =\pm 1,\pm \frac{1}{2},\pm 2,\pm 3,\pm \frac{3}{2},\pm 4,\pm 5,\pm \frac{5}{2},\pm 6,\pm 9,\pm \frac{9}{2},\pm 10,\pm 12,\pm 15,\\ \pm \frac{15}{2},\pm 18,\pm 20,\pm 30,\pm 36,\pm 45,\pm \frac{45}{2},\pm 60,\pm 90,\pm 180\end{array}$

By checking possible zeros in the given function, it can be seen that $x=5$ is a zero of the function. Now, use synthetic division as shown below:

  $\begin{array}{l}5\overline{)\begin{array}{l}2-2-499180\\ \downarrow 1040-45-180\end{array}}\\ 28-9-360\end{array}$

Because $x=5$ is a zero of the function, so the function can be written,

  $f\left(x\right)=\left(x-5\right)\left(2x^3+8x^2-9x-36\right)$ .

Solve for the cubic equation as:

  $\begin{array}{c}2x^3+8x^2-9x-36=0\\ \left(2x^3+8x^2\right)+\left(-9x-36\right)=0\\ 2x^2\left(x+4\right)-9\left(x+4\right)=0\\ \left(2x^2-9\right)\left(x+4\right)=0\end{array}$

Using zero product property, it can be written as:

  $\begin{array}{c}\left(2x^2-9\right)=0\text{or}\left(x+4\right)=0\\ x=\pm \frac{3}{\sqrt{2}}\text{or}x=-4\end{array}$

Therefore, the zeros of the function would be $-4,\pm \frac{3}{\sqrt{2}},5$ .