Chapter 2.6, Problem 6GP

Solution

To find: all real zeros of the function.

The real zeros of the function $-2,\frac{3}{2},-1+2\sqrt{2},-1-2\sqrt{2}$ .

Given information: 

The function $f\left(x\right)=2x^4+5x^3-18x^2-19x+42$ .

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

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

Calculation:

Consider the function

  $f\left(x\right)=2x^4+5x^3-18x^2-19x+42$

First, let's use the Rational Zeros theorem to find the possible rational zeros of the function.

  $\begin{array}{c}\frac{p}{q}=\frac{\text{ factors of }42}{\text{ factors of }3}\\ =\frac{\pm 1,\pm 2,\pm 3,\pm 6,\pm 7,\pm 14,\pm 21,\pm 42}{\pm 1,\pm 3}\\ =\pm 1,\pm 2,\pm 3,\pm 6,\pm 7,\pm 14,\pm 21,\pm 42,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{7}{3},\pm \frac{14}{3}\end{array}$

.

Check the values using synthetic division until a zero is found.

  $\frac{3}{2}$is a zero.

  $\begin{array}{cccccc}\frac{3}{2}& 2& 5& -18& -19& 42\\ & & 3& 12& -9& -42\\ & 2& 8& -6& -28& 0\end{array}$

Factor out a binomial using the result of the synthetic division.

Write a product of factors.

  $\begin{array}{c}\left(x-\frac{3}{2}\right)\left(2x^3+8x^2-6x-28\right)\\ \left(x-\frac{3}{2}\right)2\left(x^3+4x^2-3x-14\right)\\ \left(x-3\right)\left(x^3+4x^2-3x-14\right)\\ \left(x-3\right)\left(x^3+2x^2+2x^2+4x-7x-14\right)\\ \left(x-3\right)\left\{x^2\left(x+2\right)+2x\left(x+2\right)-7\left(x+2\right)\right\}\\ \left(x-3\right)\left(x+2\right)\left(x^2+2x-7\right)\end{array}$

Use quadratic formula for $x^2+2x-7=0$ .

  $\begin{array}{c}{x}^2+2x-7=0\\ x=-1+2\sqrt{2},-1-2\sqrt{2}\end{array}$

The real zeros of $f$ are $-2,\frac{3}{2},-1+2\sqrt{2},-1-2\sqrt{2}$ .