Chapter 2.6, Problem 30E

Solution

To find: all real zeros of the function.

The real zeros of the function $x=-4,\frac{1}{2}$ .

Given information: 

The function $g\left(x\right)=2x^4+9x^3+5x^2+3x-4$ .

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

  $g\left(x\right)=2x^4+9x^3+5x^2+3x-4$

Rational zero theorem:

Factors of constant term $4$ :

  $p=1,2,4$

Factors of leading coefficient $2$ :

  $q=1,2$

All possible combinations of $\pm \frac{p}{q}$ .

The function's zeros could be all or none of them.

Simplify and remove any duplicates.

  $\begin{array}{c}\pm \frac{p}{q}=\pm \left(1,2,4,\frac{1}{2},\frac{2}{2},\frac{4}{2}\right)\\ =\pm \left(1,2,4,\frac{1}{2}\right)\end{array}$

Test the possible rational zero combinations by plugging into $g\left(x\right)$ or with synthetic

division to see which ones generate a zero.

  $\backslash$ poly Horner scheme $\left[x=1/2\right]\left\{2x^4+9x^3+5x^2+3x-4\right\}$

  $\backslash$ poly Horner scheme $[x=-4]\left\{2x^4+9x^3+5x^2+3x-4\right\}$

Only $x=-4,\frac{1}{2}$ produce zeros. This gives us two factors of $g$ .

  $\left(x+4\right)\left(2x-1\right)=2x^2+7x-4$

Divide $g$ by the factors to find the remaining part to work on.

  

  $x^2+x+1$ has a discriminant of

  $\begin{array}{c}{b}^2-4ac=1^2-4\left(1\right)\left(1\right)\\ =-3\end{array}$

which is negative, so the other $2$ zeros are complex.

Therefore, the real zeros of the function $x=-4,\frac{1}{2}$ .