Chapter 2.6, Problem 39E

Solution

To find: which of the following is not a zero of the function.

(A) $-\frac{3}{2}$

(B) $-\frac{3}{8}$

(C) $\frac{3}{4}$

(D) $\frac{4}{5}$

The correct option is B.

Given information: 

The function $f\left(x\right)=40x^5-42x^4-107x^3+107x^2+33x-36$ .

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:

Use of option $\text{A}:$

  $\begin{array}{c}f\left(-\frac{3}{2}\right)=40{\left(-\frac{3}{2}\right)}^5-42{\left(-\frac{3}{2}\right)}^4-107{\left(-\frac{3}{2}\right)}^3+107{\left(-\frac{3}{2}\right)}^2+33\left(-\frac{3}{2}\right)-36\\ =40\left(-\frac{243}{32}\right)-42\left(\frac{81}{16}\right)-107\left(-\frac{27}{8}\right)+107\left(\frac{9}{4}\right)+33\left(-\frac{3}{2}\right)-36\\ =-303.75-212.625+361.125+240.75-49.5-36\\ =0\end{array}$

So, $-\frac{3}{2}$ is a zero of the function.

B:

  $\begin{array}{c}f\left(-\frac{3}{8}\right)=40{\left(-\frac{3}{8}\right)}^5-42{\left(-\frac{3}{8}\right)}^4-107{\left(-\frac{3}{8}\right)}^3+107{\left(-\frac{3}{8}\right)}^2+33\left(-\frac{3}{8}\right)-36\\ =40\left(-\frac{243}{32768}\right)-42\left(\frac{81}{4096}\right)-107\left(-\frac{27}{512}\right)+107\left(\frac{9}{64}\right)+33\left(-\frac{3}{8}\right)-36\\ \approx -28.8\end{array}$

So, $-\frac{3}{8}$ is not a zero of this function.

Hence, the correct option is B.