Chapter 4.5, Problem 117AYU

To determine

To check: If $\frac{1}{3}$ is a zero of $f\left(x\right)=4x^3-5x^2-3x+1$.

Answer to Problem 117AYU

$\text{Therefore }\frac{1}{3}\text{ is not a zero of }f\left(x\right)=4x^3-5x^2-3x+1$.

Explanation of Solution

Given:

$f\left(x\right)=4x^3-5x^2-3x+1$

By factor theorem, $\left(x-c\right)$ is the factor of $f\left(x\right)$ then $f\left(c\right)=0$.

To prove if $\frac{1}{3}$ is a zero of $f\left(x\right)=4x^3-5x^2-3x+1$.

Here $\text{c }=\frac{1}{3}$ finding $f\left(\frac{1}{3}\right)=4{\left(\frac{1}{3}\right)}^3-5{\left(\frac{1}{3}\right)}^2-3\left(\frac{1}{3}\right)+1=2.\frac{1}{27}+3.\frac{1}{9}-6.\frac{1}{3}+7=\frac{-11}{27}\ne 0$.

Therefore $\frac{1}{3}$ is not a zero of $f\left(x\right)=4x^3-5x^2-3x+1$.