Question

Is$\frac{1}{3}$a zero of $f\left(x\right)=4x^3-5x^2-3x+1$? Explain.

Short answer

$\frac{1}{3}$is not a zero of functionlocalid="1646143603596" $f\left(x\right)=4x^3-5x^2-3x+1$becouse$f\left(\frac{1}{3}\right)\ne 0$

Step-by-step solution

Step 1. Given data 

The given function is

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

Zero of the function is $\frac{1}{3}$

Step 2. Verification 

Substitute $\frac{1}{3}$ for x in the function

role="math" localid="1646143292347" $$\begin{gathered} f\left(x\right)=4x^3-5x^2-3x+1 \\ f\left(\frac{1}{3}\right)=4{\left(\frac{1}{3}\right)}^3-5{\left(\frac{1}{3}\right)}^2-3\left(\frac{1}{3}\right)+1 \\ f\left(\frac{1}{3}\right)=-0.04 \\ f\left(\frac{1}{3}\right)\ne 0 \end{gathered}$$

$\frac{1}{3}$is not equal to zero so $\frac{1}{3}$is not a zero of the function