Question

Is $\frac{1}{3}$a zero of localid="1646142357128" $f\left(x\right)=2x^3+3x^2-6x+7$? Explain.

Short answer

$\frac{1}{3}$ is not a zero of function$f\left(x\right)=2x^3+3x^2-6x+7$becouse$f\left(\frac{1}{3}\right)\ne 0$

Step-by-step solution

Step 1. Given data

The given function is

$f\left(x\right)=2x^3+3x^2-6x+7$

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

Step 2. Verification

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

role="math" $$\begin{gathered} f\left(x\right)=2x^3+3x^2-6x+7 \\ f\left(\frac{1}{3}\right)=2{\left(\frac{1}{3}\right)}^3+3{\left(\frac{1}{3}\right)}^2-6\left(\frac{1}{3}\right)+7 \\ f\left(\frac{1}{3}\right)=\frac{2}{27}+\frac{1}{3}-2+7 \\ f\left(\frac{1}{3}\right)=5.407 \end{gathered}$$

$f\left(\frac{1}{3}\right)$ is not equal to zero so$\frac{1}{3}$is not a zero of the function