Question

Is$\frac{3}{5}$a zero of localid="1646143732762" $f\left(x\right)=2x^6-5x^4+x^3-x+1$? Explain.

Short answer

$\frac{3}{5}$is not a zero of function $f\left(x\right)=2x^6-5x^4+x^3-x+1$becouse$f\left(\frac{3}{5}\right)\ne 0$

Step-by-step solution

Step 1. Given data 

The given function is

$f\left(x\right)=2x^6-5x^4+x^3-x+1$

Zero of the function is$\frac{3}{5}$

Step 2. Verification 

Substitute $\frac{3}{5}$ for x in the function

localid="1646143682252" $$\begin{gathered} f\left(x\right)=2x^6-5x^4+x^3-x+1 \\ f\left(\frac{3}{5}\right)=2{\left(\frac{3}{5}\right)}^6-5{\left(\frac{3}{5}\right)}^4+{\left(\frac{3}{5}\right)}^3-\left(\frac{3}{5}\right)+1 \\ f\left(\frac{3}{5}\right)=0.06 \\ f\left(\frac{3}{5}\right)\ne 0 \end{gathered}$$

$f\left(\frac{3}{5}\right)$is not equal to zero so$\frac{3}{5}$is not a zero of the function