Chapter 1, Problem 60RE

To determine

To find:Thezeros of the function $f\left(x\right)=x^3-x^2$ algebraically.

Answer to Problem 60RE

The zeros of the function $f\left(x\right)=x^3-x^2$ lies at $x=1,\pm 5$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=x^3-x^2$ .

Calculation:

For calculate the zeroes of the function, we have to equate it to zero.

  $f\left(x\right)=0$

Let the equation.

  $f\left(x\right)=x^3-x^2-25x+25x$

Calculate the zeroes of $x$ .

  $\begin{array}{c}f\left(x\right)=0\\ x^3-x^2-25x+25x=0\\ x^2\left(x-1\right)-25\left(x-1\right)=0\\ \left(x-1\right)\left(x^2-25\right)=0\end{array}$

Equate above factors to $0$ .

  $\begin{array}{c}x-1=0\\ x=1\\ x^2-25=0\\ x=\pm 5\end{array}$

If numerator is zero, the function will be zero so zero of the function lie at $8x+3=0$ .

Therefore, the zeros of the function $f\left(x\right)=x^3-x^2$ lies at $x=1,\pm 5$ .