Chapter 4.4, Problem 15Q

To determine

To show: that $x+5$ is a factor of the given polynomial and then factorize completely.

Answer to Problem 15Q

The factorized form is $x^3-2x^2-23x+60=\left(x+5\right)\left(x-4\right)\left(x-3\right)$ .

Explanation of Solution

Given information:

A polynomial is given as

  $f\left(x\right)=x^3-2x^2-23x+60$

Concept used:

The factor theorem states that “A polynomial has a factor $x-k$ is and only if $f\left(k\right)=0$ ”.

Calculation:

Consider the given polynomial.

  $f\left(x\right)=x^3-2x^2-23x+60$

If $x+5$ is a factor of the given polynomial, then $f\left(-5\right)=0$ .

Now, check whether $f\left(-5\right)=0$ or not.

  $\begin{array}{c}f\left(-5\right)={\left(-5\right)}^3-2{\left(-5\right)}^2-23\left(-5\right)+60\\ =-125-50+115+60\\ =-75+175\\ =0\end{array}$

Therefore, $x+5$ is a factor of the given polynomial $f\left(x\right)=x^3-2x^2-23x+60$ .

Now, factorize the polynomial as shown:

  $\begin{array}{c}f\left(x\right)=x^3-2x^2-23x+60\\ =x^3+5x^2-7x^2-35x+12x+60\\ =x^2\left(x+5\right)-7x\left(x+5\right)+12\left(x+5\right)\\ =\left(x+5\right)\left(x^2-7x+12\right)\end{array}$

Now, factorize $\left(x^2-7x+12\right)$ as shown:

  $\begin{array}{c}{x}^2-7x+12=x^2-3x-4x+12\\ =x\left(x-3\right)-4\left(x-3\right)\\ =\left(x-4\right)\left(x-3\right)\end{array}$

Therefore, $x^3-2x^2-23x+60=\left(x+5\right)\left(x-4\right)\left(x-3\right)$ .