Chapter 4.4, Problem 49E

To determine

To show: that $x+7$ is a factor of the given polynomial and then factorize completely.

Answer to Problem 49E

The factorized form is $x^3-37x+84=\left(x+7\right)\left(x-3\right)\left(x-4\right)$ .

Explanation of Solution

Given information:

A polynomial is given as

  $r\left(x\right)=x^3-37x+84$

Concept used:

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

Calculation:

Consider the given polynomial.

  $r\left(x\right)=x^3-37x+84$

If $x+7$ is a factor of the given polynomial, then $r\left(-7\right)=0$ .

Now, check whether $r\left(-7\right)=0$ or not.

  $\begin{array}{c}r\left(-7\right)={\left(-7\right)}^3-37\left(-7\right)+84\\ =-343+259+84\\ =-343+343\\ =0\end{array}$

Therefore, $x+7$ is a factor of the given polynomial $r\left(x\right)=x^3-37x+84$ .

Now, factorize the polynomial as shown:

  $\begin{array}{c}r\left(x\right)=x^3-37x+84\\ =x^3+7x^2-7x^2-49x+12x+84\\ =x^2\left(x+7\right)-7x\left(x+7\right)+12\left(x+7\right)\\ =\left(x+7\right)\left(x^2-7x+12\right)\end{array}$

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

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

Therefore, $x^3-37x+84=\left(x+7\right)\left(x-3\right)\left(x-4\right)$ .