Chapter 4, Problem 53RE

To determine

To calculate:

All the real zeroes of the following functions of polynomial with the help of Rational zeroes theorem and the factor $f$ over the real zeroes:

  $f\left(x\right)=x^3-3x^2-6x+8$

Answer to Problem 53RE

The real zeroes of the given polynomial function are $f\left(x\right)=\left(x-1\right)\left(x+2\right)\left(x-4\right)$ are $1,-2,4$

Explanation of Solution

Given information:

  $f\left(x\right)=x^3-3x^2-6x+8$

Formula Used:

The potential rational zeroes is given by the following expression:

  $x=\pm \frac{p}{q}$

Where,

Quotients of the factors of the last term: $p$

The factors of the leading coefficient: $q$

Calculation:

The given function is:

  $f\left(x\right)=x^3-3x^2-6x+8$

The degree of the given polynomial function is $3$ which shows that $f$ must have at most $3$ real zeroes.

Now, the values of $p,q$ are defined as:

  $\begin{array}{l}p=8\\ q=1\end{array}$

Factors of $p$ are $\pm 1,\pm 2,\pm 4,\pm 8$ and factors of $q$ is $\pm 1$ .

So the potential rational zeroes are defined as:

  $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8$

Now test the potential rational zeroes with the help of the synthetic division:

If the value of remainder is zero $f\left(1\right)=0$ then $1$ is zero of $f$ and $x-1$ is a factor.

  $\begin{array}{ccccc}1& 1& -3& -6& 8\\ & & 1& 2& -8\\ & 1& -2& -8& 0\end{array}$

Get the remainder is zero, hence the then $1$ is zero of $f$ . Factor the $f$ by using the bottom row of the synthetic division:

  $\begin{array}{l}f\left(x\right)=x^3-3x^2-6x+8\\ f\left(x\right)=\left(x-1\right)\left(x^2-2x-8\right)\end{array}$

The solution of the given polynomial equation $f\left(x\right)=x^3-3x^2-6x+8$ are the zeroes of the $q_1\left(x\right)=x^2-2x-8$ .

The potential rational zeroes of $q_1$ are $\pm 1,\pm 2,\pm 4,\pm 8$ .

Now test for $1$ by using the synthetic division:

  $\begin{array}{cccc}1& 1& -2& -8\\ & & 1& -1\\ & 1& -1& -9\end{array}$

Here, the remainder is $f\left(1\right)=9$ . That’s why $1$ is not zero of $q_1$ .

Now test for $-1$ by using the synthetic division:

  $\begin{array}{cccc}-1& 1& -2& -8\\ & & -1& 3\\ & 1& -3& -5\end{array}$

Here, the remainder is $f\left(-1\right)=-5$ . That’s why $-1$ is not zero of $q_1$ .

Now test for $2$ by using the synthetic division:

  $\begin{array}{cccc}2& 1& -2& -8\\ & & 2& 0\\ & 1& 0& -8\end{array}$

Here, the remainder is $f\left(2\right)=-8$ . That’s why $-2$ is not zero of $q_1$ .

Now test for $-2$ by using the synthetic division:

  $\begin{array}{cccc}-2& 1& -2& -8\\ & & -2& 8\\ & 1& -4& 0\end{array}$

Here, the remainder is $f\left(-2\right)=0$ . That’s why $-2$ is zero of $q_1$ and $x-\left(-2\right)=x+2$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=x^3-3x^2-6x+8\\ f\left(x\right)=\left(x-1\right)\left(x^2-2x-8\right)\\ f\left(x\right)=\left(x-1\right)\left(x+2\right)\left(x-4\right)\end{array}$

Since $x-4$ is a factor of $f$ , $4$ is zero of $f$ . The real zeroes of $f$ are $1,-2,4$ . When factored, $f$ is

  $\begin{array}{l}f\left(x\right)=x^3-3x^2-6x+8\\ f\left(x\right)=\left(x-1\right)\left(x+2\right)\left(x-4\right)\end{array}$

Hence, real zeroes are $1,-2,4$ .