Chapter 4, Problem 59RE

To determine

To calculate:

The solution set of the following function:

  $f\left(x\right)=2x^4+2x^3-11x^2+x-6$

Answer to Problem 59RE

The solution set of equation of $f\left(x\right)=2x^4+2x^3-11x^2+x-6$ is $\left\{2,-3\right\}$

Explanation of Solution

Given information:

  $f\left(x\right)=2x^4+2x^3-11x^2+x-6$

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:

  $f\left(x\right)=2x^4+2x^3-11x^2+x-6$

The degree of the given polynomial function is $4$ which shows that $f$ must have at most $4$ real zeroes.

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

  $\begin{array}{l}p=6\\ q=2\end{array}$

Factors of $p$ are $\pm 1,\pm 2$ and factors of $q$ is $\pm 1,\pm 2,\pm 3,\pm 6$ .

So the potential rational zeroes are defined as:

  $\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

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.

  $\begin{array}{cccccc}1& 2& 2& -11& 1& -6\\ & & 2& 4& -7& -6\\ & 2& 4& -7& -6& -12\end{array}$

Here, the remainder is $f\left(1\right)=-12$ . That’s why $1$ is not zero of $q_1$ .

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

  $\begin{array}{cccccc}-1& 2& 2& -11& 1& -6\\ & & -2& 0& 11& -12\\ & 2& 0& -11& 12& -18\end{array}$

Here, the remainder is $f\left(-1\right)=-18$ . That’s why $-1$ is not zero of $q_1$ .

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

  $\begin{array}{cccccc}2& 2& 2& -11& 1& -6\\ & & 4& 12& 2& 6\\ & 2& 6& 1& 3& 0\end{array}$

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

  $\begin{array}{l}f\left(x\right)=2x^4+2x^3-11x^2+x-6\\ f\left(x\right)=\left(x-2\right)\left(2x^3+6x^2+x+3\right)\end{array}$

The solution of the given polynomial equation $f\left(x\right)=2x^4+2x^3-11x^2+x-6$ are the zeroes of the $q_1\left(x\right)=2x^3+6x^2+x+3$ .

The potential rational zeroes of $q_1\left(x\right)=2x^3+6x^2+x+3$ :

  $\begin{array}{l}p=3\\ q=2\\ \frac{p}{q}=\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}\end{array}$

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

Now test for $1$ : :

  $\begin{array}{ccccc}1& 2& 6& 1& 3\\ & & 2& 8& 9\\ & 2& 8& 9& 12\end{array}$

Here, the remainder is $f\left(1\right)=12$ . That’s why $1$ is not zero of $q_1$ .

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

  $\begin{array}{ccccc}-1& 2& 6& 1& 3\\ & & -2& -4& 3\\ & 2& 4& -3& 6\end{array}$

Here, the remainder is $f\left(-1\right)=6$ . That’s why $-1$ is not zero of $q_1$ .

Now test for $3$ by using the synthetic division:

  $\begin{array}{ccccc}3& 2& 6& 1& 3\\ & & 6& 36& 111\\ & 2& 12& 37& 114\end{array}$

Here, the remainder is $f\left(3\right)=114$ . That’s why $-1$ is not zero of $q_1$ .

Now test for $-3$ by using the synthetic division:

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

Here, the remainder is $f\left(-3\right)=0$ . That’s why $-3$ is zero of $q_1$ and $x+3$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=2x^4+2x^3-11x^2+x-6\\ f\left(x\right)=\left(x-2\right)\left(2x^3+6x^2+x+3\right)\\ f\left(x\right)=\left(x-2\right)\left(x+3\right)\left(2x^2+1\right)\end{array}$

As the $2x^2+1=0$ has no real solution, $2$ and $-3$ are the only zeroes of $f$

Hence, the solution set of equation of the given polynomial is $\left\{2,-3\right\}$