Chapter 4, Problem 80RE

To determine

The complex zeroes and the factored form of the following functions is to be estimated.

  $f(x)=x^3-x^2-10x-8$

Answer to Problem 80RE

The solution of $f(x)=x^3-x^2-10x-8$ are $-1,-2,4$ , does not contain any complex root.

Explanation of Solution

Given:

  $f(x)=x^3-x^2-10x-8$

Calculation:

The given function is:

  $f(x)=x^3-x^2-10x-8$

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.

  $\begin{array}{ccccc}1& 1& -1& -10& -8\\ & & 1& 0& -10\\ & 1& 0& -10& -18\end{array}$

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

Check for $-1$

  $\begin{array}{ccccc}-1& 1& -1& -10& -8\\ & & -1& 2& 8\\ & 1& -2& -8& 0\end{array}$

Here, the remainder is $f\left(-1\right)=0$ . That’s why $-1$ is zero of is $f$ .

So, it follows:

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

So, $-1,-2,4$ is zeroes of function. There is no complex root is available for the given function.

Conclusion:

Hence, given function has no complex root.