Chapter 2, Problem 62RE

Solution

To write the function as a product of linear and irreducible quadratic factors of real coefficients

  $9x^3-3x^2-13x-1=\left(x+1\right)\left(9x^2-12x-1\right)$ .

Given:

The Function is: $9x^3-3x^2-13x-1$ .

Concept Used:

The factor theorem states that if $f(x)$ is a polynomial of degree $n\ge 1$ , and $a$ is any real number, then $\left(x-a\right)$ is a factor of $f(x)$ if $f\left(a\right)=0$ .

Calculation:

Consider, $f(x)=9x^3-3x^2-13x-1$ .

Now, put $x=-1$ in $f(x)$ .

  $\begin{array}{c}f(-1)=9{\left(-1\right)}^3-3{\left(-1\right)}^2-13\left(-1\right)-1\\ =-9-3+13-1\\ =13-13\\ =0\end{array}$

Thus, $\left(x+1\right)$ is a factor of $f(x)$ .

Now, divide $f(x)=9x^3-3x^2-13x-1$ by $\left(x+1\right)$ to find the other factors of $f(x)$ .

  

  $\begin{array}{l}9x^3+9x^2\\ \underset{¯}{--}\\ -12x^2-13x\\ -12x^2-12x\\ \underset{¯}{++}\\ -x-1\\ -x-1\\ \underset{¯}{++}\\ 0\end{array}$

Conclusion:

Therefore, function $9x^3-3x^2-13x-1$ can be expressed as:

  $9x^3-3x^2-13x-1=\left(x+1\right)\left(9x^2-12x-1\right)$ .