Chapter 2.8, Problem 10QR

Solution

(a)

To List:

The possible rational zeroes of the polynomial $3x^3-x^2-10x+8$ .

  $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3}\text{,}\pm \frac{2}{3}\text{,}\pm \frac{4}{3}\text{, and, }\pm \frac{8}{3}$.

Given:

The polynomial. $3x^3-x^2-10x+8$ .

Concepts Used:

The Rational Zeroes Theorem: A polynomial of degree $n$ , $a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_2{x}^2+a_{1}x+a_0$ , where $a_n,a_0\ne 0$ , has all its rational zeroes among the numbers obtained by the expression $\pm \frac{{\text{Factors of a}}_0}{\text{Factors of }{a}_n}$ .

Calculations:

For the given polynomial $3x^3-x^2-10x+8$ , all its rational zeroes are among the numbers obtained by the expression $\pm \frac{{\text{Factors of a}}_0=8}{\text{Factors of }{a}_3=3}$ .

Factors of $8$ are $1,2,4,8$ . Factors of $3$ are $1,3$ . The possible values of the expression $\pm \frac{{\text{Factors of a}}_0=8}{\text{Factors of }{a}_3=3}$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3}\text{,}\pm \frac{2}{3}\text{,}\pm \frac{4}{3}\text{, and, }\pm \frac{8}{3}$ .

Conclusion:

The possible rational zeroes of the polynomial $3x^3-x^2-10x+8$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3}\text{,}\pm \frac{2}{3}\text{,}\pm \frac{4}{3}\text{, and, }\pm \frac{8}{3}$ .

(b)

To Calculate:

Factor the polynomial $3x^3-x^2-10x+8$ completely.

  $\left(x+1\right)\left(x+2\right)\left(3x-4\right)$ .

Given:

The polynomial. $3x^3-x^2-10x+8$ .

Known from previous part:

The possible rational zeroes of the polynomial $3x^3-x^2-10x+8$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3}\text{,}\pm \frac{2}{3}\text{,}\pm \frac{4}{3}\text{, and, }\pm \frac{8}{3}$ .

Concepts Used:

Factor Theorem. If $a$ is a zero of polynomial $p\left(x\right)$ then $x-a$ is a factor of $p\left(x\right)$ .

Synthetic division of polynomials.

Splitting the linear term to factorize quadratic polynomials.

Calculations:

Check the possible rational zeroes $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3}\text{,}\pm \frac{2}{3}\text{,}\pm \frac{4}{3}\text{, and, }\pm \frac{8}{3}$ one by one to determine if one of them gives a factor of $3x^3-x^2-10x+8$ .

By synthetic division, check if the possible zero $-1$ gives that $x-\left(-1\right)$ is a factor of $3x^3-x^2-10x+8$ .

  $\begin{array}{cc}\overline{)-1}& \begin{array}{rrrr}\hfill 3& \hfill -1& \hfill -10& \hfill 8\\ \hfill & \hfill -3& \hfill 4& \hfill 6\\ \hfill 3& \hfill -4& \hfill -6& \hfill 14\end{array}\end{array}$

The last result is not zero. Thus $x-\left(-1\right)=x+1$ is not a factor of $3x^3-x^2-10x+8$ .

By synthetic division, check if the possible zero $1$ gives that $x-1$ is a factor of $3x^3-x^2-10x+8$ .

  $\begin{array}{rr}\hfill \overline{)1}& \hfill \begin{array}{rrrr}\hfill 3& \hfill -1& \hfill -10& \hfill 8\\ \hfill & \hfill 3& \hfill 2& \hfill -8\\ \hfill 3& \hfill 2& \hfill -8& \hfill 0\end{array}\end{array}$

The last result is zero. Thus $x-1$ is a factor of $3x^3-x^2-10x+8$ . In fact, $3x^3-x^2-10x+8=\left(x-1\right)\left(3x^2+2x-8\right)$ , where the coefficients of the quadratic polynomial are obtained from the result of synthetic division.

Factorize the polynomial further by splitting the linear term of the quadratic polynomial:

  $\begin{array}{c}3x^3-x^2-10x+8=\left(x-1\right)\left(3x^2+2x-8\right)\\ =\left(x-1\right)\left(3x^2+6x-4x-8\right)\\ =\left(x+1\right)\left(3x\left(x+2\right)-4\left(x+2\right)\right)\\ =\left(x+1\right)\left(x+2\right)\left(3x-4\right)\end{array}$

Conclusion:

The completely factorized expression for the polynomial $3x^3-x^2-10x+8$ is $\left(x+1\right)\left(x+2\right)\left(3x-4\right)$ .