Chapter 2.8, Problem 9QR

Solution

(a)

To List:

The possible rational zeroes of the polynomial $2x^3+x^2-4x-3$ .

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

Given:

The polynomial. $2x^3+x^2-4x-3$ .

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 $2x^3+x^2-4x-3$ , all its rational zeroes are among the numbers obtained by the expression $\pm \frac{{\text{Factors of a}}_0=-3}{\text{Factors of }{a}_3=2}$ .

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

Conclusion:

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

(b)

To Calculate:

Factor the polynomial $2x^3+x^2-4x-3$ completely.

  ${\left(x+1\right)}^2(2x-3)$ .

Given:

The polynomial. $2x^3+x^2-4x-3$ .

Known from previous part:

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

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

By synthetic division, check if the possible zero $-1$ gives that $x-\left(-1\right)$ is a factor of $2x^3+x^2-4x-3$ .

  $\begin{array}{cc}\overline{)-1}& \underset{¯}{\begin{array}{rrrr}\hfill 2& \hfill 1& \hfill -4& \hfill -3\\ \hfill & \hfill -2& \hfill 1& \hfill 3\end{array}}\\ & \begin{array}{rrrr}\hfill 2& \hfill -1& \hfill -3& \hfill 0\end{array}\end{array}$

The last result is zero. Thus $x-\left(-1\right)=x+1$ is a factor of $2x^3+x^2-4x-3$ . In fact, $2x^3+x^2-4x-3=\left(x+1\right)\left(2x^2-x-3\right)$ , 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:

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

Conclusion:

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