Chapter 11, Problem 3CR

To determine

The solutions of the equation $2x^3-3x^2-8x-3=0$.

Answer to Problem 3CR

Solution:

The solutions of the equation is $2x^3-3x^2-8x-3=0$ are $x=3,x=-\frac{1}{2},x=-1$.

Explanation of Solution

Given information:

The equation $2x^3-3x^2-8x-3=0$.

Explanation:

Consider the equation $2x^3-3x^2-8x-3=0$.

Find one zero of the cubic equation by graphing the function.

Steps for graphing an equation using a graphing utility:

Step 1: Press ‘ON’ key.

Step 2: Press [Y=] key. From the $Y_1=$ screen, enter the expression after prompt $2x^3-3x^2-8x-3$.

Step 3: Press $[\text{GRAPH}]$ key.

The window shows the graph of the function $y=2x^3-3x^2-8x-3$ as,

From the graph, it is observed that, $x=3$ is zero of $y=2x^3-3x^2-8x-3$.

To find the quadratic equation, use synthetic division.

$\begin{array}{ccccc}3& 2& -3& -8& -3\\ & & 6& 9& 3\\ & 2& 3& 1& 0\end{array}$

By using the synthetic division, $2x^3-3x^2-8x-3=0$ can be written as,

$\Rightarrow \left(x-3\right)\left(2x^2+3x+1\right)=0$

$\Rightarrow \left(x-3\right)\left(2x^2+2x+x+1\right)=0$

Now find the factors of the quadratic term,

$\Rightarrow \left(x-3\right)\left[\left(2x^2+2x\right)+\left(x+1\right)\right]=0$

By factoring out the common terms,

$\Rightarrow \left(x-3\right)\left[2x\left(x+1\right)+1\left(x+1\right)\right]=0$

$\Rightarrow \left(x-3\right)\left(2x+1\right)\left(x+1\right)=0$

By using the zero product property,

$\Rightarrow \left(x-3\right)=0,\left(2x+1\right)=0,\left(x+1\right)=0$

By solving each of these equations,

$\Rightarrow x=3,x=-\frac{1}{2},x=-1$

The solutions of the equation $2x^3-3x^2-8x-3=0$ are $x=3,x=-\frac{1}{2},x=-1$.