Question

In Problems $1-6$, solve each equation.

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

Short answer

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

Step-by-step solution

Step 1 Given equation is,

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

The given equation can be solved by assuming a value for $x$which is a solution of the given cubic equation in $x$.

Substitute $x=-1$in the given equation.

$$\begin{gathered} 2{\left(-1\right)}^3-3{\left(-1\right)}^2-8\left(-1\right)-3=0 \\ -2-3+8-3=0 \\ -5+5=0 \\ 0=0 \end{gathered}$$

Since$0=0$is a true statement, the value$x=-1$is a solution to the given equation.

Step 2 The given cubic equation is a product of x+1 and a quadratic equation in x.

$2x^3-3x^2-8x-3=\left(x+1\right)\left(a{x}^2+bx+c\right)$.

The quadratic equation can be obtained by dividing the cubic equation by $\left(x+1\right)$.

Step 3 The obtained equation can now be solved by using the formula.

$$\begin{gathered} x=\frac{5\pm \sqrt{{\left(-5\right)}^2-4\left(2\right)\left(-3\right)}}{2\left(2\right)} \\ =\frac{5\pm 7}{4} \\ =-\frac{1}{2},3 \end{gathered}$$

Hence the required solution set is,$\left\{-1,-\frac{1}{2},3\right\}$.