Chapter 4, Problem 62RE

To determine

To calculate: Solve given equation in real number system.

Answer to Problem 62RE

Values of $x$ are $2,-3,-1/2,2$ .

Explanation of Solution

Given information:

  $2x^4+7x^3-5x^2-28x-12=0$

Calculation:

As it is given that,

  $2x^4+7x^3-5x^2-28x-12=0$

By using rational root theorem,

Divider of $12$ are $1,2,3,4,6,12$

After putting $x=-2$ we get,

  $\begin{array}{l}2{\left(-2\right)}^4+7{\left(-2\right)}^3-5{\left(-2\right)}^2-28\left(-2\right)-12\\ =32-56-20+56-12\\ =0\end{array}$

  $\left(x+2\right)$ is root of given equation.

Divide given equation by $x+2$ ,

  $\frac{2x^4+7x^3-5x^2-28x-12}{x+2}=2x^3+3x^2+x-6$

On factorization of $2x^3+3x^2+x-6$

  $\begin{array}{l}=\left(x-2\right)\left(x+3\right)\left(2x+1\right)\\ 2x^4+7x^3-5x^2-28x-12=\left(x-2\right)\left(x+3\right)\left(2x+1\right)\left(x+2\right)\end{array}$

So, roots are,

  $\begin{array}{l}x-2=0\\ x=2\\ 2x+1=0\\ x=\frac{-1}{2}\\ x+3=0\\ x=-3\\ x+2=0\\ x=-2\end{array}$

Values of $x$ are $2,-3,-1/2,2$ .