Chapter 8.5, Problem 31WE

To determine

To find:

The solution of given equation with the help of given two indicated roots

Answer to Problem 31WE

The solution set of the given equation is $\left\{-1-\sqrt{2}i,-1+\sqrt{2}i,1,-\frac{2}{3}\right\}$

Explanation of Solution

Given information:

The equation is $3x^4+5x^3-7x^2-3x+2$

The roots of the equation are $1,-\frac{2}{3}$

Formula used:

If $a{x}^2+bx+c=0$ and $a\ne 0$ then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

To find the depressed equation, divide $P\left(x\right)=3x^4+5x^3-7x^2-3x+2$ by $x-\left(-\frac{2}{3}\right)\text{ or }x+\frac{2}{3}$

  $\begin{array}{l}-\frac{2}{3}\overline{)\begin{array}{l}\text{3 }5-7-3\text{ 2}\\ -2-2\text{ 6 }-2\end{array}}\\ \text{ 3 3 }-9\text{ 3 }\overline{)\text{0}}\end{array}$

So, $3x^4+5x^3-7x^2-3x+2=\left(x+\frac{2}{3}\right)\left(3x^3+3x^2-9x+3\right)$ and the depressed equation is $3x^3+3x^2-9x+3$

Again, to find the depressed equation, divide $P\left(x\right)=3x^3+3x^2-9x+3$ by $x-1$

  $\begin{array}{l}1\overline{)\begin{array}{l}\text{3 3 }-9\text{ 3}\\ \text{ 3 6 }-3\end{array}}\\ \text{ 3 6 }-3\overline{)\text{0}}\end{array}$

So, $3x^3+3x^2-9x+3=\left(x-1\right)\left(3x^2+6x-3\right)$ and the depressed equation is $3x^2+6x-3$

To solve the depressed equation, use the quadratic formula

  $\begin{array}{l}x=\frac{-6\pm \sqrt{{\left(6\right)}^2-4\cdot 3\cdot \left(-3\right)}}{2\cdot 3}\\ x=\frac{-6\pm \sqrt{36+36}}{6}\\ x=\frac{-3\pm 6\sqrt{2}}{6}\\ x=-1\pm \sqrt{2}i\\ x=-1+\sqrt{2}i\text{ and }-1-\sqrt{2}i\end{array}$

Therefore, the solution set of the given equation is $\left\{-1-\sqrt{2}i,-1+\sqrt{2}i,1,-\frac{2}{3}\right\}$