Chapter 8.5, Problem 29WE

To determine

To find:

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

Answer to Problem 29WE

The solution of the given equation is $x=\pm 2$

Explanation of Solution

Given information:

The equation is $x^4-3x^3-8x^2+12x+16=0$

The root of the equation is $-1,4$

Calculation:

To find the depressed equation, divide $P\left(x\right)=x^4-3x^3-8x^2+12x+16$ by $x-\left(-1\right)\text{ or }x+1$

  $\begin{array}{l}-1\overline{)\begin{array}{l}\text{1 }-3-8\text{ 12 16}\\ -1\text{ 4 4 }-16\end{array}}\\ \text{ 1 }-4-4\text{ 16 }\overline{)\text{0}}\end{array}$

So, $x^4-3x^3-8x^2+12x+16=\left(x+1\right)\left(x^3-4x^2-4x+16\right)$ and the depressed equation is $x^3-4x^2-4x+16$

Again, to find the depressed equation divide, $P\left(x\right)=x^3-4x^2-4x+16$ by $x-\left(4\right)\text{ or }x-4$

  $\begin{array}{l}4\overline{)\begin{array}{l}\text{1 }-4-4\text{ 16}\\ \text{ 4 0 }-16\end{array}}\\ \text{ 1 0 }-4\overline{)\text{0}}\end{array}$

So, the depressed equation is $x^2-4$

This equation is in the form $a^2-b^2$ , hence it can be factored as $\left(a+b\right)\left(a-b\right)$

  $x^2-4=\left(x+2\right)\left(x-2\right)$

To find the remaining factor set the above factor equal to zero and solve for $x$

  $\begin{array}{l}{x}^2=4\\ x=\pm \sqrt{4}\\ x=\pm 2\end{array}$

Therefore, the solution of the given equation is $x=\pm 2$