Chapter 2.3, Problem 51E

To determine

Determine the solutions of the polynomial equation $x^3+2x^2-3x-6=0$ . Also, verify whether $x=\sqrt{3}$ is a solution to given polynomial equation using synthetic division method.

Answer to Problem 51E

  $-2,-\sqrt{3}\text{,}\sqrt{3}$

Explanation of Solution

Given:

Polynomial equation: $x^3+2x^2-3x-6=0$

Value of x: $\sqrt{3}$

Calculation:

1. Verifying whether $x=\sqrt{3}$ is a solution to given polynomial equation or not by using synthetic division method.

Step 1: Writing coefficients of polynomial equation on the right side and value of x on the left side.

  $\begin{array}{ccccc}\sqrt{3}& 1& 2& -3& -6\\ & & & & \\ & & & & \end{array}$

Step 2: Now, adding terms in columns and multiplying the results by $\sqrt{3}$ .

  $\begin{array}{ccccc}\sqrt{3}& 1& 2& -3& -6\\ & & \sqrt{3}& 2\sqrt{3}+3& 6\\ & 1& 2+\sqrt{3}& 2\sqrt{3}& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, the remainder is zero, it means that x = $\sqrt{3}$ is the solution of given polynomial equation.

Also, quotient is $x^2+(2+\sqrt{3})x+2\sqrt{3}$ .

Hence, $x^3+2x^2-3x-6=0\iff \left(x-\sqrt{3}\right)(x^2+(2+\sqrt{3})x+2\sqrt{3})=0$

2. Finding roots of polynomial equation $x^2+(2+\sqrt{3})x+2\sqrt{3}=0.$

  $\begin{array}{l}\Rightarrow x^2+(2+\sqrt{3})x+2\sqrt{3}=0\\ \Rightarrow x^2+2x+\sqrt{3}x+2\sqrt{3}=0\\ \Rightarrow x\left(x+2\right)+\sqrt{3}\left(x+2\right)=0\\ \Rightarrow \left(x+2\right)\left(x+\sqrt{3}\right)=0\\ \Rightarrow x=-2,-\sqrt{3}\end{array}$

Conclusion:

Therefore, the solutions of given polynomial equation are $x=-2,-\sqrt{3}\text{and}\sqrt{3}.$