Chapter 2.3, Problem 51E

To determine

To show: $x=\sqrt{3}$ is a solution of the polynomial $x^3+2x^2-3x-6=0$ and using synthetic

division factor the polynomial completely.

Answer to Problem 51E

  $\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\left(x+2\right)$

Solutions: $\sqrt{3},-\sqrt{3},-2$

Explanation of Solution

Calculation:

First, find the remainder obtained on dividing $p\left(x\right)=x^3+2x^2-3x-6$ by $x-\sqrt{3}$ using synthetic division and if the remainder is 0, then $x=\sqrt{3}$ is a solution of the given polynomial.

  $\begin{array}{l}\text{ 1 2 }-3-6\\ \sqrt{3}3+2\sqrt{3}\text{ 6}\\ \sqrt{3}\overline{)\text{1 2+}\sqrt{3}\text{ 2}\sqrt{3}\text{ 0}}=R\end{array}$

Since, the remainder obtained is 0.

So, $x=\sqrt{3}$ is a solution of the given polynomial.

Factor theorem:

“For a polynomial $P\left(x\right)$ r is a zero of a polynomial if and only if $x-r$ is a factor of $P\left(x\right)$ ”.

Since $x=\sqrt{3}$ is a zero of $P\left(x\right)$ , so by factor theorem $x-\sqrt{3}$ is a factor of $P\left(x\right)$ .

Also, from the above division, the quotient obtained on division is $x^2+\left(2+\sqrt{3}\right)x+2\sqrt{3}$ .

So, the polynomial can be factored as shown below:

  $\begin{array}{c}{x}^3+2x^2-3x-6=\left(x^2+\left(2+\sqrt{3}\right)x+2\sqrt{3}\right)\left(x-\sqrt{3}\right)\\ =\left(x^2+2x+\sqrt{3}x+2\sqrt{3}\right)\left(x-\sqrt{3}\right)\\ =\left(x\left(x+2\right)+\sqrt{3}\left(x+2\right)\right)\left(x-\sqrt{3}\right)\\ =\left(x+2\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\end{array}$

Now, find the zeros of the polynomial equation $x^3+2x^2-3x-6=0$

Using above factorization, $x^3+2x^2-3x-6=0$

can be written as:

  $\begin{array}{c}{x}^3+2x^2-3x-6=0\\ \left(x+2\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\end{array}$

It is known that if $ab=0$ then either $a=0$ or $b=0$ (or both $a=0$ and $b=0$ are zero).

This gives that,

  $\begin{array}{c}x+2=0\\ x=-2\end{array}$ Or $\begin{array}{c}x+\sqrt{3}=0\\ x=-\sqrt{3}\end{array}$ Or $\begin{array}{c}x-\sqrt{3}=0\\ x=\sqrt{3}\end{array}$

Therefore, the zeros of given polynomial equation are: $-2,\sqrt{-3},\sqrt{3}$