Chapter 8, Problem 18PT

To determine

To factor the given polynomial

Answer to Problem 18PT

  $(x+1)(x-2)$

Explanation of Solution

Given:

Polynomial: $x^2-x-2$

Formula used:

Roots of the polynomial of the form $a{x}^2+bx+c=a(x-x_1)(x-x_2)$ are

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the given polynomial,

  $\Rightarrow x^2-x-2$

Applying the above formula for $a=1,b=-1,c=-2$

  $\begin{array}{l}\Rightarrow x=\frac{-(-1)\pm \sqrt{{(-1)}^2-4(1)(-2)}}{2(1)}\\ \Rightarrow x=\frac{1\pm \sqrt{1+8}}{2}\\ \Rightarrow x=\frac{1\pm \sqrt{9}}{2}\\ \Rightarrow x=\frac{1\pm 3}{2}\end{array}$

Therefore, the roots are

  $\begin{array}{l}{x}_1=\frac{1+3}{2}=\frac{4}{2}=2\\ x_2=\frac{1-3}{2}=\frac{-2}{2}=-1\end{array}$

Therefore, the polynomial can be factorised as,

  $\begin{array}{l}\Rightarrow (x-(-1))(x-2)\\ \Rightarrow (x+1)(x-2)\end{array}$

Conclusion:

Therefore, the polynomial is factored as $(x+1)(x-2).$