Chapter 8, Problem 52SGR

To determine

To factorize the given polynomial

Answer to Problem 52SGR

  $(x-3)(x+6)$

Explanation of Solution

Given:

Polynomial: $x^2+3x-18$

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,

  $\Rightarrow x^2+3x-18$

Substituting in the above formula where $a=1,b=3,c=-18$

  $\begin{array}{l}\Rightarrow x=\frac{-3\pm \sqrt{3^2-4(1)(-18)}}{2(1)}\\ \Rightarrow x=\frac{-3\pm \sqrt{9+72}}{2}\\ \Rightarrow x=\frac{-3\pm \sqrt{81}}{2}\\ \Rightarrow x=\frac{-3\pm 9}{2}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{-3+9}{2}=\frac{6}{2}=3\\ \Rightarrow x_2=\frac{-3-9}{2}=\frac{-12}{2}=-6\end{array}$

The roots of the quadratic polynomial $x^2+3x-18=(x-3)(x+6).$

Conclusion:

Therefore, the factors of the polynomial are $(x-3)(x+6).$