Chapter 8, Problem 61SGR

To determine

To factorize the given polynomial

Answer to Problem 61SGR

  $3(x+5)(x-3)$

Explanation of Solution

Given:

Polynomial: $3x^2-6x-45$

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,

  $\begin{array}{l}\Rightarrow 3x^2-6x-45\\ \Rightarrow 3(x^2-2x-15)\end{array}$

Substituting in the above formula where $a=1,b=-2,c=-15$

  $\begin{array}{l}\Rightarrow x=\frac{-2\pm \sqrt{{(-2)}^2-4(1)(-15)}}{2(1)}\\ \Rightarrow x=\frac{-2\pm \sqrt{4+60}}{2}\\ \Rightarrow x=\frac{-2\pm \sqrt{64}}{2}\\ \Rightarrow x=\frac{-2\pm 8}{2}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{-2-8}{2}=\frac{-10}{2}=-5\\ \Rightarrow x_2=\frac{-2+8}{2}=\frac{6}{2}=3\end{array}$

The roots of the quadratic polynomial $3x^2-6x-45=3(x+5)(x-3).$

Conclusion:

Therefore, the factors of the polynomial are $3(x+5)(x-3).$