Chapter 8, Problem 50SGR

To determine

To factorize the given polynomial

Answer to Problem 50SGR

  $(x+5)(x+4)$

Explanation of Solution

Given:

Polynomial: $x^2+9x+20$

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,

  $\Rightarrow x^2+9x+20$

Substituting in the above formula where $a=1,b=9,c=20$

  $\begin{array}{l}\Rightarrow x=\frac{-9\pm \sqrt{9^2-4(1)(20)}}{2(1)}\\ \Rightarrow x=\frac{-9\pm \sqrt{81-80}}{2}\\ \Rightarrow x=\frac{-9\pm \sqrt{1}}{2}\\ \Rightarrow x=\frac{-9\pm 1}{2}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{-9-1}{2}=\frac{-10}{2}=-5\\ \Rightarrow x_2=\frac{-9+1}{2}=\frac{-8}{2}=-4\end{array}$

The roots of the quadratic polynomial $x^2-8x+15=(x+5)(x+4).$

Conclusion:

Therefore, the factors of the polynomial are $(x+5)(x+4).$