Chapter 8, Problem 57SGR

To determine

To factorize the given polynomial

Answer to Problem 57SGR

  $(x+1)(x+10)$

Explanation of Solution

Given:

Polynomial: $x^2+11x+10$

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,

  $\Rightarrow x^2+11x+10$

Substituting in the above formula where $a=1,b=11,c=10$

  $\begin{array}{l}\Rightarrow x=\frac{-11\pm \sqrt{{11}^2-4(1)(10)}}{2(1)}\\ \Rightarrow x=\frac{-11\pm \sqrt{121-40}}{2}\\ \Rightarrow x=\frac{-11\pm \sqrt{81}}{2}\\ \Rightarrow x=\frac{-11\pm 9}{2}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{-11+9}{2}=\frac{-2}{2}=-1\\ \Rightarrow x_2=\frac{-11-9}{2}=\frac{-20}{2}=-10\end{array}$

The roots of the quadratic polynomial $x^2+11x+10=(x+1)(x+10).$

Conclusion:

Therefore, the factors of the polynomial are $(x+1)(x+10).$