Chapter 8, Problem 21PT

To determine

To factor the given trinomial

Answer to Problem 21PT

  $(x+1)(x+6)$

Explanation of Solution

Given:

Trinomial: $x^2+7x+6$

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 trinomial,

  $\Rightarrow x^2+7x+6$

Substitute in the above formula for $a=1,b=7,c=6$

  $\begin{array}{l}\Rightarrow x=\frac{-7\pm \sqrt{7^2-4(1)(6)}}{2(1)}\\ \Rightarrow x=\frac{-7\pm \sqrt{49-24}}{2}\\ \Rightarrow x=\frac{-7\pm \sqrt{25}}{2}\\ \Rightarrow x=\frac{-7\pm 5}{2}\end{array}$

Hence the roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{-7+5}{2}=\frac{-2}{2}=-1\\ \Rightarrow x_2=\frac{-7-5}{2}=\frac{-12}{2}=-6\end{array}$

The trinomial becomes,

  $\begin{array}{c}{x}^2+7x+6=(x-(-1))(x-(-6))\\ =(x+1)(x+6)\end{array}$

Conclusion:

Therefore, the factor of the given trinomial is $(x+1)(x+6).$