Chapter 8, Problem 59SGR

To determine

To factorize the trinomial

Answer to Problem 59SGR

  $2(x-\frac{1}{2})(x+\frac{7}{3})$

Explanation of Solution

Given:

Trinomial: $12x^2+22x-14$

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 trinomial $12x^2+22x-14$ ,

Find a common number in the given expression.

  $\begin{array}{l}\Rightarrow 12x^2+22x-14\\ \Rightarrow 2(6x^2+11x-7)\end{array}$

Substituting in the above formula where $a=6,b=11,c=-14$

  $\begin{array}{c}\Rightarrow x=\frac{-11\pm \sqrt{{11}^2-4(6)(-7}}{2(6)}\\ \Rightarrow x=\frac{-11\pm \sqrt{121+168}}{12}\\ \Rightarrow x=\frac{-11\pm \sqrt{289}}{12}\\ \Rightarrow x=\frac{-11\pm 17}{12}\end{array}$

Therefore,

  $\begin{array}{l}\Rightarrow x_1=\frac{-11+17}{12}=\frac{6}{12}=\frac{1}{2}\\ \Rightarrow x_2=\frac{-11-17}{12}=\frac{-28}{12}=\frac{-7}{3}\end{array}$

Thus, the expression becomes,

  $\begin{array}{l}\Rightarrow 2(6x^2+11x-7)\\ \Rightarrow 2(x-\frac{1}{2})(x-(-\frac{7}{3}))\\ \Rightarrow 2(x-\frac{1}{2})(x+\frac{7}{3})\end{array}$

Conclusion:

Therefore, the factors of the given trinomial are $2(x-\frac{1}{2})(x+\frac{7}{3}).$