Chapter 8, Problem 24PT

To determine

To factor the given trinomial

Answer to Problem 24PT

  $(x-\frac{1}{5})(x+\frac{2}{3})$

Explanation of Solution

Given:

Trinomial: $15x^2+7x-2$

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,

  $\Rightarrow 15x^2+7x-2$

Substitute in the above formula for $a=15,b=7,c=-2$

  $\begin{array}{l}\Rightarrow x=\frac{-7\pm \sqrt{7^2-4(15)(-2)}}{2(15)}\\ \Rightarrow x=\frac{-7\pm \sqrt{49+120}}{30}\\ \Rightarrow x=\frac{-7\pm \sqrt{169}}{30}\\ \Rightarrow x=\frac{-7\pm 13}{30}\end{array}$

Hence the roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{-7+13}{30}=\frac{6}{30}=\frac{1}{5}\\ \Rightarrow x_2=\frac{-7-13}{30}=\frac{-20}{30}=\frac{-2}{3}\end{array}$

The trinomial becomes,

  $\begin{array}{c}15x^2+7x-2=(x-\frac{1}{5})(x-(\frac{-2}{3}))\\ =(x-\frac{1}{5})(x+\frac{2}{3})\end{array}$

Conclusion:

Therefore, the factor of the given trinomial is $(x-\frac{1}{5})(x+\frac{2}{3}).$