Chapter 8, Problem 23PT

To determine

To factor the given trinomial

Answer to Problem 23PT

  $(x-\frac{6}{10})(x+\frac{5}{10})$

Explanation of Solution

Given:

Trinomial: $10x^2-x-3$

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,

  $\Rightarrow 10x^2-x-3$

Substitute in the above formula for $a=10,b=-1,c=-3$

  $\begin{array}{l}\Rightarrow x=\frac{-(-1)\pm \sqrt{{(-1)}^2-4(10)(-3)}}{2(10)}\\ \Rightarrow x=\frac{1\pm \sqrt{1+120}}{20}\\ \Rightarrow x=\frac{1\pm \sqrt{121}}{20}\\ \Rightarrow x=\frac{1\pm 11}{20}\end{array}$

Hence the roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{1+11}{20}=\frac{12}{20}=\frac{6}{10}\\ \Rightarrow x_2=\frac{1-11}{20}=\frac{-10}{20}=\frac{-5}{10}\end{array}$

The trinomial becomes,

  $\begin{array}{c}10x^2-x-3=(x-\frac{6}{10})(x-(\frac{-5}{10}))\\ =(x-\frac{6}{10})(x+\frac{5}{10})\end{array}$

Conclusion:

Therefore, the factor of the given trinomial is $(x-\frac{6}{10})(x+\frac{5}{10}).$