Chapter 8, Problem 65SGR

To determine

To factor the given polynomial.

Answer to Problem 65SGR

  $(x-5)(x+\frac{2}{3})$

Explanation of Solution

Given:

Polynomial: $3x^2-13x-10$

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,

  $\Rightarrow 3x^2-13x-10$

Substituting in the above formula where $a=3,b=-13,c=-10$

  $\begin{array}{l}\Rightarrow x=\frac{-(-13)\pm \sqrt{{(-13)}^2-4(3)(-10)}}{2(3)}\\ \Rightarrow x=\frac{13\pm \sqrt{169+120}}{6}\\ \Rightarrow x=\frac{13\pm \sqrt{289}}{6}\\ \Rightarrow x=\frac{13\pm 17}{6}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{13+17}{6}=\frac{30}{6}=5\\ \Rightarrow x_2=\frac{13-17}{6}=\frac{-4}{6}=\frac{-2}{3}\end{array}$

The roots of the quadratic polynomial $3x^2-13x-10=(x-5)(x+\frac{2}{3}).$

Conclusion:

Therefore, the factors of the polynomial are $(x-5)(x+\frac{2}{3}).$