Chapter 8, Problem 66SGR

To determine

To factorize the given polynomial

Answer to Problem 66SGR

  $(x+\frac{1}{2})(x-\frac{5}{3})$

Explanation of Solution

Given:

Polynomial: $6x^2-7x-5$

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,

  $\Rightarrow 6x^2-7x-5$

Substituting in the above formula where $a=6,b=-7,c=-5$

  $\begin{array}{l}\Rightarrow x=\frac{-(-7)\pm \sqrt{{(-7)}^2-4(6)(-5)}}{2(6)}\\ \Rightarrow x=\frac{7\pm \sqrt{49+120}}{12}\\ \Rightarrow x=\frac{7\pm \sqrt{169}}{12}\\ \Rightarrow x=\frac{7\pm 13}{12}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow x_1=\frac{7+13}{12}=\frac{20}{12}=\frac{5}{3}\\ \Rightarrow x_2=\frac{7-13}{12}=\frac{-6}{12}=\frac{-1}{2}\end{array}$

The roots of the quadratic polynomial $6x^2-7x-5=(x+\frac{1}{2})(x-\frac{5}{3}).$

Conclusion:

Therefore, the factors of the polynomial are $(x+\frac{1}{2})(x-\frac{5}{3}).$