Chapter 8, Problem 62SGR

To determine

To factorize the given polynomial

Answer to Problem 62SGR

  $(a+8)(a-\frac{3}{2})$

Explanation of Solution

Given:

Polynomial: $2a^2+13a-24$

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,

  $\Rightarrow 2a^2+13a-24$

Substituting in the above formula where $a=2,b=13,c=-24$

  $\begin{array}{l}\Rightarrow a=\frac{-13\pm \sqrt{{(13)}^2-4(2)(-24)}}{2(2)}\\ \Rightarrow a=\frac{-13\pm \sqrt{169+192}}{4}\\ \Rightarrow a=\frac{-13\pm \sqrt{361}}{4}\\ \Rightarrow a=\frac{-13\pm 19}{4}\end{array}$

The roots are,

  $\begin{array}{l}\Rightarrow a_1=\frac{-13-19}{4}=\frac{-32}{4}=-8\\ \Rightarrow a_2=\frac{-13+19}{4}=\frac{6}{4}=\frac{3}{2}\end{array}$

The roots of the quadratic polynomial $2a^2+13a-24=(a+8)(a-\frac{3}{2}).$

Conclusion:

Therefore, the factors of the polynomial are $(a+8)(a-\frac{3}{2}).$