Chapter 14.8, Problem 12PPS

To determine

To calculate: The factors of the trinomial $a^2+3a+5$ .

Answer to Problem 12PPS

The factors of the trinomial $a^2+3a+5$ are $\frac{-3\pm i\sqrt{11}}{2}$ .

Explanation of Solution

Given Information:

The trinomial expression is $a^2+3a+5$ .

Formula Used:

Roots of a quadratic equation $p{x}^2+qx+r$ .

  $f=\frac{-q\pm \sqrt{q^2-4pr}}{2p}$ .

Calculation:

The given trinomial is $a^2+3a+5$ .

Find the factors of the trinomial $a^2+3a+5$ .

In this expression it is not possible to break $3a$ into two parts whose product is equal to the product of $a^2$ and 5.

Thus, factors cannot be determined without fraction.

Find the roots of the trinomial $a^2+3a+5$ . Use the formula $f=\frac{-q\pm \sqrt{q^2-4pr}}{2p}$ .

Here $p=1,q=3,r=5$ .

Substitute 1 for p , 3 for q and 5 for r in the formula $f=\frac{-q\pm \sqrt{q^2-4pr}}{2p}$ .

  $\begin{array}{c}f=\frac{-q\pm \sqrt{q^2-4pr}}{2p}\\ =\frac{-3\pm \sqrt{3^2-4\times 1\times 5}}{2\times 1}\\ =\frac{-3\pm \sqrt{-11}}{2}\end{array}$

Since the roots of the trinomial $a^2+3a+5$ are imaginary. Thus, factors are also imaginary.

Use $i=\sqrt{-1}$ .

  $\begin{array}{c}f=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-3\pm \sqrt{-11}}{2}\\ =\frac{-3\pm i\sqrt{11}}{2}\end{array}$

Hence, the factors of the trinomial $a^2+3a+5$ are $\frac{-3\pm i\sqrt{11}}{2}$ .