Chapter 14.8, Problem 23HP

To determine

To calculate: The reason why the trinomial $x^2+x+5$ cannot be factored.

Answer to Problem 23HP

The trinomial $x^2+x+5$ cannot be factored.

Explanation of Solution

Given Information:

The trinomial expressionis $x^2+x+5$ .

Formula Used:

Roots of a quadratic equation $a{x}^2+bx+c$ .

  $f=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

The given trinomial expression is $x^2+x+5$ .

Without fraction it is not possible to split $x$ into two parts whose product is equal to the product of $x^2$ and 5.

Now, find the roots of the trinomial $x^2+x+5$ . Use the formula $f=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Here $a=1,b=1,c=5$ .

Substitute 1 for a , 1 for b and 5 for cin the formula $f=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

  $\begin{array}{c}f=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-1\pm \sqrt{1^2-4\times 1\times 5}}{2\times 1}\\ =\frac{-1\pm \sqrt{-19}}{2}\end{array}$

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

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

  $\begin{array}{c}f=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-1\pm \sqrt{-19}}{2}\\ =\frac{-1\pm i\sqrt{19}}{2}\end{array}$

Hence, the trinomial $x^2+x+5$ cannot be factored.