Question

How do you factor $5{{n}^{2}}+10n+20$?

Answer 1

Hint: We are given a quadratic equation which has to be solved by the method of factoring the equation. We shall first calculate the discriminant of the given quadratic equation and find out whether it is greater than zero or less than zero. Since, in this case the discriminant would be less than zero, thus we shall find the roots by the formula $x=\dfrac{-b\pm \sqrt{D}}{2a}$.

Complete step by step solution:
There are four methods for solving quadratic equations, namely, factoring method, completing the square method, taking square root method and the last method is solving using the various properties of polynomials.
However, we will use the method of factoring the quadratic equation which makes our calculations simpler.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $5{{n}^{2}}+10n+20$,
$\Rightarrow 5{{n}^{2}}+10n+20=0$
The discriminant of any quadratic equation are given as:
$D={{b}^{2}}-4ac$
Calculating the discriminant of the quadratic equation, we get
$\Rightarrow D={{\left( 10 \right)}^{2}}-4\left( 5 \right)\left( 20 \right)$
$\Rightarrow D=100-400$
$\Rightarrow D=-300$
Since $D<0$, thus we shall use $x=\dfrac{-b\pm \sqrt{D}}{2a}$ for finding the roots.
The roots are given as:
$n=\dfrac{-10\pm \sqrt{-300}}{2\left( 5 \right)}$
We know that $\sqrt{-1}=\iota $, thus substituting this iota, we get
$\Rightarrow n=\dfrac{-10\pm 10\sqrt{3}\iota }{10}$
$\Rightarrow n=\dfrac{-1\pm \sqrt{3}\iota }{1}$
$\Rightarrow n=-1\pm \sqrt{3}\iota $
$\Rightarrow n=-1+\iota \sqrt{3}$ and $n=-1-\iota \sqrt{3}$
Therefore, the roots of the equation are $n=-1+\iota \sqrt{3},-1-\iota \sqrt{3}$

Note: We must always remember that if the discriminant of the quadratic equation is less than zero, then that quadratic equation can never be factorized and it would always have unreal and complex roots. The unreal and complex roots always occur in pairs where both the roots are conjugates.