Question

How do you factor completely $5{{x}^{2}}+6x-8$

Answer 1

Hint: Now we are given a quadratic equation in x. We know that for any quadratic equation we can find the roots with the help of formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence we will use this formula to first find the roots of the equation. Now we know that if $\alpha $ and $\beta $ are the roots of the equation then we have $\left( x-\alpha \right)$ and $\left( x-\beta \right)$ are the factors of the given equation. Hence we can easily find the factors once, we will have the roots of the equation.

Complete step-by-step solution:
Now consider the given expression $5{{x}^{2}}+6x-8$.
Now we know that the given expression is quadratic in one variable of the form $a{{x}^{2}}+bx+c$.
Now we know that for any quadratic equation of the form $a{{x}^{2}}+bx+c=0$ the roots of the equation are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Now by comparing the given equation with the general form of quadratic equation we get a = 5, b = 6 and c = - 8.
Now we will find the roots of the equation by substituting the values of a, b and c in the formula. Hence we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4\left( 5 \right)\left( -8 \right)}}{2\left( 5 \right)} \\
 & \Rightarrow x=\dfrac{-6\pm \sqrt{160+36}}{10} \\
 & \Rightarrow x=\dfrac{-6\pm \sqrt{196}}{10} \\
 & \Rightarrow x=\dfrac{-6\pm 14}{10} \\
\end{align}$
Hence we get either $x=\dfrac{-6-14}{10}=-2$ or $x=\dfrac{-6+14}{10}=\dfrac{8}{10}$
Hence the roots of the equation are $x=-2$ or $x=\dfrac{2}{10}=\dfrac{1}{5}$.
Now we know that if $\alpha $ and $\beta $ are the roots of the equation then $x-\alpha $ and $x-\beta $ are the factors of the given equation.
Hence, the factors of the given equation are $\left( x+2 \right)$ or $\left( x-\dfrac{1}{5} \right)$
Hence $5{{x}^{2}}-6x+10=\left( x+2 \right)\left( x-\dfrac{1}{5} \right)$.

Note: Note that we can also find the roots of the equation by using the complete square method. In this we first divide the whole equation by $a$ such that the coefficient of ${{x}^{2}}$ is 1. Now we will add and subtract ${{\left( \dfrac{b}{a} \right)}^{2}}$ and then simplify the equation by using ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ .Hence we can find the roots of the equation then factors of the given equation.