Question

How do you factor$3{{x}^{2}}-2x-8$?

Answer 1

Hint: IThe above given equation is in the form of a trinomial quadratic, which means it has only three terms instead of four, but we will end up with four terms in the answer. The polynomial is of the type $a{{x}^{2}}+bx+c$, where $a$ is the coefficient of ${{x}^{2}}$ and $b$ is the coefficient of $x$ and $c$ is the constant.

Complete step by step solution:
The given equation is $3{{x}^{2}}-2x-8$, where $3$ is the coefficient of ${{x}^{2}}$, $\left( -2 \right)$ is the coefficient of $x$and $\left( -8 \right)$ is constant. There are few steps which we have to follow to solve the given equation.
First we have to choose two numbers such that the product of the coefficient of ${{x}^{2}}$and the constant term is equal to the product of those two numbers.
Here we get the two numbers are $\left( -6 \right)$ and $\left( 4 \right)$ and second the addition these two numbers must be equal to the coefficient of $x$. The product is $-24$ and addition is $-2$.
Now we can write the equation as:
$\begin{align}
  & \Rightarrow 3{{x}^{2}}-2x-8 \\
 & \Rightarrow 3{{x}^{2}}-6x+4x-8 \\
\end{align}$
Now take common from $3{{x}^{2}}-6x$ and $4x-8$ then we get
$\Rightarrow 3x\left( x-2 \right)+4\left( x-2 \right)$
Rewriting them we get
$\Rightarrow \left( 3x+4 \right)\left( x-2 \right)$
Hence we get the factor of$3{{x}^{2}}-2x-8$is $\left( 3x+4 \right)\left( x-2 \right)$.

Note: We can also find the factor of these type of equation by using another method which is $\Rightarrow {{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ , where $b$ is the coefficient of $x$ and $a$ is the coefficient of ${{x}^{2}}$ and $c$ is the constant.
Now putting the value of $b$,$a$,$c$from the given equation $3{{x}^{2}}-2x-8$, we get
$\begin{align}
  & \Rightarrow {{x}_{1,2}}=\dfrac{2\pm \sqrt{4-4\times 3\times \left( -8 \right)}}{2\times 3} \\
 & \Rightarrow {{x}_{1,2}}=\dfrac{2\pm \sqrt{4+96}}{6} \\
\end{align}$
Now $4+96$ is $100$ and $\sqrt{100}$ is $10$ put this in above equation we get
$\Rightarrow {{x}_{1,2}}=\dfrac{2\pm 10}{6}$
Now by above expressions we get the values of ${{x}_{1}}$ and${{x}_{2}}$ ,
$\Rightarrow {{x}_{1}}=\dfrac{2+10}{6}=\dfrac{12}{6}=2$ and
$\Rightarrow {{x}_{2}}=\dfrac{2-10}{6}=\dfrac{-8}{6}=\dfrac{-4}{3}$
And now we have to use the formula:
$\Rightarrow a{{x}^{2}}+bx+c=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$
So,
$\begin{align}
  & \Rightarrow 3{{x}^{2}}-2x-8=3\left( x-2 \right)\left( x+\dfrac{4}{3} \right) \\
 & \Rightarrow 3{{x}^{2}}-2x-8=\left( 3x+4 \right)\left( x-2 \right) \\
\end{align}$
Hence we got the same answer as we solved above$3{{x}^{2}}-2x-8=\left( 3x+4 \right)\left( x-2 \right)$.