Answer
The factor of the polynomial $6{{x}^{2}}+x-12$ is $\left( 2x+3 \right)\left( 3x-4 \right)$.
Work Step by Step
The factor of the polynomial $6{{x}^{2}}+x-12$ is calculated as follows.
Multiply the leading coefficient 6, and the constant term $-12$.
That is $6\cdot \left( -12 \right)=-72$.
Find a pair of factors whose product is $-72$ and whose sum is $1$.
$\begin{matrix}
Pairs of Factors of -72 & Sum of Factors \\
-1,72 & 71 \\
-2,36 & 34 \\
-3,24 & 21 \\
-4,18& 14 \\
-6,12 & 6 \\
-8,9 & 1 \\
\end{matrix}$
Therefore, the required integers are $-8$ and $9$.
So,
$6{{x}^{2}}+x-12=6{{x}^{2}}+9x-8x-12$
Group the first two terms and the last two terms.
$6{{x}^{2}}+x-12=3x\left( 2x+3 \right)-4\left( 2x+3 \right)$
Factor out$\left( 2x+3 \right)$.
$6{{x}^{2}}+x-12=\left( 2x+3 \right)\left( 3x-4 \right)$
Thus, the factor of the polynomial $6{{x}^{2}}+x-12$ is$\left( 2x+3 \right)\left( 3x-4 \right)$.
Check:
Consider the expression$\left( 2x+3 \right)\left( 3x-4 \right)$.
Use the FOIL method to solve the inner bracket;
$\begin{align}
& \left( 2x+3 \right)\left( 3x-4 \right)=2x\cdot \left( 3x \right)+2x\cdot \left( -4 \right)+\left( 3 \right)\cdot \left( 3x \right)+\left( 3 \right)\cdot \left( -4 \right) \\
& =6{{x}^{2}}-8x+9x-12 \\
& =6{{x}^{2}}+x-12
\end{align}$
This verifies the result.
