Question

How do you factor \[2{{x}^{2}}-5x-12\]?

Answer 1

Hint: In the given question, we are given a quadratic equation in the form \[a{{x}^{2}}+bx+c\] and we have to find its factor. The product of the zeroes is \[ac=2\times (-12)=-24\] and the sum is \[b=-5\]. We have to find the factors which satisfy these conditions. So, we have, \[24=2\times 12=4\times 6=8\times 3\]. We can see that 8 and 3 satisfy the given conditions with appropriate signs. Hence, we will have the factors of the given quadratic equation.

Complete step by step solution:
According to the given question, we are given with a quadratic equation and we are asked to find its factors.
The given quadratic equation is in the form \[a{{x}^{2}}+bx+c\]. So the product of the zeroes is \[ac\] and the sum is \[b\].
The equation we have is,
\[2{{x}^{2}}-5x-12\]----(1)
The product of the zeroes is \[ac=2\times (-12)=-24\] and the sum of the zeroes is \[b=-5\].
So, the possible factors are,
\[24=2\times 12=4\times 6=8\times 3\]
Using 8 and 3 in the equation (1), we get,
\[= 2{{x}^{2}}-(8-3)x-12\]-----(2)
Opening up the brackets in the above equation, we get,
\[= 2{{x}^{2}}-8x+3x-12\]-----(3)
In the terms \[2{{x}^{2}}\] and \[8x\], we will take \[2x\] out as common and in the terms \[3x\] and \[12\], we will take 3 out as common and we will have the expression as,
\[= 2x\left( x-4 \right)+3\left( x-4 \right)\]-----(4)
From the equation (4), we will take the term \[\left( x-4 \right)\] out as common and we get,
\[= \left( x-4 \right)302\left( 2x+3 \right)\]
Therefore, the factors of the given quadratic equation \[2{{x}^{2}}-5x-12=\left( x-4 \right)\left( 2x+3 \right)\].

Note: The factors of a quadratic equation mean to find the compact form of writing the given quadratic equation or in simple words, writing the given quadratic equation as a product of two monomials. Do not confuse the terms factors with zeroes of an equation. Zeroes refers to the value of the function’s variables which when substituted in the function gives a value zero.