How do you factor $6{x^2} - x - 2$?

Hint:In order to determine the factors of the above quadratic question use the Splitting up the middle

Complete step by step solution:
Given a quadratic equation, $6{x^2} - x - 2$let it be $f(x)$
$f(x) = 6{x^2} - x - 2$

Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 6

b becomes -1

And c becomes -2

To find the quadratic factorization we’ll use splitting up the middle term method

So first calculate the product of coefficient of ${x^2}$and the constant term which comes to be $ - 2 \times 6 = - 12$

Now the second Step is to find the 2 factors of the number 2 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors results in the value of constant .

So if we factorize -12 ,the answer comes to be -4 and 3 as $ - 4 + 3 = - 1$ that is the middle term and $3 \times ( - 4) = - 12$ which is perfectly equal to the constant value.

Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(x) = 6{x^2} + 3x - 4x - 2$

Now taking common from the first 2 terms and last 2 terms
$f(x) = 3x(2x + 1) - 2(2x + 1)$

Finding the common binomial parenthesis, the equation becomes
$f(x) = (3x - 2)(2x + 1)$

Hence , We have successfully factorized our quadratic equation.

Therefore the factors are$(3x - 2)$ and $(2x + 1)$

Alternative: You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as

$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$

x1,x2 are root to quadratic equation $a{x^2} + bx + c$

Hence the factors will be $(x - x1)\,and\,(x - x2)\,$.

Note: 1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.