Question

How do you factor $ 6{x^2} - 5x - 6 $ ?

Answer 1

Hint: In order to determine the factors of the above quadratic equation use the Splitting up the middle term method or also called mid-term factorization method. In which the middle term would be splitted into two parts and similar values will be taken common and equated to zero to obtain the roots.

Complete step-by-step answer:
Given a quadratic equation $ 6{x^2} - 5x - 6 $ , let it be $ f(x) $
 $ f(x) = 6{x^2} - 5x - 6 $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
 $ a $ becomes $ 6 $
 $ b $ becomes $ - 5 $
And $ c $ becomes $ - 6 $
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
 $ = 6 \times \left( { - 6} \right) = \left( { - 36} \right) $
Now the second Step is to find the $ 2 $ factors of the number $ - 36 $ such that 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 $ - 36 $ , the answer comes to be $ - 9 $ and $ 4 $ as $ - 9 + 4 = - 5 $ that is the middle term. and $ - 9 \times 4 = - 36 $ 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} - 9x + 4x - 6 $
Now taking $ 3x $ common from the first $ 2 $ terms and $ 2 $ from last $ 2 $ terms, we get:
 $ f(x) = 3x(2x - 3) + 2(2x - 3) $
Finding the common binomial parenthesis, the equation becomes
 $ f(x) = (3x + 2)(2x - 3) $
Hence, we have successfully factorized our quadratic equation.
Therefore, the factors are $ \left( {3x + 2} \right) $ and $ \left( {2x - 3} \right) $ .
So, the correct answer is “(3x + 2)(2x - 3)”.

Note: You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
 $ x_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} $ and $ x_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} $
 $ x_1 $ , $ x_2 $ are root to quadratic equation $ a{x^2} + bx + c $
 Hence the factors will be $ (x - x_1)\,and\,(x - x_2)\, $ .