Question

How do you factor $ 8{x^3} + 4{x^2} - 18x - 9 $ ?

Answer 1

Hint: The given question is a cubic equation . We can solve it first by finding a factor like say $ 0, - 1,2 $ but in this case it is not possible. Since there exists not common roots of this equation therefore we will first see that the ratio between the first and second terms is the same as that of the third and fourth terms which will help us to easily factorize the given cubic equation by grouping the given cubic equation into two parts. And then try to break the equation into parts and get our desired factors.

Complete step-by-step solution:
The ratio between the first and second terms is the same as that of the ration between the third and fourth terms in the given cubic equations and therefore we will write as,
 $ 8{x^3} + 4{x^2} - 18x - 9 $
 $ = (8{x^3} + 4{x^2}) - (18x + 9) $
Now we will take $ 4{x^2} $ common from first part and $ - 9 $ common from the second part of the equation,
 $ = 4{x^2}(2x + 1) - 9(2x + 1) $
 $ = (4{x^2} - 9)(2x + 1) $
Now we will factorize the first part of this above equation because we need linear factors but it is of the form of a quadratic factor which will be broken into the further parts,
 $ = ({(2x)^2} - {3^2})(2x + 1) $
Now we will use the identity in the first part of the equation which is given by,
 $ {a^2} - {b^2} = (a - b)(a + b) $
Thus our equation becomes in the form of factors like,
\[ = \left( {2x - 3} \right)\left( {2x + 3} \right)\left( {2x + 1} \right)\]

Thus the following three factors are there \[\left( {2x - 3} \right),\left( {2x + 3} \right),\left( {2x + 1} \right)\] for the given equation $ 8{x^3} + 4{x^2} - 18x - 9 $

Note: Always remember the identity of difference of squares which is given by
 $ {a^2} - {b^2} = (a - b)(a + b) $
It will be helpful in many questions. And also we have few more algebraic identities that will be helpful while solving the above kind of problems.
i) $ (a+b)^2 = a^2+b^2+2ab $
ii) $ (a-b)^2 = a^2+b^2+2ab $