Question

How do you factor $4{{x}^{3}}+2{{x}^{2}}-2x-1$ ?
(a) Factor by grouping
(b) Zero putting
(c) Guessing the factors
(d) None of the above

Answer 1

Hint: To find the factor of the given equation $4{{x}^{3}}+2{{x}^{2}}-2x-1$, we will try to factorize it by grouping them among terms. We will start off with grouping the terms from which we can take some factors common like $4{{x}^{3}},2{{x}^{2}}$and 2x,1 together. Then by taking the proper terms common we can get the needed answer and factorization.

Complete step-by-step answer:
We have our given equation as, $4{{x}^{3}}+2{{x}^{2}}-2x-1$and we are to factorize this equation.
So, to start with,
$4{{x}^{3}}+2{{x}^{2}}-2x-1$
As we are trying to factorize this with grouping, we will check which one of the terms are similar and in the next step we will take common terms from the terms. We will take $2{{x}^{2}}$ from the first two terms and -1 from the last two terms.
$\Rightarrow 2{{x}^{2}}(2x+1)-1(2x+1)$
Bringing them together,
$\Rightarrow (2{{x}^{2}}-1)(2x+1)$
One has to determine all the terms that were multiplied to obtain the given polynomial. Then try to factor every term that you got in the first step and this continues until you cannot factor further. When you can’t perform any more factoring, it is said that the polynomial factored completely.
Hence, again, we can factorize $2{{x}^{2}}-1$ as ${{(\sqrt{2}x)}^{2}}-{{(1)}^{2}}$ ,
So, using, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Now we have,
$\Rightarrow (\sqrt{2}x+1)(\sqrt{2}x-1)(2x+1)$
We can also leave this at the stage of $(2{{x}^{2}}-1)(2x+1)$ as $\sqrt{2}$ is not an integer. And when we factorize something we will always try to find the factors with integer coefficients.
Hence the solution is, (a) Factor by grouping.

Note: A polynomial $4{{x}^{3}}+2{{x}^{2}}-2x-1$ can be written as a product of two or more polynomials of degree less than or equal to that of it. Each polynomial involved in the product will be a factor of it. Monomials can be factorized in the same way as integers, just by writing the monomial as the product of its constituent prime factors. In the case of monomials, these prime factors can be integers as well as other monomials which cannot be factorized further.