Question

How do you factor $2{{x}^{3}}+{{x}^{2}}-6x-3$ ?

Answer 1

Hint: First let us take-out common terms from the first two terms and last two terms. Then write them together as the product of sums form. Now among the factorized terms, find the expression which can be further factorized and then write that also in the product of sums form. Now write all of them together and represent them as the factors of the given expression.

Complete step by step solution:
The given polynomial which must be factorized is $2{{x}^{3}}+{{x}^{2}}-6x-3$
The polynomial is of degree $3$
To factorize this polynomial, we shall,
Now, firstly let us take the common terms out of the first two terms.
We consider two terms together because there are four terms in total and to factorize, we need to group them into two halves.
$\Rightarrow {{x}^{2}}\left( 2x+1 \right)-6x-3$
Now secondly take the common terms out of the last two terms.
$\Rightarrow {{x}^{2}}\left( 2x+1 \right)-3\left( 2x+1 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( {{x}^{2}}-3 \right)\left( 2x+1 \right)$
Now we can see that there is still another polynomial left that can be factored further.
We can write ${{x}^{2}}-3$ as ${{\left( x \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}$
And since ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can factorize it as,
$\Rightarrow \left( x+\sqrt{3} \right)\left( x-\sqrt{3} \right)$
Now writing it all together we get,
$\Rightarrow \left( 2x+1 \right)\left( x+\sqrt{3} \right)\left( x-\sqrt{3} \right)$

Hence the factors for the polynomial $2{{x}^{3}}+{{x}^{2}}-6x-3$ are $\left( 2x+1 \right)\left( x+\sqrt{3} \right)\left( x-\sqrt{3} \right)$

Note: The process of factorization is the reverse multiplication. In the above question, we have multiplied three linear line equations to get a $3$ degree equation (a polynomial of degree $3$ ) expression using the distributive law. Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression. The number of roots is decided by the degree of the polynomial.