Question

How do you factor $2{{x}^{4}}-8{{x}^{3}}-11{{x}^{2}}-4x-6$ ?

Answer 1

Hint: First to factorize this given polynomial we must split the middle term which is $-11{{x}^{2}}$ . To split the middle term $-11{{x}^{2}}$, we use the sum and product form. Now take-out common terms from the first three terms which will be $2{{x}^{2}}$ and last three terms which will be $1$ after the splitting of the middle term. Then write them together as the product of sums form. Now represent them as the factors of the given expression.

Complete step by step solution:
The given polynomial which must be factorized is $2{{x}^{4}}-8{{x}^{3}}-11{{x}^{2}}-4x-6$
To factorize this polynomial, we must first split the middle term.
For this, we use the product sum form.
If the polynomial’s general form is $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e$
Then to split the middle term, we use the product and sum formula.
Which is,
The product of the middle terms after splitting must be $a\times e$
And the sum of the middle terms must be $c$
For our polynomial of degree $4$, We do not consider the $3,1\;$ degree terms.
Here $a=2;b=-11;c=-6$
The product of the terms is $2\times -6=-12$
The sum of the terms is $-11$
Therefore, the terms can be $-12{{x}^{2}},{{x}^{2}}$
On substituting back, we get,
$2{{x}^{4}}-8{{x}^{3}}-12{{x}^{2}}+{{x}^{2}}-4x-6$
The polynomial is of degree $4$
Now, firstly let us take the common terms out of the first three terms.
We consider three terms together because there are six terms in total and to factorize, we need to group them into two halves.
$\Rightarrow 2{{x}^{2}}\left( {{x}^{2}}-4x-6 \right)+{{x}^{2}}-4x-6$
Now secondly take the common terms out of the last two terms.
Since there is nothing common, we take $1$ as the factor common.
$\Rightarrow 2{{x}^{2}}\left( {{x}^{2}}-4x-6 \right)+1\left( {{x}^{2}}-4x-6 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( 2{{x}^{2}}+1 \right)\left( {{x}^{2}}-4x-6 \right)$
The given equations cannot be further factorized so we leave it as it is.
Now writing it all together we get,
$\Rightarrow \left( 2{{x}^{2}}+1 \right)\left( {{x}^{2}}-4x-6 \right)$

Hence the factors for the polynomial $2{{x}^{4}}-8{{x}^{3}}-11{{x}^{2}}-4x-6$ are $\left( 2{{x}^{2}}+1 \right)\left( {{x}^{2}}-4x-6 \right)$

Note: Whenever there are an odd number of terms in the expression, always factorize the middle term. If there are even numbers of terms in the expression, group them together in halves and then find the common factor amongst them and take it out.