Question

How do you factor $2{{x}^{3}}+10{{x}^{2}}+4x+20$ ?

Answer 1

Hint: The given polynomial has even number of terms. Hence, we do not need to split the middle term. Instead, we should just take the common factors out from the first two which will be $2{{x}^{2}}$ and the last two terms whose common factor will be $4$ . Now represent the equation in product of sums form and then the terms will then be the factors of the expression.

Complete step by step solution:
The given polynomial which must be factorized is $2{{x}^{3}}+10{{x}^{2}}+4x+20$
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.
$\Rightarrow 2{{x}^{2}}\left( x+5 \right)+4x+20$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 2{{x}^{2}}\left( x+5 \right)+4\left( x+5 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( 2{{x}^{2}}+4 \right)\left( x+5 \right)$
Now we can see that there is no other polynomial left that can be factored further.
Now writing it all together we get,
$\Rightarrow \left( 2{{x}^{2}}+4 \right)\left( x+5 \right)$

Hence the factors for the polynomial $2{{x}^{3}}+10{{x}^{2}}+4x+20$ are $\left( 2{{x}^{2}}+4 \right)\left( x+5 \right)$.

Note: We could always check our solution by reverse multiplying all the factors.
The factors were, $\left( 2{{x}^{2}}+4 \right)\left( x+5 \right)$
On multiplying them we get,
$\Rightarrow \left( 2{{x}^{2}}+4 \right)\left( x+5 \right)$
$\Rightarrow \left( 2{{x}^{2}}\left( x+5 \right)+4\left( x+5 \right) \right)$
On opening the brackets and multiplying the contents we get,
$\Rightarrow \left( 2{{x}^{3}}+10{{x}^{2}}+4x+20 \right)$
The above equation is our question.
Hence our solution is correct.