Question

How do you factor completely: $2{{x}^{3}}+14{{x}^{2}}+4x+28$?

Answer 1

Hint: In this question we have a polynomial equation in the $3rd$ degree. Since the equation is not in the form of the quadratic equation, we cannot solve the equation directly by splitting the middle term or using the quadratic formula therefore, we will solve the expression by taking out the common factors and simplifying the expression to get the required solution.

Complete step by step solution:
We have the given expression as:
$\Rightarrow 2{{x}^{3}}+14{{x}^{2}}+4x+28$
Since the term $2$ is common in all the terms, we can take it out as common as:
$\Rightarrow 2\left( {{x}^{3}}+7{{x}^{2}}+2x+14 \right)$
Now, we can see that the term ${{x}^{2}}$ is common in the first two terms in the expression, we can it out as common as:
$\Rightarrow 2\left( {{x}^{2}}\left( x+7 \right)+2x+14 \right)$
We can also see that the term $2$ is common in the last two terms therefore, on taking it out as common, we get:
$\Rightarrow 2\left( {{x}^{2}}\left( x+7 \right)+2\left( x+7 \right) \right)$
Now the term $\left( x+7 \right)$is common in the expression therefore on taking it out as common, we get:
$\Rightarrow 2\left( {{x}^{2}}+2 \right)\left( x+7 \right)$
Since the term ${{x}^{2}}+2$ cannot be factored further, the above expression is the final factored form of the expression therefore, we can write:
$2{{x}^{3}}+14{{x}^{2}}+4x+28=2\left( {{x}^{2}}+2 \right)\left( x+7 \right)$, which is the required solution.

Note:
In the given question we were able to factor the term directly by taking out common terms in the expression. If there are no common terms which are present in the expression the expression can be solved as a quadratic equation by splitting the middle term or using the quadratic formula. It is to be remembered that the general form of a quadratic equation is $a{{x}^{2}}+bx+c$.