# How do you factor by grouping: 3x^2 + 3 + x^3 + x?

##### 2 Answers
Mar 13, 2018

The factored form of the expression is $\left({x}^{2} + 1\right) \left(x + 3\right)$.

#### Explanation:

First, rearrange the terms. Then, factor ${x}^{2}$ out from the first two terms and $1$ out from the last two terms. Lastly, combine the two factors to get an answer:

$\textcolor{w h i t e}{=} 3 {x}^{2} + 3 + {x}^{3} + x$

$= {x}^{3} + 3 {x}^{2} + x + 3$

$= \textcolor{red}{{x}^{2}} \cdot x + \textcolor{red}{{x}^{2}} \cdot 3 + x + 3$

$= \textcolor{red}{{x}^{2}} \left(x + 3\right) + x + 3$

$= \textcolor{red}{{x}^{2}} \left(x + 3\right) + \textcolor{b l u e}{1} \cdot x + \textcolor{b l u e}{1} \cdot 3$

$= \textcolor{red}{{x}^{2}} \left(x + 3\right) + \textcolor{b l u e}{1} \left(x + 3\right)$

$= \left(\textcolor{red}{{x}^{2}} + \textcolor{b l u e}{1}\right) \left(x + 3\right)$

Mar 13, 2018

This does it: $\left(x + 3\right) \left({x}^{2} + 1\right)$

#### Explanation:

I am not sure about the "by grouping" method, but I stumbled on a way to factor that expression.

I noticed that it could be written $3 \cdot \left({x}^{2} + 1\right) + x \cdot \left({x}^{2} + 1\right)$.

From there I put the 3 and the x inside a second set of parentheses and saw that

$3 {x}^{2} + 3 + {x}^{3} + x = \left(3 + x\right) \left({x}^{2} + 1\right)$

So $\left(3 + x\right) \mathmr{and} \left({x}^{2} + 1\right)$ are the factors.

A preferable way to write it would probably be $\left(x + 3\right) \left({x}^{2} + 1\right)$.

I hope this helps,
Steve