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

2 Answers
Mar 13, 2018

Answer:

The factored form of the expression is #(x^2+1)(x+3)#.

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:

#color(white)=3x^2+3+x^3+x#

#=x^3+3x^2+x+3#

#=color(red)(x^2)*x+color(red)(x^2)*3+x+3#

#=color(red)(x^2)(x+3)+x+3#

#=color(red)(x^2)(x+3)+color(blue)1*x+color(blue)1*3#

#=color(red)(x^2)(x+3)+color(blue)1(x+3)#

#=(color(red)(x^2)+color(blue)1)(x+3)#

Mar 13, 2018

Answer:

This does it: #(x+3)(x^2 + 1)#

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*(x^2 +1) + x*(x^2 +1)#.

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

#3x^2 + 3 + x^3+ x = (3+x)(x^2 + 1)#

So #(3+x)and(x^2 + 1)# are the factors.

A preferable way to write it would probably be #(x+3)(x^2 + 1)#.

I hope this helps,
Steve