How do you factor the expression #x^3 + 3x^2 + 4x + 12#?

2 Answers
Apr 26, 2018

Answer:

#(x+3)(x+2i)(x-2i)#

Explanation:

#color(blue)"factor by grouping"#

#=color(red)(x^2)(x+3)color(red)(+4)(x+3)#

#"take out the "color(blue)"common factor "(x+3)#

#=(x+3)(color(red)(x^2+4))#

#"we can factor "x^2+4" by solving "x^2+4=0#

#x^2+4=0rArrx^2=-4rArrx=+-2i#

#rArrx^2+4=(x+2i)(x-2i)#

#rArrx^3+3x^2+4x+12=(x+3)(x+2i)(x-2i)#

Apr 26, 2018

Answer:

#(x+3)*(x^2+4)#

Explanation:

#x^3+3x^2+4x+12#
If you are asked to completely factor an expression with four terms such as this, check to see if you are able to FACTOR BY GROUPING . If you can, this will be a simple process!

First, split the expression into two, find and factor out their greatest common factor in each term, and add them back together.

The common factor of the first half (#x^3+3x^2#) is #x^2# while the common factor of the second half (#4x+12#) is #4#.
Now factor out and add back the expressions into:
#x^2*(x+3)+4*(x+3)#

This is not your final answer, however.

Note that the the two halves of the expression share a common factor, #x+3#. This means that you may FACTOR BY GROUPING . You can factor out #x+3# (by reversing the distributive property of multiplication) from the expression into:
#(x+3)*(x^2+4)#

Neither of these two terms can be factored no more. Therefore, this is your final answer!

Remember that the presence of a common factor at the second step indicates that you may FACTOR BY GROUPING to easily answer this question.