How do you factor completely: #2x^3 + 14x^2 + 6x + 42#?

1 Answer
Jul 17, 2015

Answer:

#2(x+7)(x^2+3)#

Explanation:

Notice that, if you factor out #2x^2# from the first two members and factor out #6# from the last two members of this equation, you will have the same in both remainders - #(x+7)#:
#2x^3+14x^2+6x+42=2x^2(x+7)+6(x+7)#

Now we can use the distributive law and factor out #(x+7)# getting:
#(x+7)(2x^2+6)=2(x+7)(x^2+3)#