How do you factor the trinomial #v^2 + 13v + 42#?

2 Answers
May 2, 2016

#v^2+13x+42 = (v+6)(v+7)#

Explanation:

In general we have:

#(v+a)(v+b) = v^2+(a+b)v+ab#

Note that:

#6+7=13#

#6xx7 = 42#

So (putting #a=6# and #b=7#) we find:

#v^2+13x+42 = (v+6)(v+7)#

May 2, 2016

#v^2+13v+42=(v+6)*(v+7)#

Explanation:

Although there are many ways to solve this type of problem, we can try some simple inspection first. We know that we are looking for a solution in the form:

#v^2+13v+42=(v+p)*(v+q)#

We can expand the right-hand side to get:

#(v+p)*(v+q)= v^2+(p+q)v+pq#

So, by inspection we know that we need:

#(p+q)=13# and #pq=42#

Let's start by factoring the product, as this will give us all of the possible integer solutions for #p# and #q#

#42 = 2*3*7#

We notice that #2*3=6# and that #6+7 = 13# so we have our solution where

#p=6# and #q=7# resulting in the factorization:

#v^2+13v+42=(v+6)*(v+7)#