How do you factor #7n ^ { 2} + 45n + 18#?

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Answer 1 · 2018-06-07T18:18:35

#(n-6)(n-3/7)#

After dividing by #7# we get
#n^2+45/7n+18/7=0#
using the quadratic Formula

#n_{1,2}=-45/14pm\sqrt(1521/196)#
this is
#n_{1,2}=45/14pm39/14#
so we get
#n_1=6#
#n_2=3/7#
and we get
the product as
#(n-6)(n-3/7)#

Answer 2 · 2018-06-07T18:25:29

#(7n+3)(n+6)#

We can start by factoring by grouping. Here, we will rewrite the #b# term as the sum of two terms. We get

#7n^2+color(blue)(42n+3n)+18#

Notice, what I have in blue is the same as #45n#, so I didn't change the value of the expression.

#color(darkblue)(7n^2+42n)+color(purple)(3n+18)#

I can factor a #7n# out of the dark blue terms, and a #3# out of the purple term. Doing this, we get

#7ncolor(red)((n+6))+3color(red)((n+6))#

Both terms have a #n+6# in common, so we can factor that out. Doing this, we get

#(7n+3)(n+6)#

as our final answer.

Hope this helps!