#m2+14m+33=#?

2 Answers
Meave60
Jun 25, 2018

#m^2+14m+33=(m+3)(m+11)#

#m=-3,-11#

Explanation:

Given:

#m^2+14m+33=#

If you are asking for the factorization, find two numbers that when added equal #14# and when multiplied equal #33#. The numbers #3# and #11# meet the requirements.

#m^2+14m+33=(m+3)(m+11)#

If you are asking to solve for #m#, set the equation equal to #0#.

#m^2+14m+33=0#

#(m+3)(m+11)=0#

Set each binomial equal to #0# and solve.

#m+3=0#

#m=-3#

#m+11=0#

#m=-11#

#m=-3,-11#

KillerBunny
Jun 25, 2018

#m=-3# or #m=-11#

Explanation:

I'm assuming you want to solve

#m^2+14m+33=0#

To do so we may use the quadratic formula, but in this case there is a shorter, easier way.

Everytime you have an equation like

#x^2-sx+p=0#

you can solve it by looking for two numbers #x_1#, #x_2# such that #x_1+x_2=s# and #x_1x_2=p#

In this case, #s=-14# and #p=33#.

If we want two numbers to give #33# when multiplied, we can only choose #3# and #11# or #-3# and #-11#. Since we want the sum to be #-14#, the correct choice is #-3# and #-11#, which are the two solutions.

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