# m2+14m+33=?

Jun 25, 2018

${m}^{2} + 14 m + 33 = \left(m + 3\right) \left(m + 11\right)$

$m = - 3 , - 11$

#### Explanation:

Given:

${m}^{2} + 14 m + 33 =$

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.

${m}^{2} + 14 m + 33 = \left(m + 3\right) \left(m + 11\right)$

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

${m}^{2} + 14 m + 33 = 0$

$\left(m + 3\right) \left(m + 11\right) = 0$

Set each binomial equal to $0$ and solve.

$m + 3 = 0$

$m = - 3$

$m + 11 = 0$

$m = - 11$

$m = - 3 , - 11$

Jun 25, 2018

$m = - 3$ or $m = - 11$

#### Explanation:

I'm assuming you want to solve

${m}^{2} + 14 m + 33 = 0$

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

${x}^{2} - s x + p = 0$

you can solve it by looking for two numbers ${x}_{1}$, ${x}_{2}$ such that ${x}_{1} + {x}_{2} = s$ and ${x}_{1} {x}_{2} = p$

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.