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m^{2}+12m+11
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=12 ab=1\times 11=11
Factor the expression by grouping. First, the expression needs to be rewritten as m^{2}+am+bm+11. To find a and b, set up a system to be solved.
a=1 b=11
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
\left(m^{2}+m\right)+\left(11m+11\right)
Rewrite m^{2}+12m+11 as \left(m^{2}+m\right)+\left(11m+11\right).
m\left(m+1\right)+11\left(m+1\right)
Factor out m in the first and 11 in the second group.
\left(m+1\right)\left(m+11\right)
Factor out common term m+1 by using distributive property.
m^{2}+12m+11=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
m=\frac{-12±\sqrt{12^{2}-4\times 11}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
m=\frac{-12±\sqrt{144-4\times 11}}{2}
Square 12.
m=\frac{-12±\sqrt{144-44}}{2}
Multiply -4 times 11.
m=\frac{-12±\sqrt{100}}{2}
Add 144 to -44.
m=\frac{-12±10}{2}
Take the square root of 100.
m=-\frac{2}{2}
Now solve the equation m=\frac{-12±10}{2} when ± is plus. Add -12 to 10.
m=-1
Divide -2 by 2.
m=-\frac{22}{2}
Now solve the equation m=\frac{-12±10}{2} when ± is minus. Subtract 10 from -12.
m=-11
Divide -22 by 2.
m^{2}+12m+11=\left(m-\left(-1\right)\right)\left(m-\left(-11\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -11 for x_{2}.
m^{2}+12m+11=\left(m+1\right)\left(m+11\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.