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2\left(2m^{3}+5m^{2}-42m\right)
Factor out 2.
m\left(2m^{2}+5m-42\right)
Consider 2m^{3}+5m^{2}-42m. Factor out m.
a+b=5 ab=2\left(-42\right)=-84
Consider 2m^{2}+5m-42. Factor the expression by grouping. First, the expression needs to be rewritten as 2m^{2}+am+bm-42. To find a and b, set up a system to be solved.
-1,84 -2,42 -3,28 -4,21 -6,14 -7,12
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -84.
-1+84=83 -2+42=40 -3+28=25 -4+21=17 -6+14=8 -7+12=5
Calculate the sum for each pair.
a=-7 b=12
The solution is the pair that gives sum 5.
\left(2m^{2}-7m\right)+\left(12m-42\right)
Rewrite 2m^{2}+5m-42 as \left(2m^{2}-7m\right)+\left(12m-42\right).
m\left(2m-7\right)+6\left(2m-7\right)
Factor out m in the first and 6 in the second group.
\left(2m-7\right)\left(m+6\right)
Factor out common term 2m-7 by using distributive property.
2m\left(2m-7\right)\left(m+6\right)
Rewrite the complete factored expression.