How do you factor #4m^3+9m^2-36m-81#?

1 Answer
George C.
Nov 22, 2016

#4m^3+9m^2-36m-81 = (m-3)(m+3)(4m+9)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We will use this with #a=m# and #b=3#, but first...

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms, so this cubic will factor by grouping:

#4m^3+9m^2-36m-81 = (4m^3+9m^2)-(36m+81)#

#color(white)(4m^3+9m^2-36m-81) = m^2(4m+9)-9(4m+9)#

#color(white)(4m^3+9m^2-36m-81) = (m^2-9)(4m+9)#

#color(white)(4m^3+9m^2-36m-81) = (m^2-3^2)(4m+9)#

#color(white)(4m^3+9m^2-36m-81) = (m-3)(m+3)(4m+9)#

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