How do you factor #c^3 - 2c^2 - 8c#?

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Answer 1 · 2015-10-19T23:28:54

The factorization of #c^3 - 2c^2 - 8c# is

#c (c + 2)(c - 4)#

To factor #c^3 - 2c^2 - 8c#
begin by factoring out the variable #c# as it is a factor of each unit of the tri-nomial.

#c(c^2 - 2c - 8)#

Next, factor the tri-nomial #(c^2 - 2c - 8)#
by finding the factor of 8 that will subtract to get 2.

1 and 8 2 and 4

Since the second sign of the tri-nomial is subtraction the factors must have different signs.

#-4(2) = -8#
#-4 +2 = -2#

Now factor the tri-nomial into two binomials.

#c^2 - 2c - 8#
#(c + 2)(c - 4)#

Now complete the factors.

#c (c + 2)(c - 4)#

Here is a video on factoring.

Answer 2 · 2015-10-19T23:29:39

#c * (c+2) * (c-4)#

Your starting expression looks like this

#c^3 - 2c^2 - 8c#

Notice that you can use #c# as a common factor for all three terms. This will get you

#c * (c^2 - 2c - 8)#

Now focus on the paranthesis. Notice that you can rewrite that expression as

#c^2 - 2c - 8 = c^2 +2c - 4c - 8#

#=c * (c+2) * (-4) * (c + 2)#

#= (c + 2) * (c - 4)#

The expression can thus be factored as

#c^3 - 2c^2 - 8c = color(green)(c * (c+2) * (c-4))#