How do you factor #f^3+2f^2-64f-128#?

1 Answer
Dec 3, 2017

Answer:

Required factors are : #(f+8)(f-8)(f+2)#

Explanation:

The problem is to factor

#f^3 + 2f^2 - 64f - 128#

Observe the groupings collected for further simplification:

#color(green)((f^3 + 2f^2)##color(blue)(- 64f - 128)# #color(red)(Expression.1)#

The group #color(green)((f^3 + 2f^2)# can be factored as

#color(green)(f^2(f + 2))# #" "color(red)(Res.1)#

The group #color(blue)(- 64f - 128)# can be factored as

#color(blue)(-64(f+2))# #" "color(red)(Res.2)#

Using our intermediate results #color(red)(Res.1)# and #color(red)(Res.2)# we can write our #color(red)(Expression.1)# as

#color(green)(f^2(f+2)-64(f+2))#

We can now factor them as

#(f^2 - 64)(f+2)# #color(red)(Expression.2)#

Next, we rewrite #(f^2 - 64)# as

#(f^2 - 8^2)#

Using the factoring rule "Difference of Squares" we get

#(f+8)(f - 8)#

Using the above result and our #color(red)(Expression.2)#
we get all of our required factors:

#color(blue)((f+8)(f-8)(f+2))#