How do you factor #(z-4)^3-4(z-4) #?

1 Answer
Apr 7, 2016

Answer:

#(z-4)^3 - 4(z-4)=(z-4)(z-6)(z-2)#

Explanation:

Separate out the common #(z-4)# factor then use the difference of squares identity:

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

with #a = (z-4)# and #b=2# as follows:

#(z-4)^3 - 4(z-4)#

#=(z-4)((z-4)^2-4)#

#=(z-4)((z-4)^2-2^2)#

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

#=(z-4)(z-6)(z-2)#