How do you factor #z^3-9x^2+27z-27#?

2 Answers
Oct 3, 2015

Answer:

Assuming that the #x# should have been a #z#...

Notice that #z^3# and #-27 = (-3)^3# are both perfect cubes.

Expand #(z-3)^3# and find #(z-3)^3 = z^3-9z^2+27x-27#

Explanation:

In the general case we have:

#(a+b)^3 = a^3+3a^2b+3ab^2+b^3#

In our case, we notice that #z^3# and #-27 = (-3)^3# are both perfect cubes.

So try #a=z# and #b = -3# to find:

#(z-3)^3 = z^3 + 3z^2(-3) + 3z(-3)^2 + (-3)^3#

#=z^3-9z^2+27z-27#

Oct 3, 2015

Answer:

#z^3-9z^2+27z-27= color(blue)( (z-3)(z-3)(z-3))#

Explanation:

Let #f(z) = z^3-9z^2+27z-27#

We can set #f(z)=0# and use the rational roots theorem to find the factors.

If #z# is a root, then #z# must be a factor of #(-27)/1 = -27#.

∴ Possible values of #z# are #±1, ±3, ±9, ±27#.

We can test a possible root by plugging in the value and seeing if #f(z)# becomes zero.

We find that #3# is a root because

#f(3) = 3^3-9×3^2+27×3-27 = 27-81+81-27 = 0#.

If #3# is a root, then #z-3# is a factor of #f(z)#.

We can use synthetic division to divide #f(z)# by #z-3#.

#3|1color(white)(l)-9color(white)(Xll)27color(white)(l)-27#
#color(white)(1)|color(white)(XXl)3color(white)(l)-18color(white)(Xll)27#
#" "stackrel("—————————————)#
#color(white)(Xll)1color(white)(l)-6color(white)(XXl)9color(white)(XXl)color(red)(0)#

We get #z^2-6z+9# with a remainder of zero.

So #f(z) = (z-3)( z^2-6z+9)#.

We can factor the quadratic to get

#z^2-6z+9 = (z-3)(z-3)#.

So #f(z) = (z-3) (z-3)(z-3)#

#z^3-9z^2+27z-27= (z-3)(z-3)(z-3)#