How do you factor z^3-9x^2+27z-27z3−9x2+27z−27?

2 Answers
George C.
Oct 3, 2015

Assuming that the xx should have been a zz...

Notice that z^3z3 and -27 = (-3)^3−27=(−3)3 are both perfect cubes.

Expand (z-3)^3(z−3)3 and find (z-3)^3 = z^3-9z^2+27x-27(z−3)3=z3−9z2+27x−27

Explanation:

In the general case we have:

(a+b)^3 = a^3+3a^2b+3ab^2+b^3(a+b)3=a3+3a2b+3ab2+b3

In our case, we notice that z^3z3 and -27 = (-3)^3−27=(−3)3 are both perfect cubes.

So try a=za=z and b = -3b=−3 to find:

(z-3)^3 = z^3 + 3z^2(-3) + 3z(-3)^2 + (-3)^3(z−3)3=z3+3z2(−3)+3z(−3)2+(−3)3

=z^3-9z^2+27z-27=z3−9z2+27z−27

Ernest Z.
Oct 3, 2015

z^3-9z^2+27z-27= color(blue)( (z-3)(z-3)(z-3))z3−9z2+27z−27=(z−3)(z−3)(z−3)

Explanation:

Let f(z) = z^3-9z^2+27z-27f(z)=z3−9z2+27z−27

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

If zz is a root, then zz must be a factor of (-27)/1 = -27−271=−27.

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

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

We find that 33 is a root because

f(3) = 3^3-9×3^2+27×3-27 = 27-81+81-27 = 0f(3)=33−9×32+27×3−27=27−81+81−27=0.

If 33 is a root, then z-3z−3 is a factor of f(z)f(z).

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

3|1color(white)(l)-9color(white)(Xll)27color(white)(l)-273∣1l−9Xll27l−27
color(white)(1)|color(white)(XXl)3color(white)(l)-18color(white)(Xll)271∣XXl3l−18Xll27
" "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)

Related questions
See all questions in Factoring Completely
Impact of this question
2042 views around the world
You can reuse this answer
Creative Commons License