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

Oct 3, 2015

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

Notice that ${z}^{3}$ and $- 27 = {\left(- 3\right)}^{3}$ are both perfect cubes.

Expand ${\left(z - 3\right)}^{3}$ and find ${\left(z - 3\right)}^{3} = {z}^{3} - 9 {z}^{2} + 27 x - 27$

#### Explanation:

In the general case we have:

${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$

In our case, we notice that ${z}^{3}$ and $- 27 = {\left(- 3\right)}^{3}$ are both perfect cubes.

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

${\left(z - 3\right)}^{3} = {z}^{3} + 3 {z}^{2} \left(- 3\right) + 3 z {\left(- 3\right)}^{2} + {\left(- 3\right)}^{3}$

$= {z}^{3} - 9 {z}^{2} + 27 z - 27$

Oct 3, 2015

${z}^{3} - 9 {z}^{2} + 27 z - 27 = \textcolor{b l u e}{\left(z - 3\right) \left(z - 3\right) \left(z - 3\right)}$

#### Explanation:

Let $f \left(z\right) = {z}^{3} - 9 {z}^{2} + 27 z - 27$

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

If $z$ is a root, then $z$ must be a factor of $\frac{- 27}{1} = - 27$.

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

We can test a possible root by plugging in the value and seeing if $f \left(z\right)$ 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 $f \left(z\right)$.

We can use synthetic division to divide $f \left(z\right)$ by $z - 3$.

$3 | 1 \textcolor{w h i t e}{l} - 9 \textcolor{w h i t e}{X l l} 27 \textcolor{w h i t e}{l} - 27$
$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{X X l} 3 \textcolor{w h i t e}{l} - 18 \textcolor{w h i t e}{X l l} 27$
" "stackrel("—————————————)
$\textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{l} - 6 \textcolor{w h i t e}{X X l} 9 \textcolor{w h i t e}{X X l} \textcolor{red}{0}$

We get ${z}^{2} - 6 z + 9$ with a remainder of zero.

So $f \left(z\right) = \left(z - 3\right) \left({z}^{2} - 6 z + 9\right)$.

We can factor the quadratic to get

${z}^{2} - 6 z + 9 = \left(z - 3\right) \left(z - 3\right)$.

So $f \left(z\right) = \left(z - 3\right) \left(z - 3\right) \left(z - 3\right)$

${z}^{3} - 9 {z}^{2} + 27 z - 27 = \left(z - 3\right) \left(z - 3\right) \left(z - 3\right)$