# How do you find all zeros of f(t)=t^3-4t^2+4t?

Jan 8, 2017

The zeros are: $0 , 2 , 2$

#### Explanation:

Ignoring signs, notice that the sequence of coefficients is: $1 \textcolor{w h i t e}{,} 4 \textcolor{w h i t e}{,} 4$

You probably know that $144 = {12}^{2}$ and similarly we find:

${t}^{2} - 4 t + 4 = {\left(t - 2\right)}^{2}$

(the sequence of coefficients, ignoring signs, of $\left(t - 2\right)$ being $1 \textcolor{w h i t e}{,} 2$)

Multiply by $t$ and you get the example function, so:

$f \left(t\right) = {t}^{3} - 4 {t}^{2} + 4 t = t {\left(t - 2\right)}^{2}$

which has zeros:

$t = 0 \text{ }$ (with multiplicity $1$)

$t = 2 \text{ }$ (with multiplicity $2$)