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

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Answer 1 · 2017-01-08T20:26:46

The zeros are: $0, 2, 2$ $0, 2, 2$

Ignoring signs, notice that the sequence of coefficients is: $1, 4, 4$ $1color(white)(,)4color(white)(,)4$

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

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

(the sequence of coefficients, ignoring signs, of $(t - 2)$ $(t-2)$ being $1, 2$ $1color(white)(,)2$ )

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

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

which has zeros:

$t = 0$ $t = 0" "$ (with multiplicity $1$ $1$ )

$t = 2$ $t = 2" "$ (with multiplicity $2$ $2$ )

How do you find all zeros of f(t)=t3−4t2+4tf(t)=t^3-4t^2+4t?