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

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Answer 1 · 2017-01-08T20:26:46

The zeros are: #0, 2, 2#

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

You probably know that #144 = 12^2# and similarly we find:

#t^2-4t+4 = (t-2)^2#

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

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

#f(t) = t^3-4t^2+4t = t(t-2)^2#

which has zeros:

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

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