How do you find all the zeros of $f(x) = 9x^3 - 45x^2 + 36x$ ?

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Answer 1 · 2016-04-09T10:28:44

$x = 0, 1, 4$

Factorise $f(x)$ by taking out $9x$ , the greatest common factor, and going from there,

$f(x) = 9x^3 - 45x^2 + 36x$
$= 9x(x^2 - 5x + 4)$
$= 9x(x - 4)(x - 1)$

One of these three parts has to equal $0$ for the whole thing to be $0$ , so you'll end up with three answers.

$9x = 0$
$x = 0/9 = 0$

$x - 4 = 0$
$x = 4$

$x - 1 = 0$
$x = 1$

and you've got your answers.

How do you find all the zeros of f(x) = 9x^3 - 45x^2 + 36xf(x) = 9x^3 - 45x^2 + 36x?