How do you find all zeros of the function $f (x) = x^{3} - 6 x^{2} + 9 x$ $f(x) = x^3 - 6x^2 + 9x$ ?

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Answer 1 · 2016-03-02T16:03:46

$f (x) = 0 at x = 0; x = + 3$ $f(x)=0" at "x=0" ; "x=+3$

Notice that $x$ $x$ is in each term so we can do this:

$f (x) = x (x^{2} - 6 x + 9)$ $f(x)=x(x^2-6x+9)$

Notice that $3 \times 3 = 9 and that 3 + 3 = 6$ $3xx3 =9" and that "3+3=6$

But we have $- 6$ $-6$ so it has to be $- 3 - 3 = - 6$ $-3-3=-6$

$f (x) = x {(x - 3)}^{2}$ $f(x)=x(x-3)^2$

so $f (x) = 0 at x = 0; x = + 3$ $f(x)=0" at "x=0" ; "x=+3$

Tony B

How do you find all zeros of the function f(x)=x3−6x2+9xf(x) = x^3 - 6x^2 + 9x?