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

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

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Answer 1 · 2016-04-09T10:28:44

$x = 0, 1, 4$ $x = 0, 1, 4$

Explanation:

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

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

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

$9 x = 0$ $9x = 0$
$x = \frac{0}{9} = 0$ $x = 0/9 = 0$

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

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

and you've got your answers.

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

1 Answer

Explanation:

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How do you find all the zeros of f(x)=9x3−45x2+36xf(x) = 9x^3 - 45x^2 + 36x?

1 Answer

Explanation:

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