Find All Complex Solutions x^3-9x^2+20x=0

$x^{3} - 9 x^{2} + 20 x = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 9 x^{2} + 20 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 9 x^{2} + 20 x = 0$

Factor $x$ out of $- 9 x^{2}$ .

$x \cdot x^{2} + x (- 9 x) + 20 x = 0$

Factor $x$ out of $20 x$ .

$x \cdot x^{2} + x (- 9 x) + x \cdot 20 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 9 x)$ .

$x (x^{2} - 9 x) + x \cdot 20 = 0$

Factor $x$ out of $x (x^{2} - 9 x) + x \cdot 20$ .

$x (x^{2} - 9 x + 20) = 0$

Factor.

Factor $x^{2} - 9 x + 20$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $20$ and whose sum is $- 9$ .

$- 5, - 4$

Write the factored form using these integers.

$x ((x - 5) (x - 4)) = 0$

Remove unnecessary parentheses.

$x (x - 5) (x - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 5 = 0$

$x - 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $x (x - 5) (x - 4) = 0$ true.

$x = 0, 5, 4$

Find All Complex Solutions x^3-9x^2+20x=0

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