Solve by Factoring x^3-5x^2=36x

$x^{3} - 5 x^{2} = 36 x$

Step 1

Subtract $36 x$ from both sides of the equation.

$x^{3} - 5 x^{2} - 36 x = 0$

Step 2

Factor $x$ out of $x^{3} - 5 x^{2} - 36 x$ .

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

$x \cdot x^{2} - 5 x^{2} - 36 x = 0$

Factor $x$ out of $- 5 x^{2}$ .

$x \cdot x^{2} + x (- 5 x) - 36 x = 0$

Factor $x$ out of $- 36 x$ .

$x \cdot x^{2} + x (- 5 x) + x \cdot - 36 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 5 x)$ .

$x (x^{2} - 5 x) + x \cdot - 36 = 0$

Factor $x$ out of $x (x^{2} - 5 x) + x \cdot - 36$ .

$x (x^{2} - 5 x - 36) = 0$

Step 3

Factor.

Factor $x^{2} - 5 x - 36$ 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 $- 36$ and whose sum is $- 5$ .

$- 9, 4$

Write the factored form using these integers.

$x ((x - 9) (x + 4)) = 0$

Remove unnecessary parentheses.

$x (x - 9) (x + 4) = 0$

Step 4

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

$x = 0$

$x - 9 = 0$

$x + 4 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Step 7

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8

The final solution is all the values that make $x (x - 9) (x + 4) = 0$ true.

$x = 0, 9, - 4$

Solve by Factoring x^3-5x^2=36x

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