Find the Roots (Zeros) f(x)=x^3-13x^2+40x

$f (x) = x^{3} - 13 x^{2} + 40 x$

Step 1

Set $x^{3} - 13 x^{2} + 40 x$ equal to $0$ .

$x^{3} - 13 x^{2} + 40 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 13 x^{2} + 40 x$ .

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

$x \cdot x^{2} - 13 x^{2} + 40 x = 0$

Factor $x$ out of $- 13 x^{2}$ .

$x \cdot x^{2} + x (- 13 x) + 40 x = 0$

Factor $x$ out of $40 x$ .

$x \cdot x^{2} + x (- 13 x) + x \cdot 40 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 13 x)$ .

$x (x^{2} - 13 x) + x \cdot 40 = 0$

Factor $x$ out of $x (x^{2} - 13 x) + x \cdot 40$ .

$x (x^{2} - 13 x + 40) = 0$

Factor.

Factor $x^{2} - 13 x + 40$ 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 $40$ and whose sum is $- 13$ .

$- 8, - 5$

Write the factored form using these integers.

$x ((x - 8) (x - 5)) = 0$

Remove unnecessary parentheses.

$x (x - 8) (x - 5) = 0$

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

$x = 0$

$x - 8 = 0$

$x - 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

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.

$x = 5$

The final solution is all the values that make $x (x - 8) (x - 5) = 0$ true.

$x = 0, 8, 5$

Step 3

Find the Roots (Zeros) f(x)=x^3-13x^2+40x

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