Find the Roots (Zeros) f(x)=x^3-2x^2-25x+50

$f (x) = x^{3} - 2 x^{2} - 25 x + 50$

Step 1

Set $x^{3} - 2 x^{2} - 25 x + 50$ equal to $0$ .

$x^{3} - 2 x^{2} - 25 x + 50 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} - 2 x^{2}) - 25 x + 50 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 2) - 25 (x - 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 2$ .

$(x - 2) (x^{2} - 25) = 0$

Rewrite $25$ as $5^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$(x - 2) ((x + 5) (x - 5)) = 0$

Remove unnecessary parentheses.

$(x - 2) (x + 5) (x - 5) = 0$

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

$x - 2 = 0$

$x + 5 = 0$

$x - 5 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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 $(x - 2) (x + 5) (x - 5) = 0$ true.

$x = 2, - 5, 5$

Step 3