Find the Roots (Zeros) (x-7)(x^2-5x)

$(x - 7) (x^{2} - 5 x)$

Set $(x - 7) (x^{2} - 5 x)$ equal to $0$ .

$(x - 7) (x^{2} - 5 x) = 0$

Solve for $x$ .

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

$x - 7 = 0$

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

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set $x^{2} - 5 x$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5 x$ equal to $0$ .

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

Solve $x^{2} - 5 x = 0$ for $x$ .

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

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

$x \cdot x - 5 x = 0$

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$x (x - 5) = 0$

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

$x = 0$

$x - 5 = 0$

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.

$x = 5$

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

$x = 0, 5$

The final solution is all the values that make $(x - 7) (x^{2} - 5 x) = 0$ true.

$x = 7, 0, 5$