Find the Roots (Zeros) p(x)=4x(x-3)(x+1)(2x-5)

$p (x) = 4 x (x - 3) (x + 1) (2 x - 5)$

Set $4 x (x - 3) (x + 1) (2 x - 5)$ equal to $0$ .

$4 x (x - 3) (x + 1) (2 x - 5) = 0$

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 $0$ .

$x = 0$

$x - 3 = 0$

$x + 1 = 0$

$2 x - 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Solve $2 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

The final solution is all the values that make $4 x (x - 3) (x + 1) (2 x - 5) = 0$ true.

$x = 0, 3, - 1, \frac{5}{2}$