Find the Roots (Zeros) y=(2x+5)(x-3)^2

$y = (2 x + 5) {(x - 3)}^{2}$

Step 1

Set $(2 x + 5) {(x - 3)}^{2}$ equal to $0$ .

$(2 x + 5) {(x - 3)}^{2} = 0$

Step 2

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 $0$ .

$2 x + 5 = 0$

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

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 $x$ .

Subtract $5$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{2}$ equal to $0$ .

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

Solve ${(x - 3)}^{2} = 0$ for $x$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

Step 3