Identify the Zeros and Their Multiplicities y=-(x+3)(x+1)^2(2x-5)

$y = - (x + 3) {(x + 1)}^{2} (2 x - 5)$

Step 1

Set $- (x + 3) {(x + 1)}^{2} (2 x - 5)$ equal to $0$ .

$- (x + 3) {(x + 1)}^{2} (2 x - 5) = 0$

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 $0$ .

$x + 3 = 0$

${(x + 1)}^{2} = 0$

$2 x - 5 = 0$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 1)}^{2}$ equal to $0$ .

${(x + 1)}^{2} = 0$

Solve ${(x + 1)}^{2} = 0$ for $x$ .

Set the $x + 1$ 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 $- (x + 3) {(x + 1)}^{2} (2 x - 5) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - 3$ (Multiplicity of $1$ )

$x = - 1$ (Multiplicity of $2$ )

$x = \frac{5}{2}$ (Multiplicity of $1$ )

$x = - 3$ (Multiplicity of $1$ )

$x = - 1$ (Multiplicity of $2$ )

$x = \frac{5}{2}$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities y=-(x+3)(x+1)^2(2x-5)

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