Identify the Zeros and Their Multiplicities f(x)=x(5-x)(7-x)

$f (x) = x (5 - x) (7 - x)$

Set $x (5 - x) (7 - x)$ equal to $0$ .

$x (5 - x) (7 - x) = 0$

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 $0$ .

$x = 0$

$5 - x = 0$

$7 - x = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$5 - x = 0$

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

Subtract $5$ from both sides of the equation.

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

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

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

$7 - x = 0$

Solve $7 - x = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$- x = - 7$

Divide each term in $- x = - 7$ by $- 1$ and simplify.

Divide each term in $- x = - 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 7}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 7}{- 1}$

Simplify the right side.

Divide $- 7$ by $- 1$ .

$x = 7$

The final solution is all the values that make $x (5 - x) (7 - x) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 0$ (Multiplicity of $1$ )

$x = 5$ (Multiplicity of $1$ )

$x = 7$ (Multiplicity of $1$ )

$x = 0$ (Multiplicity of $1$ )

$x = 5$ (Multiplicity of $1$ )

$x = 7$ (Multiplicity of $1$ )

Identify the Zeros and Their Multiplicities f(x)=x(5-x)(7-x)

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