Identify the Zeros and Their Multiplicities x^3-34x^2+289x

$x^{3} - 34 x^{2} + 289 x$

Step 1

Set $x^{3} - 34 x^{2} + 289 x$ equal to $0$ .

$x^{3} - 34 x^{2} + 289 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 34 x^{2} + 289 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 34 x^{2} + 289 x = 0$

Factor $x$ out of $- 34 x^{2}$ .

$x \cdot x^{2} + x (- 34 x) + 289 x = 0$

Factor $x$ out of $289 x$ .

$x \cdot x^{2} + x (- 34 x) + x \cdot 289 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 34 x)$ .

$x (x^{2} - 34 x) + x \cdot 289 = 0$

Factor $x$ out of $x (x^{2} - 34 x) + x \cdot 289$ .

$x (x^{2} - 34 x + 289) = 0$

Factor using the perfect square rule.

Rewrite $289$ as $17^{2}$ .

$x (x^{2} - 34 x + 17^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$34 x = 2 \cdot x \cdot 17$

Rewrite the polynomial.

$x (x^{2} - 2 \cdot x \cdot 17 + 17^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 17$ .

$x {(x - 17)}^{2} = 0$

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

$x = 0$

${(x - 17)}^{2} = 0$

Set $x$ equal to $0$ .

$x = 0$

Set ${(x - 17)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 17)}^{2}$ equal to $0$ .

${(x - 17)}^{2} = 0$

Solve ${(x - 17)}^{2} = 0$ for $x$ .

Set the $x - 17$ equal to $0$ .

$x - 17 = 0$

Add $17$ to both sides of the equation.

$x = 17$

The final solution is all the values that make $x {(x - 17)}^{2} = 0$ true. The multiplicity of a root is the number of times the root appears.

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

$x = 17$ (Multiplicity of $2$ )

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

$x = 17$ (Multiplicity of $2$ )

Step 3

Identify the Zeros and Their Multiplicities x^3-34x^2+289x

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