Identify the Zeros and Their Multiplicities x^3+x^2-42x

$x^{3} + x^{2} - 42 x$

Step 1

Set $x^{3} + x^{2} - 42 x$ equal to $0$ .

$x^{3} + x^{2} - 42 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} + x^{2} - 42 x$ .

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

$x \cdot x^{2} + x^{2} - 42 x = 0$

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

$x \cdot x^{2} + x \cdot x - 42 x = 0$

Factor $x$ out of $- 42 x$ .

$x \cdot x^{2} + x \cdot x + x \cdot - 42 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot x$ .

$x (x^{2} + x) + x \cdot - 42 = 0$

Factor $x$ out of $x (x^{2} + x) + x \cdot - 42$ .

$x (x^{2} + x - 42) = 0$

Factor.

Factor $x^{2} + x - 42$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 42$ and whose sum is $1$ .

$- 6, 7$

Write the factored form using these integers.

$x ((x - 6) (x + 7)) = 0$

Remove unnecessary parentheses.

$x (x - 6) (x + 7) = 0$

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

$x = 0$

$x - 6 = 0$

$x + 7 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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

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

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $x (x - 6) (x + 7) = 0$ true. The multiplicity of a root is the number of times the root appears.

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

$x = 6$ (Multiplicity of $1$ )

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

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

$x = 6$ (Multiplicity of $1$ )

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

Step 3

Identify the Zeros and Their Multiplicities x^3+x^2-42x

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