Identify the Zeros and Their Multiplicities x^3+4x^2-36x-144

$x^{3} + 4 x^{2} - 36 x - 144$

Step 1

Set $x^{3} + 4 x^{2} - 36 x - 144$ equal to $0$ .

$x^{3} + 4 x^{2} - 36 x - 144 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} + 4 x^{2}) - 36 x - 144 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x + 4) - 36 (x + 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$(x + 4) (x^{2} - 36) = 0$

Rewrite $36$ as $6^{2}$ .

$(x + 4) (x^{2} - 6^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 6$ .

$(x + 4) ((x + 6) (x - 6)) = 0$

Remove unnecessary parentheses.

$(x + 4) (x + 6) (x - 6) = 0$

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

$x + 4 = 0$

$x + 6 = 0$

$x - 6 = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

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.

$x = 6$

The final solution is all the values that make $(x + 4) (x + 6) (x - 6) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - 4$ (Multiplicity of $1$ )

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

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

$x = - 4$ (Multiplicity of $1$ )

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

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

Step 3

Identify the Zeros and Their Multiplicities x^3+4x^2-36x-144

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