Identify the Zeros and Their Multiplicities 2x^4+12x^3+16x^2-12x-18

$2 x^{4} + 12 x^{3} + 16 x^{2} - 12 x - 18$

Step 1

Set $2 x^{4} + 12 x^{3} + 16 x^{2} - 12 x - 18$ equal to $0$ .

$2 x^{4} + 12 x^{3} + 16 x^{2} - 12 x - 18 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2$ out of $2 x^{4} + 12 x^{3} + 16 x^{2} - 12 x - 18$ .

Factor $2$ out of $2 x^{4}$ .

$2 (x^{4}) + 12 x^{3} + 16 x^{2} - 12 x - 18 = 0$

Factor $2$ out of $12 x^{3}$ .

$2 (x^{4}) + 2 (6 x^{3}) + 16 x^{2} - 12 x - 18 = 0$

Factor $2$ out of $16 x^{2}$ .

$2 (x^{4}) + 2 (6 x^{3}) + 2 (8 x^{2}) - 12 x - 18 = 0$

Factor $2$ out of $- 12 x$ .

$2 (x^{4}) + 2 (6 x^{3}) + 2 (8 x^{2}) + 2 (- 6 x) - 18 = 0$

Factor $2$ out of $- 18$ .

$2 x^{4} + 2 (6 x^{3}) + 2 (8 x^{2}) + 2 (- 6 x) + 2 \cdot - 9 = 0$

Factor $2$ out of $2 x^{4} + 2 (6 x^{3})$ .

$2 (x^{4} + 6 x^{3}) + 2 (8 x^{2}) + 2 (- 6 x) + 2 \cdot - 9 = 0$

Factor $2$ out of $2 (x^{4} + 6 x^{3}) + 2 (8 x^{2})$ .

$2 (x^{4} + 6 x^{3} + 8 x^{2}) + 2 (- 6 x) + 2 \cdot - 9 = 0$

Factor $2$ out of $2 (x^{4} + 6 x^{3} + 8 x^{2}) + 2 (- 6 x)$ .

$2 (x^{4} + 6 x^{3} + 8 x^{2} - 6 x) + 2 \cdot - 9 = 0$

Factor $2$ out of $2 (x^{4} + 6 x^{3} + 8 x^{2} - 6 x) + 2 \cdot - 9$ .

$2 (x^{4} + 6 x^{3} + 8 x^{2} - 6 x - 9) = 0$

Regroup terms.

$2 (6 x^{3} - 6 x + x^{4} + 8 x^{2} - 9) = 0$

Factor $6 x$ out of $6 x^{3} - 6 x$ .

Factor $6 x$ out of $6 x^{3}$ .

$2 (6 x (x^{2}) - 6 x + x^{4} + 8 x^{2} - 9) = 0$

Factor $6 x$ out of $- 6 x$ .

$2 (6 x (x^{2}) + 6 x (- 1) + x^{4} + 8 x^{2} - 9) = 0$

Factor $6 x$ out of $6 x (x^{2}) + 6 x (- 1)$ .

$2 (6 x (x^{2} - 1) + x^{4} + 8 x^{2} - 9) = 0$

Rewrite $1$ as $1^{2}$ .

$2 (6 x (x^{2} - 1^{2}) + x^{4} + 8 x^{2} - 9) = 0$

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 $b = 1$ .

$2 (6 x ((x + 1) (x - 1)) + x^{4} + 8 x^{2} - 9) = 0$

Remove unnecessary parentheses.

$2 (6 x (x + 1) (x - 1) + x^{4} + 8 x^{2} - 9) = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$2 (6 x (x + 1) (x - 1) + {(x^{2})}^{2} + 8 x^{2} - 9) = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$2 (6 x (x + 1) (x - 1) + u^{2} + 8 u - 9) = 0$

Factor $u^{2} + 8 u - 9$ 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 $- 9$ and whose sum is $8$ .

$- 1, 9$

Write the factored form using these integers.

$2 (6 x (x + 1) (x - 1) + (u - 1) (u + 9)) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$2 (6 x (x + 1) (x - 1) + (x^{2} - 1) (x^{2} + 9)) = 0$

Rewrite $1$ as $1^{2}$ .

$2 (6 x (x + 1) (x - 1) + (x^{2} - 1^{2}) (x^{2} + 9)) = 0$

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 $b = 1$ .

$2 (6 x (x + 1) (x - 1) + (x + 1) (x - 1) (x^{2} + 9)) = 0$

Factor $(x + 1) (x - 1)$ out of $6 x (x + 1) (x - 1) + (x + 1) (x - 1) (x^{2} + 9)$ .

Factor $(x + 1) (x - 1)$ out of $6 x (x + 1) (x - 1)$ .

$2 ((x + 1) (x - 1) (6 x) + (x + 1) (x - 1) (x^{2} + 9)) = 0$

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) (6 x) + (x + 1) (x - 1) (x^{2} + 9)$ .

$2 ((x + 1) (x - 1) (6 x + x^{2} + 9)) = 0$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$2 ((x + 1) (x - 1) (6 u + u^{2} + 9)) = 0$

Factor using the perfect square rule.

Rearrange terms.

$2 ((x + 1) (x - 1) (u^{2} + 6 u + 9)) = 0$

Rewrite $9$ as $3^{2}$ .

$2 ((x + 1) (x - 1) (u^{2} + 6 u + 3^{2})) = 0$

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

$6 u = 2 \cdot u \cdot 3$

Rewrite the polynomial.

$2 ((x + 1) (x - 1) (u^{2} + 2 \cdot u \cdot 3 + 3^{2})) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 3$ .

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

Factor.

Replace all occurrences of $u$ with $x$ .

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

Remove unnecessary parentheses.

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

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

$x + 1 = 0$

$x - 1 = 0$

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

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

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

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

Solve ${(x + 3)}^{2} = 0$ for $x$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $2 (x + 1) (x - 1) {(x + 3)}^{2} = 0$ true. The multiplicity of a root is the number of times the root appears.

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

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

$x = - 3$ (Multiplicity of $2$ )

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

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

$x = - 3$ (Multiplicity of $2$ )

Step 3

Identify the Zeros and Their Multiplicities 2x^4+12x^3+16x^2-12x-18

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