Examples

y = x^{3} + 9 x^{2} + 27 x + 27

$y = x^{3} + 9 x^{2} + 27 x + 27$

Step 1

Set

x^{3} + 9 x^{2} + 27 x + 27

$x^{3} + 9 x^{2} + 27 x + 27$ equal to

0

$0$ .

x^{3} + 9 x^{2} + 27 x + 27 = 0

$x^{3} + 9 x^{2} + 27 x + 27 = 0$

Step 2

Solve for

x

$x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Regroup terms.

x^{3} + 27 + 9 x^{2} + 27 x = 0

$x^{3} + 27 + 9 x^{2} + 27 x = 0$

Step 2.1.2

Rewrite

27

$27$ as

3^{3}

$3^{3}$ .

x^{3} + 3^{3} + 9 x^{2} + 27 x = 0

$x^{3} + 3^{3} + 9 x^{2} + 27 x = 0$

Step 2.1.3

Since both terms are perfect cubes, factor using the sum of cubes formula,

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where

a = x

$a = x$ and

b = 3

$b = 3$ .

(x + 3) (x^{2} - x \cdot 3 + 3^{2}) + 9 x^{2} + 27 x = 0

$(x + 3) (x^{2} - x \cdot 3 + 3^{2}) + 9 x^{2} + 27 x = 0$

Step 2.1.4

Simplify.

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Step 2.1.4.1

Multiply

3

$3$ by

- 1

$- 1$ .

(x + 3) (x^{2} - 3 x + 3^{2}) + 9 x^{2} + 27 x = 0

$(x + 3) (x^{2} - 3 x + 3^{2}) + 9 x^{2} + 27 x = 0$

Step 2.1.4.2

Raise

3

$3$ to the power of

2

$2$ .

(x + 3) (x^{2} - 3 x + 9) + 9 x^{2} + 27 x = 0

$(x + 3) (x^{2} - 3 x + 9) + 9 x^{2} + 27 x = 0$

Step 2.1.5

Factor

$9 x$ out of

$9 x^{2} + 27 x$ .

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Step 2.1.5.1

Factor

$9 x$ out of

$9 x^{2}$ .

$(x + 3) (x^{2} - 3 x + 9) + 9 x (x) + 27 x = 0$

Step 2.1.5.2

Factor

$9 x$ out of

$27 x$ .

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

Step 2.1.5.3

Factor

$9 x$ out of

$9 x (x) + 9 x (3)$ .

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

Step 2.1.6

Factor

$x + 3$ out of

$(x + 3) (x^{2} - 3 x + 9) + 9 x (x + 3)$ .

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Step 2.1.6.1

Factor

$x + 3$ out of

$9 x (x + 3)$ .

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

Step 2.1.6.2

Factor

$x + 3$ out of

$(x + 3) (x^{2} - 3 x + 9) + (x + 3) (9 x)$ .

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

Step 2.1.7

Add

$- 3 x$ and

$9 x$ .

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

Step 2.1.8

Factor using the perfect square rule.

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Step 2.1.8.1

Rewrite

$9$ as

$3^{2}$ .

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

Step 2.1.8.2

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

$6 x = 2 \cdot x \cdot 3$

Step 2.1.8.3

Rewrite the polynomial.

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

Step 2.1.8.4

Factor using the perfect square trinomial rule

$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where

$a = x$ and

$b = 3$ .

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

Step 2.2

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$x + 3 = 0$

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

Step 2.3

Set

$x + 3$ equal to

$0$ and solve for

$x$ .

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Step 2.3.1

Set

$x + 3$ equal to

$0$ .

$x + 3 = 0$

Step 2.3.2

Subtract

$3$ from both sides of the equation.

$x = - 3$

Step 2.4

The final solution is all the values that make

$(x + 3) {(x + 3)}^{2} = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - 3$ (Multiplicity of

$3$ )

$x = - 3$ (Multiplicity of

$3$ )

Step 3

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