Algebra Examples

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

Step 1

Regroup terms.

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

Step 2

Rewrite $27$ as $3^{3}$ .

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

Step 3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 3$ .

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

Step 4

Simplify.

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Step 4.1

Multiply $3$ by $- 1$ .

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

Step 4.2

Raise $3$ to the power of $2$ .

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

Step 5

Factor $9 x$ out of $9 x^{2} + 27 x$ .

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Step 5.1

Factor $9 x$ out of $9 x^{2}$ .

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

Step 5.2

Factor $9 x$ out of $27 x$ .

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

Step 5.3

Factor $9 x$ out of $9 x (x) + 9 x (3)$ .

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

Step 6

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

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Step 6.1

Factor $x + 3$ out of $9 x (x + 3)$ .

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

Step 6.2

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

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

Step 7

Add $- 3 x$ and $9 x$ .

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

Step 8

Factor using the perfect square rule.

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Step 8.1

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

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

Step 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 8.3

Rewrite the polynomial.

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

Step 8.4

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

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