Pre-Algebra Examples

$216 x^{6} + 19 x^{3} - 1$

Step 1

Rewrite $x^{6}$ as ${(x^{3})}^{2}$ .

$216 {(x^{3})}^{2} + 19 x^{3} - 1$

Step 2

Let $u = x^{3}$ . Substitute $u$ for all occurrences of $x^{3}$ .

$216 u^{2} + 19 u - 1$

Step 3

Factor by grouping.

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Step 3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 216 \cdot - 1 = - 216$ and whose sum is $b = 19$ .

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Step 3.1.1

Factor $19$ out of $19 u$ .

$216 u^{2} + 19 (u) - 1$

Step 3.1.2

Rewrite $19$ as $- 8$ plus $27$

$216 u^{2} + (- 8 + 27) u - 1$

Step 3.1.3

Apply the distributive property.

$216 u^{2} - 8 u + 27 u - 1$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$(216 u^{2} - 8 u) + 27 u - 1$

Step 3.2.2

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

$8 u (27 u - 1) + 1 (27 u - 1)$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $27 u - 1$ .

$(27 u - 1) (8 u + 1)$

Step 4

Replace all occurrences of $u$ with $x^{3}$ .

$(27 x^{3} - 1) (8 x^{3} + 1)$

Step 5

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

$({(3 x)}^{3} - 1) (8 x^{3} + 1)$

Step 6

Rewrite $1$ as $1^{3}$ .

$({(3 x)}^{3} - 1^{3}) (8 x^{3} + 1)$

Step 7

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 3 x$ and $b = 1$ .

$(3 x - 1) ({(3 x)}^{2} + 3 x \cdot 1 + 1^{2}) (8 x^{3} + 1)$

Step 8

Simplify.

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Step 8.1

Apply the product rule to $3 x$ .

$(3 x - 1) (3^{2} x^{2} + 3 x \cdot 1 + 1^{2}) (8 x^{3} + 1)$

Step 8.2

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

$(3 x - 1) (9 x^{2} + 3 x \cdot 1 + 1^{2}) (8 x^{3} + 1)$

Step 8.3

Multiply $3$ by $1$ .

$(3 x - 1) (9 x^{2} + 3 x + 1^{2}) (8 x^{3} + 1)$

Step 8.4

One to any power is one.

$(3 x - 1) (9 x^{2} + 3 x + 1) (8 x^{3} + 1)$

Step 9

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

$(3 x - 1) (9 x^{2} + 3 x + 1) ({(2 x)}^{3} + 1)$

Step 10

Rewrite $1$ as $1^{3}$ .

$(3 x - 1) (9 x^{2} + 3 x + 1) ({(2 x)}^{3} + 1^{3})$

Step 11

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 $a = 2 x$ and $b = 1$ .

$(3 x - 1) (9 x^{2} + 3 x + 1) ((2 x + 1) ({(2 x)}^{2} - (2 x) \cdot 1 + 1^{2}))$

Step 12

Factor.

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Step 12.1

Simplify.

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Step 12.1.1

Apply the product rule to $2 x$ .

$(3 x - 1) (9 x^{2} + 3 x + 1) ((2 x + 1) (2^{2} x^{2} - (2 x) \cdot 1 + 1^{2}))$

Step 12.1.2

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

$(3 x - 1) (9 x^{2} + 3 x + 1) ((2 x + 1) (4 x^{2} - (2 x) \cdot 1 + 1^{2}))$

Step 12.1.3

Multiply $2$ by $- 1$ .

$(3 x - 1) (9 x^{2} + 3 x + 1) ((2 x + 1) (4 x^{2} - 2 x \cdot 1 + 1^{2}))$

Step 12.1.4

Multiply $- 2$ by $1$ .

$(3 x - 1) (9 x^{2} + 3 x + 1) ((2 x + 1) (4 x^{2} - 2 x + 1^{2}))$

Step 12.1.5

One to any power is one.

$(3 x - 1) (9 x^{2} + 3 x + 1) ((2 x + 1) (4 x^{2} - 2 x + 1))$

Step 12.2

Remove unnecessary parentheses.

$(3 x - 1) (9 x^{2} + 3 x + 1) (2 x + 1) (4 x^{2} - 2 x + 1)$