Pre-Algebra Examples

$\frac{9 k - 3}{1 - 9 k^{2}} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 1

Factor $3$ out of $9 k - 3$ .

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Step 1.1

Factor $3$ out of $9 k$ .

$\frac{3 (3 k) - 3}{1 - 9 k^{2}} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 1.2

Factor $3$ out of $- 3$ .

$\frac{3 (3 k) + 3 (- 1)}{1 - 9 k^{2}} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 1.3

Factor $3$ out of $3 (3 k) + 3 (- 1)$ .

$\frac{3 (3 k - 1)}{1 - 9 k^{2}} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 2

Simplify the denominator.

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Step 2.1

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

$\frac{3 (3 k - 1)}{1^{2} - 9 k^{2}} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 2.2

Rewrite $9 k^{2}$ as ${(3 k)}^{2}$ .

$\frac{3 (3 k - 1)}{1^{2} - {(3 k)}^{2}} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 2.3

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

$\frac{3 (3 k - 1)}{(1 + 3 k) (1 - (3 k))} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 2.4

Multiply $3$ by $- 1$ .

$\frac{3 (3 k - 1)}{(1 + 3 k) (1 - 3 k)} \cdot \frac{9 k^{2} + 6 k + 1}{6}$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $9 k^{2}$ as ${(3 k)}^{2}$ .

$\frac{3 (3 k - 1)}{(1 + 3 k) (1 - 3 k)} \cdot \frac{{(3 k)}^{2} + 6 k + 1}{6}$

Step 3.2

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

$\frac{3 (3 k - 1)}{(1 + 3 k) (1 - 3 k)} \cdot \frac{{(3 k)}^{2} + 6 k + 1^{2}}{6}$

Step 3.3

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

$6 k = 2 \cdot (3 k) \cdot 1$

Step 3.4

Rewrite the polynomial.

$\frac{3 (3 k - 1)}{(1 + 3 k) (1 - 3 k)} \cdot \frac{{(3 k)}^{2} + 2 \cdot (3 k) \cdot 1 + 1^{2}}{6}$

Step 3.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 3 k$ and $b = 1$ .

$\frac{3 (3 k - 1)}{(1 + 3 k) (1 - 3 k)} \cdot \frac{{(3 k + 1)}^{2}}{6}$

Step 4

Simplify terms.

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Step 4.1

Combine.

$\frac{3 (3 k - 1) {(3 k + 1)}^{2}}{(1 + 3 k) (1 - 3 k) \cdot 6}$

Step 4.2

Cancel the common factor of $3$ and $6$ .

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Step 4.2.1

Factor $3$ out of $3 (3 k - 1) {(3 k + 1)}^{2}$ .

$\frac{3 ((3 k - 1) {(3 k + 1)}^{2})}{(1 + 3 k) (1 - 3 k) \cdot 6}$

Step 4.2.2

Cancel the common factors.

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Step 4.2.2.1

Factor $3$ out of $(1 + 3 k) (1 - 3 k) \cdot 6$ .

$\frac{3 ((3 k - 1) {(3 k + 1)}^{2})}{3 ((1 + 3 k) (1 - 3 k) \cdot 2)}$

Step 4.2.2.2

Cancel the common factor.

$\frac{3 ((3 k - 1) {(3 k + 1)}^{2})}{3 ((1 + 3 k) (1 - 3 k) \cdot 2)}$

Step 4.2.2.3

Rewrite the expression.

$\frac{(3 k - 1) {(3 k + 1)}^{2}}{(1 + 3 k) (1 - 3 k) \cdot 2}$

Step 4.3

Cancel the common factor of $3 k - 1$ and $1 - 3 k$ .

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Step 4.3.1

Factor $- 1$ out of $3 k$ .

$\frac{(- (- 3 k) - 1) {(3 k + 1)}^{2}}{(1 + 3 k) (1 - 3 k) \cdot 2}$

Step 4.3.2

Rewrite $- 1$ as $- 1 (1)$ .

$\frac{(- (- 3 k) - 1 (1)) {(3 k + 1)}^{2}}{(1 + 3 k) (1 - 3 k) \cdot 2}$

Step 4.3.3

Factor $- 1$ out of $- (- 3 k) - 1 (1)$ .

$\frac{- (- 3 k + 1) {(3 k + 1)}^{2}}{(1 + 3 k) (1 - 3 k) \cdot 2}$

Step 4.3.4

Rewrite $- (- 3 k + 1)$ as $- 1 (- 3 k + 1)$ .

$\frac{- 1 (- 3 k + 1) {(3 k + 1)}^{2}}{(1 + 3 k) (1 - 3 k) \cdot 2}$

Step 4.3.5

Reorder terms.

$\frac{- 1 (- 3 k + 1) {(3 k + 1)}^{2}}{(1 + 3 k) (- 3 k + 1) \cdot 2}$

Step 4.3.6

Cancel the common factor.

$\frac{- 1 (- 3 k + 1) {(3 k + 1)}^{2}}{(1 + 3 k) (- 3 k + 1) \cdot 2}$

Step 4.3.7

Rewrite the expression.

$\frac{- 1 {(3 k + 1)}^{2}}{(1 + 3 k) \cdot 2}$

Step 4.4

Cancel the common factor of ${(3 k + 1)}^{2}$ and $1 + 3 k$ .

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Step 4.4.1

Reorder terms.

$\frac{- 1 {(3 k + 1)}^{2}}{(3 k + 1) \cdot 2}$

Step 4.4.2

Factor $3 k + 1$ out of $- 1 {(3 k + 1)}^{2}$ .

$\frac{(3 k + 1) (- 1 (3 k + 1))}{(3 k + 1) \cdot 2}$

Step 4.4.3

Cancel the common factors.

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Step 4.4.3.1

Factor $3 k + 1$ out of $(3 k + 1) \cdot 2$ .

$\frac{(3 k + 1) (- 1 (3 k + 1))}{(3 k + 1) (2)}$

Step 4.4.3.2

Cancel the common factor.

$\frac{(3 k + 1) (- 1 (3 k + 1))}{(3 k + 1) \cdot 2}$

Step 4.4.3.3

Rewrite the expression.

$\frac{- 1 (3 k + 1)}{2}$

Step 4.5

Move the negative in front of the fraction.

$- \frac{3 k + 1}{2}$