Basic Math Examples

$\frac{k^{2} - 2 k - 35}{k^{2} + 8 + 15} \cdot \frac{k^{3} - 9 k}{k^{3} - 27}$

Step 1

Factor $k^{2} - 2 k - 35$ using the AC method.

Tap for more steps...

Step 1.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 1.2

Write the factored form using these integers.

$\frac{(k - 7) (k + 5)}{k^{2} + 8 + 15} \cdot \frac{k^{3} - 9 k}{k^{3} - 27}$

Step 2

Add $8$ and $15$ .

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k^{3} - 9 k}{k^{3} - 27}$

Step 3

Simplify the numerator.

Tap for more steps...

Step 3.1

Factor $k$ out of $k^{3} - 9 k$ .

Tap for more steps...

Step 3.1.1

Factor $k$ out of $k^{3}$ .

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k \cdot k^{2} - 9 k}{k^{3} - 27}$

Step 3.1.2

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

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k \cdot k^{2} + k \cdot - 9}{k^{3} - 27}$

Step 3.1.3

Factor $k$ out of $k \cdot k^{2} + k \cdot - 9$ .

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k^{2} - 9)}{k^{3} - 27}$

Step 3.2

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

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k^{2} - 3^{2})}{k^{3} - 27}$

Step 3.3

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

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k + 3) (k - 3)}{k^{3} - 27}$

Step 4

Simplify the denominator.

Tap for more steps...

Step 4.1

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

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k + 3) (k - 3)}{k^{3} - 3^{3}}$

Step 4.2

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 $a = k$ and $b = 3$ .

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k + 3) (k - 3)}{(k - 3) (k^{2} + k \cdot 3 + 3^{2})}$

Step 4.3

Simplify.

Tap for more steps...

Step 4.3.1

Move $3$ to the left of $k$ .

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k + 3) (k - 3)}{(k - 3) (k^{2} + 3 \cdot k + 3^{2})}$

Step 4.3.2

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

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k + 3) (k - 3)}{(k - 3) (k^{2} + 3 k + 9)}$

Step 5

Simplify terms.

Tap for more steps...

Step 5.1

Cancel the common factor of $k - 3$ .

Tap for more steps...

Step 5.1.1

Cancel the common factor.

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k + 3) (k - 3)}{(k - 3) (k^{2} + 3 k + 9)}$

Step 5.1.2

Rewrite the expression.

$\frac{(k - 7) (k + 5)}{k^{2} + 23} \cdot \frac{k (k + 3)}{k^{2} + 3 k + 9}$

Step 5.2

Multiply $\frac{(k - 7) (k + 5)}{k^{2} + 23}$ by $\frac{k (k + 3)}{k^{2} + 3 k + 9}$ .

$\frac{(k - 7) (k + 5) (k (k + 3))}{(k^{2} + 23) (k^{2} + 3 k + 9)}$

Step 5.3

Reorder factors in $\frac{(k - 7) (k + 5) k (k + 3)}{(k^{2} + 23) (k^{2} + 3 k + 9)}$ .

$\frac{k (k - 7) (k + 5) (k + 3)}{(k^{2} + 23) (k^{2} + 3 k + 9)}$