Basic Math Examples

$(\frac{4 k^{2} - 4 k - 15}{k^{2} + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k}) \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 1

Factor by grouping.

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Step 1.1

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 $a \cdot c = 4 \cdot - 15 = - 60$ and whose sum is $b = - 4$ .

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Step 1.1.1

Factor $- 4$ out of $- 4 k$ .

$\frac{4 k^{2} - 4 (k) - 15}{k^{2} + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 1.1.2

Rewrite $- 4$ as $6$ plus $- 10$

$\frac{4 k^{2} + (6 - 10) k - 15}{k^{2} + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 1.1.3

Apply the distributive property.

$\frac{4 k^{2} + 6 k - 10 k - 15}{k^{2} + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$\frac{(4 k^{2} + 6 k) - 10 k - 15}{k^{2} + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 1.2.2

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

$\frac{2 k (2 k + 3) - 5 (2 k + 3)}{k^{2} + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 1.3

Factor the polynomial by factoring out the greatest common factor, $2 k + 3$ .

$\frac{(2 k + 3) (2 k - 5)}{k^{2} + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 2

Factor $k$ out of $k^{2} + 7 k$ .

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Step 2.1

Factor $k$ out of $k^{2}$ .

$\frac{(2 k + 3) (2 k - 5)}{k \cdot k + 7 k} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 2.2

Factor $k$ out of $7 k$ .

$\frac{(2 k + 3) (2 k - 5)}{k \cdot k + k \cdot 7} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 2.3

Factor $k$ out of $k \cdot k + k \cdot 7$ .

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k^{3} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 3

Simplify the denominator.

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Step 3.1

Factor $k$ out of $k^{3} + 2 k^{2} - 35 k$ .

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Step 3.1.1

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

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k \cdot k^{2} + 2 k^{2} - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 3.1.2

Factor $k$ out of $2 k^{2}$ .

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k \cdot k^{2} + k (2 k) - 35 k} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 3.1.3

Factor $k$ out of $- 35 k$ .

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k \cdot k^{2} + k (2 k) + k \cdot - 35} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 3.1.4

Factor $k$ out of $k \cdot k^{2} + k (2 k)$ .

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k^{2} + 2 k) + k \cdot - 35} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 3.1.5

Factor $k$ out of $k (k^{2} + 2 k) + k \cdot - 35$ .

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k^{2} + 2 k - 35)} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 3.2

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

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Step 3.2.1

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 $2$ .

$- 5, 7$

Step 3.2.2

Write the factored form using these integers.

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k ((k - 5) (k + 7))} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{k^{2} - 13 k + 42}{2 k^{2} - 9 k - 18}$

Step 4

Factor $k^{2} - 13 k + 42$ using the AC method.

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Step 4.1

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 $42$ and whose sum is $- 13$ .

$- 7, - 6$

Step 4.2

Write the factored form using these integers.

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{(k - 7) (k - 6)}{2 k^{2} - 9 k - 18}$

Step 5

Factor by grouping.

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Step 5.1

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 $a \cdot c = 2 \cdot - 18 = - 36$ and whose sum is $b = - 9$ .

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Step 5.1.1

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

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{(k - 7) (k - 6)}{2 k^{2} - 9 (k) - 18}$

Step 5.1.2

Rewrite $- 9$ as $3$ plus $- 12$

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{(k - 7) (k - 6)}{2 k^{2} + (3 - 12) k - 18}$

Step 5.1.3

Apply the distributive property.

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{(k - 7) (k - 6)}{2 k^{2} + 3 k - 12 k - 18}$

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{(k - 7) (k - 6)}{(2 k^{2} + 3 k) - 12 k - 18}$

Step 5.2.2

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

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{(k - 7) (k - 6)}{k (2 k + 3) - 6 (2 k + 3)}$

Step 5.3

Factor the polynomial by factoring out the greatest common factor, $2 k + 3$ .

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \div \frac{2 k - 5}{k (k - 5) (k + 7)} \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 6

To divide by a fraction, multiply by its reciprocal.

$\frac{(2 k + 3) (2 k - 5)}{k (k + 7)} \cdot \frac{k (k - 5) (k + 7)}{2 k - 5} \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7

Reduce the expression by cancelling the common factors.

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Step 7.1

Cancel the common factor of $2 k - 5$ .

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Step 7.1.1

Factor $2 k - 5$ out of $(2 k + 3) (2 k - 5)$ .

$\frac{(2 k - 5) (2 k + 3)}{k (k + 7)} \cdot \frac{k (k - 5) (k + 7)}{2 k - 5} \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7.1.2

Cancel the common factor.

$\frac{(2 k - 5) (2 k + 3)}{k (k + 7)} \cdot \frac{k (k - 5) (k + 7)}{2 k - 5} \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7.1.3

Rewrite the expression.

$\frac{2 k + 3}{k (k + 7)} (k (k - 5) (k + 7)) \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7.2

Cancel the common factor of $k (k + 7)$ .

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Step 7.2.1

Factor $k (k + 7)$ out of $k (k - 5) (k + 7)$ .

$\frac{2 k + 3}{k (k + 7)} (k (k + 7) (k - 5)) \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7.2.2

Cancel the common factor.

$\frac{2 k + 3}{k (k + 7)} (k (k + 7) (k - 5)) \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7.2.3

Rewrite the expression.

$(2 k + 3) (k - 5) \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7.3

Cancel the common factor of $2 k + 3$ .

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Step 7.3.1

Cancel the common factor.

$(2 k + 3) (k - 5) \cdot \frac{(k - 7) (k - 6)}{(2 k + 3) (k - 6)}$

Step 7.3.2

Rewrite the expression.

$(k - 5) \cdot \frac{(k - 7) (k - 6)}{k - 6}$

Step 7.4

Cancel the common factor of $k - 6$ .

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Step 7.4.1

Cancel the common factor.

$(k - 5) \cdot \frac{(k - 7) (k - 6)}{k - 6}$

Step 7.4.2

Divide $k - 7$ by $1$ .

$(k - 5) \cdot (k - 7)$

Step 8

Expand $(k - 5) (k - 7)$ using the FOIL Method.

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Step 8.1

Apply the distributive property.

$k (k - 7) - 5 (k - 7)$

Step 8.2

Apply the distributive property.

$k \cdot k + k \cdot - 7 - 5 (k - 7)$

Step 8.3

Apply the distributive property.

$k \cdot k + k \cdot - 7 - 5 k - 5 \cdot - 7$

Step 9

Simplify and combine like terms.

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Step 9.1

Simplify each term.

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Step 9.1.1

Multiply $k$ by $k$ .

$k^{2} + k \cdot - 7 - 5 k - 5 \cdot - 7$

Step 9.1.2

Move $- 7$ to the left of $k$ .

$k^{2} - 7 \cdot k - 5 k - 5 \cdot - 7$

Step 9.1.3

Multiply $- 5$ by $- 7$ .

$k^{2} - 7 k - 5 k + 35$

Step 9.2

Subtract $5 k$ from $- 7 k$ .

$k^{2} - 12 k + 35$