Finite Math Examples

$\frac{\frac{\frac{12 y^{2} - 7 y - 12}{12 y^{2} + 25 y + 12}}{12 y^{2} - 25 y + 12}}{16 y^{2} - 24 y + 9}$

Step 1

Multiply the numerator by the reciprocal of the denominator.

$\frac{\frac{12 y^{2} - 7 y - 12}{12 y^{2} + 25 y + 12}}{12 y^{2} - 25 y + 12} \cdot \frac{1}{16 y^{2} - 24 y + 9}$

Step 2

Multiply the numerator by the reciprocal of the denominator.

$\frac{12 y^{2} - 7 y - 12}{12 y^{2} + 25 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

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 $a \cdot c = 12 \cdot - 12 = - 144$ and whose sum is $b = - 7$ .

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Step 3.1.1

Factor $- 7$ out of $- 7 y$ .

$\frac{12 y^{2} - 7 (y) - 12}{12 y^{2} + 25 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 3.1.2

Rewrite $- 7$ as $9$ plus $- 16$

$\frac{12 y^{2} + (9 - 16) y - 12}{12 y^{2} + 25 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 3.1.3

Apply the distributive property.

$\frac{12 y^{2} + 9 y - 16 y - 12}{12 y^{2} + 25 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

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.

$\frac{(12 y^{2} + 9 y) - 16 y - 12}{12 y^{2} + 25 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 3.2.2

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

$\frac{3 y (4 y + 3) - 4 (4 y + 3)}{12 y^{2} + 25 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $4 y + 3$ .

$\frac{(4 y + 3) (3 y - 4)}{12 y^{2} + 25 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 4

Factor by grouping.

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Step 4.1

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 $a \cdot c = 12 \cdot 12 = 144$ and whose sum is $b = 25$ .

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Step 4.1.1

Factor $25$ out of $25 y$ .

$\frac{(4 y + 3) (3 y - 4)}{12 y^{2} + 25 (y) + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 4.1.2

Rewrite $25$ as $9$ plus $16$

$\frac{(4 y + 3) (3 y - 4)}{12 y^{2} + (9 + 16) y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 4.1.3

Apply the distributive property.

$\frac{(4 y + 3) (3 y - 4)}{12 y^{2} + 9 y + 16 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 4.2

Factor out the greatest common factor from each group.

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Step 4.2.1

Group the first two terms and the last two terms.

$\frac{(4 y + 3) (3 y - 4)}{(12 y^{2} + 9 y) + 16 y + 12} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 4.2.2

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

$\frac{(4 y + 3) (3 y - 4)}{3 y (4 y + 3) + 4 (4 y + 3)} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 4.3

Factor the polynomial by factoring out the greatest common factor, $4 y + 3$ .

$\frac{(4 y + 3) (3 y - 4)}{(4 y + 3) (3 y + 4)} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 5

Cancel the common factor of $4 y + 3$ .

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Step 5.1

Cancel the common factor.

$\frac{(4 y + 3) (3 y - 4)}{(4 y + 3) (3 y + 4)} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 5.2

Rewrite the expression.

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{12 y^{2} - 25 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 6

Factor by grouping.

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Step 6.1

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 $a \cdot c = 12 \cdot 12 = 144$ and whose sum is $b = - 25$ .

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Step 6.1.1

Factor $- 25$ out of $- 25 y$ .

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{12 y^{2} - 25 (y) + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 6.1.2

Rewrite $- 25$ as $- 9$ plus $- 16$

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{12 y^{2} + (- 9 - 16) y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 6.1.3

Apply the distributive property.

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{12 y^{2} - 9 y - 16 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 6.2

Factor out the greatest common factor from each group.

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Step 6.2.1

Group the first two terms and the last two terms.

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{(12 y^{2} - 9 y) - 16 y + 12} \frac{1}{16 y^{2} - 24 y + 9}$

Step 6.2.2

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

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{3 y (4 y - 3) - 4 (4 y - 3)} \frac{1}{16 y^{2} - 24 y + 9}$

Step 6.3

Factor the polynomial by factoring out the greatest common factor, $4 y - 3$ .

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{(4 y - 3) (3 y - 4)} \frac{1}{16 y^{2} - 24 y + 9}$

Step 7

Simplify terms.

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Step 7.1

Cancel the common factor of $3 y - 4$ .

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Step 7.1.1

Factor $3 y - 4$ out of $(4 y - 3) (3 y - 4)$ .

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{(3 y - 4) (4 y - 3)} \frac{1}{16 y^{2} - 24 y + 9}$

Step 7.1.2

Cancel the common factor.

$\frac{3 y - 4}{3 y + 4} \cdot \frac{1}{(3 y - 4) (4 y - 3)} \frac{1}{16 y^{2} - 24 y + 9}$

Step 7.1.3

Rewrite the expression.

$\frac{1}{3 y + 4} \cdot \frac{1}{4 y - 3} \frac{1}{16 y^{2} - 24 y + 9}$

Step 7.2

Multiply $\frac{1}{3 y + 4}$ by $\frac{1}{4 y - 3}$ .

$\frac{1}{(3 y + 4) (4 y - 3)} \cdot \frac{1}{16 y^{2} - 24 y + 9}$

Step 8

Factor using the perfect square rule.

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Step 8.1

Rewrite $16 y^{2}$ as ${(4 y)}^{2}$ .

$\frac{1}{(3 y + 4) (4 y - 3)} \cdot \frac{1}{{(4 y)}^{2} - 24 y + 9}$

Step 8.2

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

$\frac{1}{(3 y + 4) (4 y - 3)} \cdot \frac{1}{{(4 y)}^{2} - 24 y + 3^{2}}$

Step 8.3

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

$24 y = 2 \cdot (4 y) \cdot 3$

Step 8.4

Rewrite the polynomial.

$\frac{1}{(3 y + 4) (4 y - 3)} \cdot \frac{1}{{(4 y)}^{2} - 2 \cdot (4 y) \cdot 3 + 3^{2}}$

Step 8.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 4 y$ and $b = 3$ .

$\frac{1}{(3 y + 4) (4 y - 3)} \cdot \frac{1}{{(4 y - 3)}^{2}}$

Step 9

Combine.

$\frac{1 \cdot 1}{(3 y + 4) (4 y - 3) {(4 y - 3)}^{2}}$

Step 10

Multiply $4 y - 3$ by ${(4 y - 3)}^{2}$ by adding the exponents.

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Step 10.1

Move ${(4 y - 3)}^{2}$ .

$\frac{1 \cdot 1}{(3 y + 4) ({(4 y - 3)}^{2} (4 y - 3))}$

Step 10.2

Multiply ${(4 y - 3)}^{2}$ by $4 y - 3$ .

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Step 10.2.1

Raise $4 y - 3$ to the power of $1$ .

$\frac{1 \cdot 1}{(3 y + 4) ({(4 y - 3)}^{2} {(4 y - 3)}^{1})}$

Step 10.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{1 \cdot 1}{(3 y + 4) {(4 y - 3)}^{2 + 1}}$

Step 10.3

Add $2$ and $1$ .

$\frac{1 \cdot 1}{(3 y + 4) {(4 y - 3)}^{3}}$

Step 11

Multiply $1$ by $1$ .

$\frac{1}{(3 y + 4) {(4 y - 3)}^{3}}$