Simplify ((16a^2-24a+9)/(4a^2+17a-15))÷((16a^2-9)/(4a^2+11a+6))

$\frac{16 a^{2} - 24 a + 9}{4 a^{2} + 17 a - 15} \div \frac{16 a^{2} - 9}{4 a^{2} + 11 a + 6}$

To divide by a fraction, multiply by its reciprocal.

$\frac{16 a^{2} - 24 a + 9}{4 a^{2} + 17 a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Factor using the perfect square rule.

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

$\frac{{(4 a)}^{2} - 24 a + 9}{4 a^{2} + 17 a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

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

$\frac{{(4 a)}^{2} - 24 a + 3^{2}}{4 a^{2} + 17 a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

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

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

Rewrite the polynomial.

$\frac{{(4 a)}^{2} - 2 \cdot (4 a) \cdot 3 + 3^{2}}{4 a^{2} + 17 a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

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

$\frac{{(4 a - 3)}^{2}}{4 a^{2} + 17 a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Factor by grouping.

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 $b = 17$ .

Factor $17$ out of $17 a$ .

$\frac{{(4 a - 3)}^{2}}{4 a^{2} + 17 (a) - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Rewrite $17$ as $- 3$ plus $20$

$\frac{{(4 a - 3)}^{2}}{4 a^{2} + (- 3 + 20) a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Apply the distributive property.

$\frac{{(4 a - 3)}^{2}}{4 a^{2} - 3 a + 20 a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{{(4 a - 3)}^{2}}{(4 a^{2} - 3 a) + 20 a - 15} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

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

$\frac{{(4 a - 3)}^{2}}{a (4 a - 3) + 5 (4 a - 3)} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

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

$\frac{{(4 a - 3)}^{2}}{(4 a - 3) (a + 5)} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Cancel the common factor of ${(4 a - 3)}^{2}$ and $4 a - 3$ .

Factor $4 a - 3$ out of ${(4 a - 3)}^{2}$ .

$\frac{(4 a - 3) (4 a - 3)}{(4 a - 3) (a + 5)} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Cancel the common factors.

Cancel the common factor.

$\frac{(4 a - 3) (4 a - 3)}{(4 a - 3) (a + 5)} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Rewrite the expression.

$\frac{4 a - 3}{a + 5} \cdot \frac{4 a^{2} + 11 a + 6}{16 a^{2} - 9}$

Factor by grouping.

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 $a \cdot c = 4 \cdot 6 = 24$ and whose sum is $b = 11$ .

Factor $11$ out of $11 a$ .

$\frac{4 a - 3}{a + 5} \cdot \frac{4 a^{2} + 11 (a) + 6}{16 a^{2} - 9}$

Rewrite $11$ as $3$ plus $8$

$\frac{4 a - 3}{a + 5} \cdot \frac{4 a^{2} + (3 + 8) a + 6}{16 a^{2} - 9}$

Apply the distributive property.

$\frac{4 a - 3}{a + 5} \cdot \frac{4 a^{2} + 3 a + 8 a + 6}{16 a^{2} - 9}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{4 a - 3}{a + 5} \cdot \frac{(4 a^{2} + 3 a) + 8 a + 6}{16 a^{2} - 9}$

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

$\frac{4 a - 3}{a + 5} \cdot \frac{a (4 a + 3) + 2 (4 a + 3)}{16 a^{2} - 9}$

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

$\frac{4 a - 3}{a + 5} \cdot \frac{(4 a + 3) (a + 2)}{16 a^{2} - 9}$

Simplify the denominator.

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

$\frac{4 a - 3}{a + 5} \cdot \frac{(4 a + 3) (a + 2)}{{(4 a)}^{2} - 9}$

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

$\frac{4 a - 3}{a + 5} \cdot \frac{(4 a + 3) (a + 2)}{{(4 a)}^{2} - 3^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 a$ and $b = 3$ .

$\frac{4 a - 3}{a + 5} \cdot \frac{(4 a + 3) (a + 2)}{(4 a + 3) (4 a - 3)}$

Simplify terms.

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

Factor $4 a - 3$ out of $(4 a + 3) (4 a - 3)$ .

$\frac{4 a - 3}{a + 5} \cdot \frac{(4 a + 3) (a + 2)}{(4 a - 3) (4 a + 3)}$

Cancel the common factor.

$\frac{4 a - 3}{a + 5} \cdot \frac{(4 a + 3) (a + 2)}{(4 a - 3) (4 a + 3)}$

Rewrite the expression.

$\frac{1}{a + 5} \cdot \frac{(4 a + 3) (a + 2)}{4 a + 3}$

Multiply $\frac{1}{a + 5}$ by $\frac{(4 a + 3) (a + 2)}{4 a + 3}$ .

$\frac{(4 a + 3) (a + 2)}{(a + 5) (4 a + 3)}$

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

Cancel the common factor.

$\frac{(4 a + 3) (a + 2)}{(a + 5) (4 a + 3)}$

Rewrite the expression.

$\frac{a + 2}{a + 5}$

Simplify ((16a^2-24a+9)/(4a^2+17a-15))÷((16a^2-9)/(4a^2+11a+6))

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