Simplify ((2x^2+5x+3)/(x^2-3x-4))÷((4x^2+2x-6)/(x^2-8x+16))

$\frac{2 x^{2} + 5 x + 3}{x^{2} - 3 x - 4} \div \frac{4 x^{2} + 2 x - 6}{x^{2} - 8 x + 16}$

Step 1

To divide by a fraction, multiply by its reciprocal.

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

Step 2

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

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

Rewrite $5$ as $2$ plus $3$

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

Apply the distributive property.

$\frac{2 x^{2} + 2 x + 3 x + 3}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 8 x + 16}{4 x^{2} + 2 x - 6}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$\frac{2 x (x + 1) + 3 (x + 1)}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 8 x + 16}{4 x^{2} + 2 x - 6}$

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$\frac{(x + 1) (2 x + 3)}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 8 x + 16}{4 x^{2} + 2 x - 6}$

Step 3

Factor $x^{2} - 3 x - 4$ using the AC method.

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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$\frac{(x + 1) (2 x + 3)}{(x - 4) (x + 1)} \cdot \frac{x^{2} - 8 x + 16}{4 x^{2} + 2 x - 6}$

Step 4

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{(x + 1) (2 x + 3)}{(x - 4) (x + 1)} \cdot \frac{x^{2} - 8 x + 16}{4 x^{2} + 2 x - 6}$

Rewrite the expression.

$\frac{2 x + 3}{x - 4} \cdot \frac{x^{2} - 8 x + 16}{4 x^{2} + 2 x - 6}$

Step 5

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$\frac{2 x + 3}{x - 4} \cdot \frac{x^{2} - 8 x + 4^{2}}{4 x^{2} + 2 x - 6}$

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$\frac{2 x + 3}{x - 4} \cdot \frac{x^{2} - 2 \cdot x \cdot 4 + 4^{2}}{4 x^{2} + 2 x - 6}$

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

$\frac{2 x + 3}{x - 4} \cdot \frac{{(x - 4)}^{2}}{4 x^{2} + 2 x - 6}$

Step 6

Simplify the denominator.

Factor $2$ out of $4 x^{2} + 2 x - 6$ .

Factor $2$ out of $4 x^{2}$ .

$\frac{2 x + 3}{x - 4} \cdot \frac{{(x - 4)}^{2}}{2 (2 x^{2}) + 2 x - 6}$

Factor $2$ out of $2 x$ .

$\frac{2 x + 3}{x - 4} \cdot \frac{{(x - 4)}^{2}}{2 (2 x^{2}) + 2 (x) - 6}$

Factor $2$ out of $- 6$ .

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

Factor $2$ out of $2 (2 x^{2}) + 2 x$ .

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

Factor $2$ out of $2 (2 x^{2} + x) + 2 \cdot - 3$ .

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

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

Rewrite $1$ as $- 2$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

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

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

Step 7

Reduce the expression by cancelling the common factors.

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

Factor $2 x + 3$ out of $2 (x - 1) (2 x + 3)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{x - 4} \cdot \frac{{(x - 4)}^{2}}{2 (x - 1)}$

Cancel the common factor of $x - 4$ .

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

$\frac{1}{x - 4} \cdot \frac{(x - 4) (x - 4)}{2 (x - 1)}$

Cancel the common factor.

$\frac{1}{x - 4} \cdot \frac{(x - 4) (x - 4)}{2 (x - 1)}$

Rewrite the expression.

$\frac{x - 4}{2 (x - 1)}$

Simplify ((2x^2+5x+3)/(x^2-3x-4))÷((4x^2+2x-6)/(x^2-8x+16))

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