Simplify (2x^2+7x+3)/(x-4)*(x^2-16)/(x^2+8x+15)

$\frac{2 x^{2} + 7 x + 3}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

Step 1

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 $b = 7$ .

Factor $7$ out of $7 x$ .

$\frac{2 x^{2} + 7 (x) + 3}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

Rewrite $7$ as $1$ plus $6$

$\frac{2 x^{2} + (1 + 6) x + 3}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

Apply the distributive property.

$\frac{2 x^{2} + 1 x + 6 x + 3}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

Multiply $x$ by $1$ .

$\frac{2 x^{2} + x + 6 x + 3}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x^{2} + x) + 6 x + 3}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

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

$\frac{x (2 x + 1) + 3 (2 x + 1)}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$\frac{(2 x + 1) (x + 3)}{x - 4} \cdot \frac{x^{2} - 16}{x^{2} + 8 x + 15}$

Step 2

Simplify the numerator.

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

$\frac{(2 x + 1) (x + 3)}{x - 4} \cdot \frac{x^{2} - 4^{2}}{x^{2} + 8 x + 15}$

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

$\frac{(2 x + 1) (x + 3)}{x - 4} \cdot \frac{(x + 4) (x - 4)}{x^{2} + 8 x + 15}$

Step 3

Factor $x^{2} + 8 x + 15$ 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 $15$ and whose sum is $8$ .

$3, 5$

Write the factored form using these integers.

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

Step 4

Simplify terms.

Cancel the common factor of $x + 3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{2 x + 1}{x - 4} \cdot \frac{(x + 4) (x - 4)}{x + 5}$

Cancel the common factor of $x - 4$ .

Factor $x - 4$ out of $(x + 4) (x - 4)$ .

$\frac{2 x + 1}{x - 4} \cdot \frac{(x - 4) (x + 4)}{x + 5}$

Cancel the common factor.

$\frac{2 x + 1}{x - 4} \cdot \frac{(x - 4) (x + 4)}{x + 5}$

Rewrite the expression.

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

Multiply $2 x + 1$ by $\frac{x + 4}{x + 5}$ .

$\frac{(2 x + 1) (x + 4)}{x + 5}$

Simplify (2x^2+7x+3)/(x-4)*(x^2-16)/(x^2+8x+15)

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