Simplify (14x^2+3x-2)/(5x+15)*(x+3)/(49x^2-4)

$\frac{14 x^{2} + 3 x - 2}{5 x + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

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 $a \cdot c = 14 \cdot - 2 = - 28$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$\frac{14 x^{2} + 3 (x) - 2}{5 x + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

Rewrite $3$ as $- 4$ plus $7$

$\frac{14 x^{2} + (- 4 + 7) x - 2}{5 x + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

Apply the distributive property.

$\frac{14 x^{2} - 4 x + 7 x - 2}{5 x + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(14 x^{2} - 4 x) + 7 x - 2}{5 x + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

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

$\frac{2 x (7 x - 2) + 1 (7 x - 2)}{5 x + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

Factor the polynomial by factoring out the greatest common factor, $7 x - 2$ .

$\frac{(7 x - 2) (2 x + 1)}{5 x + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

Step 2

Factor $5$ out of $5 x + 15$ .

Factor $5$ out of $5 x$ .

$\frac{(7 x - 2) (2 x + 1)}{5 (x) + 15} \cdot \frac{x + 3}{49 x^{2} - 4}$

Factor $5$ out of $15$ .

$\frac{(7 x - 2) (2 x + 1)}{5 x + 5 \cdot 3} \cdot \frac{x + 3}{49 x^{2} - 4}$

Factor $5$ out of $5 x + 5 \cdot 3$ .

$\frac{(7 x - 2) (2 x + 1)}{5 (x + 3)} \cdot \frac{x + 3}{49 x^{2} - 4}$

Step 3

Simplify the denominator.

Rewrite $49 x^{2}$ as ${(7 x)}^{2}$ .

$\frac{(7 x - 2) (2 x + 1)}{5 (x + 3)} \cdot \frac{x + 3}{{(7 x)}^{2} - 4}$

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

$\frac{(7 x - 2) (2 x + 1)}{5 (x + 3)} \cdot \frac{x + 3}{{(7 x)}^{2} - 2^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7 x$ and $b = 2$ .

$\frac{(7 x - 2) (2 x + 1)}{5 (x + 3)} \cdot \frac{x + 3}{(7 x + 2) (7 x - 2)}$

Step 4

Simplify terms.

Cancel the common factor of $7 x - 2$ .

Factor $7 x - 2$ out of $(7 x + 2) (7 x - 2)$ .

$\frac{(7 x - 2) (2 x + 1)}{5 (x + 3)} \cdot \frac{x + 3}{(7 x - 2) (7 x + 2)}$

Cancel the common factor.

$\frac{(7 x - 2) (2 x + 1)}{5 (x + 3)} \cdot \frac{x + 3}{(7 x - 2) (7 x + 2)}$

Rewrite the expression.

$\frac{2 x + 1}{5 (x + 3)} \cdot \frac{x + 3}{7 x + 2}$

Cancel the common factor of $x + 3$ .

Factor $x + 3$ out of $5 (x + 3)$ .

$\frac{2 x + 1}{(x + 3) \cdot 5} \cdot \frac{x + 3}{7 x + 2}$

Cancel the common factor.

$\frac{2 x + 1}{(x + 3) \cdot 5} \cdot \frac{x + 3}{7 x + 2}$

Rewrite the expression.

$\frac{2 x + 1}{5} \cdot \frac{1}{7 x + 2}$

Multiply $\frac{2 x + 1}{5}$ by $\frac{1}{7 x + 2}$ .

$\frac{2 x + 1}{5 (7 x + 2)}$

Simplify (14x^2+3x-2)/(5x+15)*(x+3)/(49x^2-4)

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