Simplify (4x+2)/(12x^2+14x+4)

$\frac{4 x + 2}{12 x^{2} + 14 x + 4}$

Step 1

Cancel the common factor of $4 x + 2$ and $12 x^{2} + 14 x + 4$ .

Factor $2$ out of $4 x$ .

$\frac{2 (2 x) + 2}{12 x^{2} + 14 x + 4}$

Factor $2$ out of $2$ .

$\frac{2 (2 x) + 2 (1)}{12 x^{2} + 14 x + 4}$

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

$\frac{2 (2 x + 1)}{12 x^{2} + 14 x + 4}$

Cancel the common factors.

Factor $2$ out of $12 x^{2}$ .

$\frac{2 (2 x + 1)}{2 (6 x^{2}) + 14 x + 4}$

Factor $2$ out of $14 x$ .

$\frac{2 (2 x + 1)}{2 (6 x^{2}) + 2 (7 x) + 4}$

Factor $2$ out of $2 (6 x^{2}) + 2 (7 x)$ .

$\frac{2 (2 x + 1)}{2 (6 x^{2} + 7 x) + 4}$

Factor $2$ out of $4$ .

$\frac{2 (2 x + 1)}{2 (6 x^{2} + 7 x) + 2 (2)}$

Factor $2$ out of $2 (6 x^{2} + 7 x) + 2 (2)$ .

$\frac{2 (2 x + 1)}{2 (6 x^{2} + 7 x + 2)}$

Cancel the common factor.

$\frac{2 (2 x + 1)}{2 (6 x^{2} + 7 x + 2)}$

Rewrite the expression.

$\frac{2 x + 1}{6 x^{2} + 7 x + 2}$

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 $a \cdot c = 6 \cdot 2 = 12$ and whose sum is $b = 7$ .

Factor $7$ out of $7 x$ .

$\frac{2 x + 1}{6 x^{2} + 7 (x) + 2}$

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

$\frac{2 x + 1}{6 x^{2} + (3 + 4) x + 2}$

Apply the distributive property.

$\frac{2 x + 1}{6 x^{2} + 3 x + 4 x + 2}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{2 x + 1}{(6 x^{2} + 3 x) + 4 x + 2}$

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

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

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

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

Step 3

Cancel the common factor of $2 x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{3 x + 2}$

Simplify (4x+2)/(12x^2+14x+4)

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