Simplify (2c^2+9c)/(4c^2-1)-5/(4c^2-1)

$\frac{2 c^{2} + 9 c}{4 c^{2} - 1} - \frac{5}{4 c^{2} - 1}$

Combine the numerators over the common denominator.

$\frac{2 c^{2} + 9 c - 5}{4 c^{2} - 1}$

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 9$ .

Factor $9$ out of $9 c$ .

$\frac{2 c^{2} + 9 (c) - 5}{4 c^{2} - 1}$

Rewrite $9$ as $- 1$ plus $10$

$\frac{2 c^{2} + (- 1 + 10) c - 5}{4 c^{2} - 1}$

Apply the distributive property.

$\frac{2 c^{2} - 1 c + 10 c - 5}{4 c^{2} - 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 c^{2} - 1 c) + 10 c - 5}{4 c^{2} - 1}$

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

$\frac{c (2 c - 1) + 5 (2 c - 1)}{4 c^{2} - 1}$

Factor the polynomial by factoring out the greatest common factor, $2 c - 1$ .

$\frac{(2 c - 1) (c + 5)}{4 c^{2} - 1}$

Simplify the denominator.

Rewrite $4 c^{2}$ as ${(2 c)}^{2}$ .

$\frac{(2 c - 1) (c + 5)}{{(2 c)}^{2} - 1}$

Rewrite $1$ as $1^{2}$ .

$\frac{(2 c - 1) (c + 5)}{{(2 c)}^{2} - 1^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 c$ and $b = 1$ .

$\frac{(2 c - 1) (c + 5)}{(2 c + 1) (2 c - 1)}$

Cancel the common factor of $2 c - 1$ .

Cancel the common factor.

$\frac{(2 c - 1) (c + 5)}{(2 c + 1) (2 c - 1)}$

Rewrite the expression.

$\frac{c + 5}{2 c + 1}$

Simplify (2c^2+9c)/(4c^2-1)-5/(4c^2-1)

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