Simplify ((2c^2-5c-3)/(c^2+4c-21))÷(2c+1)

$\frac{2 c^{2} - 5 c - 3}{c^{2} + 4 c - 21} \div (2 c + 1)$

Step 1

Rewrite the division as a fraction.

$\frac{\frac{2 c^{2} - 5 c - 3}{c^{2} + 4 c - 21}}{2 c + 1}$

Step 2

Multiply the numerator by the reciprocal of the denominator.

$\frac{2 c^{2} - 5 c - 3}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

Step 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 $b = - 5$ .

Factor $- 5$ out of $- 5 c$ .

$\frac{2 c^{2} - 5 (c) - 3}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

Rewrite $- 5$ as $1$ plus $- 6$

$\frac{2 c^{2} + (1 - 6) c - 3}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

Apply the distributive property.

$\frac{2 c^{2} + 1 c - 6 c - 3}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

Multiply $c$ by $1$ .

$\frac{2 c^{2} + c - 6 c - 3}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 c^{2} + c) - 6 c - 3}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

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

$\frac{c (2 c + 1) - 3 (2 c + 1)}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

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

$\frac{(2 c + 1) (c - 3)}{c^{2} + 4 c - 21} \cdot \frac{1}{2 c + 1}$

Step 4

Factor $c^{2} + 4 c - 21$ 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 $- 21$ and whose sum is $4$ .

$- 3, 7$

Write the factored form using these integers.

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

Step 5

Reduce the expression by cancelling the common factors.

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{c - 3}{(c - 3) (c + 7)}$

Cancel the common factor of $c - 3$ .

Cancel the common factor.

$\frac{c - 3}{(c - 3) (c + 7)}$

Rewrite the expression.

$\frac{1}{c + 7}$

Simplify ((2c^2-5c-3)/(c^2+4c-21))÷(2c+1)

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