Simplify ((2x^2+3x+1)/(x^2+2x-15))÷((x^2+6x+5)/(2x^2-7x+3))

$\frac{2 x^{2} + 3 x + 1}{x^{2} + 2 x - 15} \div \frac{x^{2} + 6 x + 5}{2 x^{2} - 7 x + 3}$

To divide by a fraction, multiply by its reciprocal.

$\frac{2 x^{2} + 3 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$\frac{2 x^{2} + 3 (x) + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Rewrite $3$ as $1$ plus $2$

$\frac{2 x^{2} + (1 + 2) x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Apply the distributive property.

$\frac{2 x^{2} + 1 x + 2 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Multiply $x$ by $1$ .

$\frac{2 x^{2} + x + 2 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x^{2} + x) + 2 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

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

$\frac{x (2 x + 1) + 1 (2 x + 1)}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

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

$\frac{(2 x + 1) (x + 1)}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Factor $x^{2} + 2 x - 15$ 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 $2$ .

$- 3, 5$

Write the factored form using these integers.

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

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 $- 7 x$ .

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{2 x^{2} - 7 (x) + 3}{x^{2} + 6 x + 5}$

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

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{2 x^{2} + (- 1 - 6) x + 3}{x^{2} + 6 x + 5}$

Apply the distributive property.

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{2 x^{2} - 1 x - 6 x + 3}{x^{2} + 6 x + 5}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{(2 x^{2} - 1 x) - 6 x + 3}{x^{2} + 6 x + 5}$

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

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{x (2 x - 1) - 3 (2 x - 1)}{x^{2} + 6 x + 5}$

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

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{(2 x - 1) (x - 3)}{x^{2} + 6 x + 5}$

Factor $x^{2} + 6 x + 5$ 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 $5$ and whose sum is $6$ .

$1, 5$

Write the factored form using these integers.

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

Simplify terms.

Cancel the common factor of $x + 1$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Raise $x + 5$ to the power of $1$ .

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

Raise $x + 5$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Simplify ((2x^2+3x+1)/(x^2+2x-15))÷((x^2+6x+5)/(2x^2-7x+3))

Related Questions

Support LCMGCF.com by disabling your adblocker.