Simplify ((6n^2-23n+7)/(8n^3-28n^2))÷((1-9n^2)/(2n^2-10n))

$\frac{6 n^{2} - 23 n + 7}{8 n^{3} - 28 n^{2}} \div \frac{1 - 9 n^{2}}{2 n^{2} - 10 n}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{6 n^{2} - 23 n + 7}{8 n^{3} - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

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 $a \cdot c = 6 \cdot 7 = 42$ and whose sum is $b = - 23$ .

Factor $- 23$ out of $- 23 n$ .

$\frac{6 n^{2} - 23 (n) + 7}{8 n^{3} - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Rewrite $- 23$ as $- 2$ plus $- 21$

$\frac{6 n^{2} + (- 2 - 21) n + 7}{8 n^{3} - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Apply the distributive property.

$\frac{6 n^{2} - 2 n - 21 n + 7}{8 n^{3} - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(6 n^{2} - 2 n) - 21 n + 7}{8 n^{3} - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

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

$\frac{2 n (3 n - 1) - 7 (3 n - 1)}{8 n^{3} - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Factor the polynomial by factoring out the greatest common factor, $3 n - 1$ .

$\frac{(3 n - 1) (2 n - 7)}{8 n^{3} - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Step 3

Simplify terms.

Factor $4 n^{2}$ out of $8 n^{3} - 28 n^{2}$ .

Factor $4 n^{2}$ out of $8 n^{3}$ .

$\frac{(3 n - 1) (2 n - 7)}{4 n^{2} (2 n) - 28 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Factor $4 n^{2}$ out of $- 28 n^{2}$ .

$\frac{(3 n - 1) (2 n - 7)}{4 n^{2} (2 n) + 4 n^{2} (- 7)} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Factor $4 n^{2}$ out of $4 n^{2} (2 n) + 4 n^{2} (- 7)$ .

$\frac{(3 n - 1) (2 n - 7)}{4 n^{2} (2 n - 7)} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

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

Cancel the common factor.

$\frac{(3 n - 1) (2 n - 7)}{4 n^{2} (2 n - 7)} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Rewrite the expression.

$\frac{3 n - 1}{4 n^{2}} \cdot \frac{2 n^{2} - 10 n}{1 - 9 n^{2}}$

Factor $2 n$ out of $2 n^{2} - 10 n$ .

Factor $2 n$ out of $2 n^{2}$ .

$\frac{3 n - 1}{4 n^{2}} \cdot \frac{2 n (n) - 10 n}{1 - 9 n^{2}}$

Factor $2 n$ out of $- 10 n$ .

$\frac{3 n - 1}{4 n^{2}} \cdot \frac{2 n (n) + 2 n (- 5)}{1 - 9 n^{2}}$

Factor $2 n$ out of $2 n (n) + 2 n (- 5)$ .

$\frac{3 n - 1}{4 n^{2}} \cdot \frac{2 n (n - 5)}{1 - 9 n^{2}}$

Step 4

Simplify the denominator.

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

$\frac{3 n - 1}{4 n^{2}} \cdot \frac{2 n (n - 5)}{1^{2} - 9 n^{2}}$

Rewrite $9 n^{2}$ as ${(3 n)}^{2}$ .

$\frac{3 n - 1}{4 n^{2}} \cdot \frac{2 n (n - 5)}{1^{2} - {(3 n)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 3 n$ .

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

Multiply $3$ by $- 1$ .

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

Step 5

Simplify terms.

Cancel the common factor of $2 n$ .

Factor $2 n$ out of $4 n^{2}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{3 n - 1}{2 n}$ by $\frac{n - 5}{(1 + 3 n) (1 - 3 n)}$ .

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

Cancel the common factor of $3 n - 1$ and $1 - 3 n$ .

Factor $- 1$ out of $3 n$ .

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

Rewrite $- 1$ as $- 1 (1)$ .

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

Factor $- 1$ out of $- (- 3 n) - 1 (1)$ .

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

Rewrite $- (- 3 n + 1)$ as $- 1 (- 3 n + 1)$ .

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

Reorder terms.

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Simplify ((6n^2-23n+7)/(8n^3-28n^2))÷((1-9n^2)/(2n^2-10n))

Related Questions

Support LCMGCF.com by disabling your adblocker.