Algebra Examples

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

Step 1

Factor by grouping.

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Step 1.1

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 $b = 5$ .

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Step 1.1.1

Factor $5$ out of $5 n$ .

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

Step 1.1.2

Rewrite $5$ as $2$ plus $3$

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

Step 1.1.3

Apply the distributive property.

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

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

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

Step 1.2.2

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

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

Step 1.3

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

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

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $9$ as $3^{2}$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Multiply $2$ by $- 1$ .

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

Step 3

Simplify terms.

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Step 3.1

Cancel the common factor of $n + 1$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Cancel the common factor of $2 n + 3$ and $3 + 2 n$ .

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Step 3.2.1

Reorder terms.

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 3.3

Multiply $2 n - 3$ by $\frac{1}{3 - 2 n}$ .

$\frac{2 n - 3}{3 - 2 n}$

Step 3.4

Cancel the common factor of $2 n - 3$ and $3 - 2 n$ .

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Step 3.4.1

Factor $- 1$ out of $2 n$ .

$\frac{- (- 2 n) - 3}{3 - 2 n}$

Step 3.4.2

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

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

Step 3.4.3

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

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

Step 3.4.4

Rewrite $- (- 2 n + 3)$ as $- 1 (- 2 n + 3)$ .

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

Step 3.4.5

Reorder terms.

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

Step 3.4.6

Cancel the common factor.

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

Step 3.4.7

Divide $- 1$ by $1$ .

$- 1$