Basic Math Examples

$\frac{18 n^{2} - 98}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 1

Factor $2$ out of $18 n^{2} - 98$ .

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Step 1.1

Factor $2$ out of $18 n^{2}$ .

$\frac{2 (9 n^{2}) - 98}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 1.2

Factor $2$ out of $- 98$ .

$\frac{2 (9 n^{2}) + 2 (- 49)}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 1.3

Factor $2$ out of $2 (9 n^{2}) + 2 (- 49)$ .

$\frac{2 (9 n^{2} - 49)}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 2

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

$\frac{2 ({(3 n)}^{2} - 49)}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 3

Rewrite $49$ as $7^{2}$ .

$\frac{2 ({(3 n)}^{2} - 7^{2})}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 4

Factor.

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Step 4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 n$ and $b = 7$ .

$\frac{2 ((3 n + 7) (3 n - 7))}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 4.2

Remove unnecessary parentheses.

$\frac{2 (3 n + 7) (3 n - 7)}{3 n^{2} - 10 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 5

Factor by grouping.

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Step 5.1

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 $a \cdot c = 3 \cdot 7 = 21$ and whose sum is $b = - 10$ .

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Step 5.1.1

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

$\frac{2 (3 n + 7) (3 n - 7)}{3 n^{2} - 10 (n) + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 5.1.2

Rewrite $- 10$ as $- 3$ plus $- 7$

$\frac{2 (3 n + 7) (3 n - 7)}{3 n^{2} + (- 3 - 7) n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 5.1.3

Apply the distributive property.

$\frac{2 (3 n + 7) (3 n - 7)}{3 n^{2} - 3 n - 7 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

$\frac{2 (3 n + 7) (3 n - 7)}{(3 n^{2} - 3 n) - 7 n + 7} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 5.2.2

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

$\frac{2 (3 n + 7) (3 n - 7)}{3 n (n - 1) - 7 (n - 1)} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 5.3

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

$\frac{2 (3 n + 7) (3 n - 7)}{(n - 1) (3 n - 7)} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 6

Reduce the expression $\frac{2 (3 n + 7) (3 n - 7)}{(n - 1) (3 n - 7)}$ by cancelling the common factors.

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Step 6.1

Cancel the common factor.

$\frac{2 (3 n + 7) (3 n - 7)}{(n - 1) (3 n - 7)} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 6.2

Rewrite the expression.

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{n^{2} - 11 n + 28}{4 n^{2} + 8 n - 5}$

Step 7

Factor $n^{2} - 11 n + 28$ using the AC method.

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Step 7.1

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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

Step 7.2

Write the factored form using these integers.

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{(n - 7) (n - 4)}{4 n^{2} + 8 n - 5}$

Step 8

Factor by grouping.

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Step 8.1

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 $a \cdot c = 4 \cdot - 5 = - 20$ and whose sum is $b = 8$ .

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Step 8.1.1

Factor $8$ out of $8 n$ .

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{(n - 7) (n - 4)}{4 n^{2} + 8 (n) - 5}$

Step 8.1.2

Rewrite $8$ as $- 2$ plus $10$

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{(n - 7) (n - 4)}{4 n^{2} + (- 2 + 10) n - 5}$

Step 8.1.3

Apply the distributive property.

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{(n - 7) (n - 4)}{4 n^{2} - 2 n + 10 n - 5}$

Step 8.2

Factor out the greatest common factor from each group.

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Step 8.2.1

Group the first two terms and the last two terms.

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{(n - 7) (n - 4)}{(4 n^{2} - 2 n) + 10 n - 5}$

Step 8.2.2

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

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{(n - 7) (n - 4)}{2 n (2 n - 1) + 5 (2 n - 1)}$

Step 8.3

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

$\frac{2 (3 n + 7)}{n - 1} \cdot \frac{(n - 7) (n - 4)}{(2 n - 1) (2 n + 5)}$