Algebra Examples

$\frac{j^{2} + 7 j + 6}{12 j^{3} - 14 j^{2} - 10 j}$

Step 1

Factor $j^{2} + 7 j + 6$ using the AC method.

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Step 1.1

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 $6$ and whose sum is $7$ .

$1, 6$

Step 1.2

Write the factored form using these integers.

$\frac{(j + 1) (j + 6)}{12 j^{3} - 14 j^{2} - 10 j}$

Step 2

Simplify the denominator.

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Step 2.1

Factor $2 j$ out of $12 j^{3} - 14 j^{2} - 10 j$ .

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Step 2.1.1

Factor $2 j$ out of $12 j^{3}$ .

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2}) - 14 j^{2} - 10 j}$

Step 2.1.2

Factor $2 j$ out of $- 14 j^{2}$ .

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2}) + 2 j (- 7 j) - 10 j}$

Step 2.1.3

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

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2}) + 2 j (- 7 j) + 2 j (- 5)}$

Step 2.1.4

Factor $2 j$ out of $2 j (6 j^{2}) + 2 j (- 7 j)$ .

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2} - 7 j) + 2 j (- 5)}$

Step 2.1.5

Factor $2 j$ out of $2 j (6 j^{2} - 7 j) + 2 j (- 5)$ .

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2} - 7 j - 5)}$

Step 2.2

Factor by grouping.

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Step 2.2.1

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 $a \cdot c = 6 \cdot - 5 = - 30$ and whose sum is $b = - 7$ .

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Step 2.2.1.1

Factor $- 7$ out of $- 7 j$ .

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2} - 7 (j) - 5)}$

Step 2.2.1.2

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

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2} + (3 - 10) j - 5)}$

Step 2.2.1.3

Apply the distributive property.

$\frac{(j + 1) (j + 6)}{2 j (6 j^{2} + 3 j - 10 j - 5)}$

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$\frac{(j + 1) (j + 6)}{2 j ((6 j^{2} + 3 j) - 10 j - 5)}$

Step 2.2.2.2

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

$\frac{(j + 1) (j + 6)}{2 j (3 j (2 j + 1) - 5 (2 j + 1))}$

Step 2.2.3

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

$\frac{(j + 1) (j + 6)}{2 j ((2 j + 1) (3 j - 5))}$

$\frac{(j + 1) (j + 6)}{2 j (2 j + 1) (3 j - 5)}$