Algebra Examples

$6 x^{3} + 14 x^{2} - 12 x - 28$

Step 1

Factor $2$ out of $6 x^{3} + 14 x^{2} - 12 x - 28$ .

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Step 1.1

Factor $2$ out of $6 x^{3}$ .

$2 (3 x^{3}) + 14 x^{2} - 12 x - 28$

Step 1.2

Factor $2$ out of $14 x^{2}$ .

$2 (3 x^{3}) + 2 (7 x^{2}) - 12 x - 28$

Step 1.3

Factor $2$ out of $- 12 x$ .

$2 (3 x^{3}) + 2 (7 x^{2}) + 2 (- 6 x) - 28$

Step 1.4

Factor $2$ out of $- 28$ .

$2 (3 x^{3}) + 2 (7 x^{2}) + 2 (- 6 x) + 2 (- 14)$

Step 1.5

Factor $2$ out of $2 (3 x^{3}) + 2 (7 x^{2})$ .

$2 (3 x^{3} + 7 x^{2}) + 2 (- 6 x) + 2 (- 14)$

Step 1.6

Factor $2$ out of $2 (3 x^{3} + 7 x^{2}) + 2 (- 6 x)$ .

$2 (3 x^{3} + 7 x^{2} - 6 x) + 2 (- 14)$

Step 1.7

Factor $2$ out of $2 (3 x^{3} + 7 x^{2} - 6 x) + 2 (- 14)$ .

$2 (3 x^{3} + 7 x^{2} - 6 x - 14)$

Step 2

Factor out the greatest common factor from each group.

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Step 2.1

Group the first two terms and the last two terms.

$2 ((3 x^{3} + 7 x^{2}) - 6 x - 14)$

Step 2.2

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

$2 (x^{2} (3 x + 7) - 2 (3 x + 7))$

Step 3

Factor.

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Step 3.1

Factor the polynomial by factoring out the greatest common factor, $3 x + 7$ .

$2 ((3 x + 7) (x^{2} - 2))$

Step 3.2

Remove unnecessary parentheses.

$2 (3 x + 7) (x^{2} - 2)$

Step 4

Since the polynomial can be factored, it is not prime.

Not prime