Factor by Grouping 8x^4+8x^3+27x+27

$8 x^{4} + 8 x^{3} + 27 x + 27$

Step 1

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(8 x^{4} + 8 x^{3}) + 27 x + 27$

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

$8 x^{3} (x + 1) + 27 (x + 1)$

Step 2

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

$(x + 1) (8 x^{3} + 27)$

Step 3

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

$(x + 1) ({(2 x)}^{3} + 27)$

Step 4

Rewrite $27$ as $3^{3}$ .

$(x + 1) ({(2 x)}^{3} + 3^{3})$

Step 5

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

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

Step 6

Factor.

Simplify.

Apply the product rule to $2 x$ .

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

Raise $2$ to the power of $2$ .

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

Multiply $2$ by $- 1$ .

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

Multiply $3$ by $- 2$ .

$(x + 1) ((2 x + 3) (4 x^{2} - 6 x + 3^{2}))$

Raise $3$ to the power of $2$ .

$(x + 1) ((2 x + 3) (4 x^{2} - 6 x + 9))$

Remove unnecessary parentheses.

$(x + 1) (2 x + 3) (4 x^{2} - 6 x + 9)$

Factor by Grouping 8x^4+8x^3+27x+27

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