Pre-Algebra Examples

$x^{28} + x^{15} + x^{13} + 1$

Step 1

Factor out the greatest common factor from each group.

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Step 1.1

Group the first two terms and the last two terms.

$(x^{28} + x^{15}) + x^{13} + 1$

Step 1.2

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

$x^{15} (x^{13} + 1) + 1 (x^{13} + 1)$

Step 2

Factor the polynomial by factoring out the greatest common factor, $x^{13} + 1$ .

$(x^{13} + 1) (x^{15} + 1)$

Step 3

Rewrite $x^{15}$ as ${(x^{5})}^{3}$ .

$(x^{13} + 1) ({(x^{5})}^{3} + 1)$

Step 4

Rewrite $1$ as $1^{3}$ .

$(x^{13} + 1) ({(x^{5})}^{3} + 1^{3})$

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 $a = x^{5}$ and $b = 1$ .

$(x^{13} + 1) ((x^{5} + 1) ({(x^{5})}^{2} - x^{5} \cdot 1 + 1^{2}))$

Step 6

Factor.

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Step 6.1

Simplify.

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Step 6.1.1

Multiply the exponents in ${(x^{5})}^{2}$ .

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Step 6.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$(x^{13} + 1) ((x^{5} + 1) (x^{5 \cdot 2} - x^{5} \cdot 1 + 1^{2}))$

Step 6.1.1.2

Multiply $5$ by $2$ .

$(x^{13} + 1) ((x^{5} + 1) (x^{10} - x^{5} \cdot 1 + 1^{2}))$

Step 6.1.2

Multiply $- 1$ by $1$ .

$(x^{13} + 1) ((x^{5} + 1) (x^{10} - x^{5} + 1^{2}))$

Step 6.1.3

One to any power is one.

$(x^{13} + 1) ((x^{5} + 1) (x^{10} - x^{5} + 1))$

Step 6.2

Remove unnecessary parentheses.

$(x^{13} + 1) (x^{5} + 1) (x^{10} - x^{5} + 1)$