Finite Math Examples

$x^{5} - 27 x^{2} y^{3}$

Step 1

Factor $x^{2}$ out of $x^{5} - 27 x^{2} y^{3}$ .

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Step 1.1

Factor $x^{2}$ out of $x^{5}$ .

$x^{2} x^{3} - 27 x^{2} y^{3}$

Step 1.2

Factor $x^{2}$ out of $- 27 x^{2} y^{3}$ .

$x^{2} x^{3} + x^{2} (- 27 y^{3})$

Step 1.3

Factor $x^{2}$ out of $x^{2} x^{3} + x^{2} (- 27 y^{3})$ .

$x^{2} (x^{3} - 27 y^{3})$

Step 2

Rewrite $27 y^{3}$ as ${(3 y)}^{3}$ .

$x^{2} (x^{3} - {(3 y)}^{3})$

Step 3

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

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

Step 4

Factor.

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Step 4.1

Simplify.

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Step 4.1.1

Multiply $3$ by $- 1$ .

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

Step 4.1.2

Rewrite using the commutative property of multiplication.

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

Step 4.1.3

Apply the product rule to $3 y$ .

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

Step 4.1.4

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

$x^{2} ((x - 3 y) (x^{2} + 3 x y + 9 y^{2}))$

Step 4.2

Remove unnecessary parentheses.

$x^{2} (x - 3 y) (x^{2} + 3 x y + 9 y^{2})$