Problem 81 - Algebra and Trigonometry

Question

Factor completely, or state that the polynomial is prime. $$y^{5}-81 y$$

Step-by-Step Solution

Answer

The fully factored form of the polynomial \(y^{5}-81y\) is \(y(y-3)(y+3)(y^{2} + 9)\).

Step 1: Identify common factors

In the polynomial \(y^{5}-81y\), the common factor is \(y\).

Step 2: Factor out the common factor

Factoring out \(y\) gives \(y(y^{4}-81)\).

Step 3: Factor the difference of squares

The expression within the parenthesis, \(y^{4}-81\), can be further factored. It is a difference of squares which can be factored into \((y^{2} - 9)(y^{2} + 9)\).

Step 4: Continue factoring

The expression within the first parenthesis, \(y^{2} - 9\), is also a difference of squares. It can be factored into \((y-3)(y+3)\). The other term, \(y^{2} + 9\), cannot be factored further using real numbers.

Step 5: Write down the fully factored polynomial

Putting all the factors together, the fully factored polynomial is \(y(y-3)(y+3)(y^{2} + 9)\).