Problem 82 - College Algebra Essentials

Question

Factor completely, or state that the polynomial is prime. $$y^{5}-16 y$$

Step-by-Step Solution

Answer

The factored form of \(y^{5}-16 y\) is \(y(y - 2)(y + 2)(y^{2} + 4)\).

Step 1: Identify the common factor

Firstly, identify the common factor in both terms. Here, the common factor is \(y\).

Step 2: Factor out the common factor

Next, factor out the common factor (\(y\)). This leaves you with \(y(y^{4} - 16)\).

Step 3: Factor the quadratic portion

Try to factor the remaining part, \(y^{4} - 16\), which is the difference of two squares. This can further be factored into \((y^{2} - 4)(y^{2} + 4)\).

Step 4: Factoring completely

Finally, the term \(y^{2} - 4\) is also a difference of squares and can be further factored into \((y - 2)(y + 2)\). The term \(y^{2} + 4\) cannot be further factored.