Problem 82 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify common factors

First, identify the common factors in all terms of the polynomial. In this case, \(y\) can be factored out of each term in the polynomial \(y^{5} - 16y\).

Step 2: Factor out the common factors

After identifying the common factors, factor them out of the polynomial. The given polynomial reduces to \(y(y^{4} - 16)\) when \(y\) is factored out.

Step 3: Recognize Difference of Squares

The second part of the factored polynomial, \(y^{4} - 16\), can be seen as a difference of two squares where \(a = y^{2}\) and \(b = 4\). The formula for factoring difference of squares is \(a^{2} - b^{2} = (a+b)(a-b)\).

Step 4: Factor the Difference of Squares

Apply the formula to the difference of squares, \(y^{4} - 16\). This gives us \((y^{2} + 4)(y^{2} - 4)\). Note that \(y^{2} - 4\) can be further factored as it is also a difference of squares.

Step 5: Further Factor the Difference of Squares

Further factor \(y^{2} - 4\) into \((y + 2)(y - 2)\). Finally, substitute these back into the original equation.

Step 6: Write the Completely Factored Polynomial

With all terms factored, the original polynomial is completely factored as \(y(y^{2} + 4)(y + 2)(y - 2)\).