Problem 84 - Algebra and Trigonometry

Question

Factor completely, or state that the polynomial is prime. $$48 y^{4}-3 y^{2}$$

Step-by-Step Solution

Answer

The polynomial \(48 y^{4}-3 y^{2}\) factored completely is \(3y^{2}(4y - 1)(4y + 1)\)

Step 1: Identify Greatest Common Factor

First, find the greatest common factor (GCF) of the coefficients of the polynomial. The GCF of 48 and 3 is 3, and the GCF of the powers of y is \(y^{2}\). Hence, \(3y^{2}\) can be factored out.

Step 2: Factor out the GCF

Taking \(3y^{2}\) as a common factor out, the polynomial can be written as: \(3y^{2}(16y^{2} - 1)\)

Step 3: Factor the Difference of Squares

Inside the parentheses, we have a difference of squares, \(16y^{2} - 1\), which could be factored further using the formula \(a^{2} - b^{2} = (a - b)(a + b)\). Applying this rule, we get: \(3y^{2}((4y - 1)(4y + 1))\)