Problem 83 - Algebra and Trigonometry

Question

Factor completely, or state that the polynomial is prime. $$20 y^{4}-45 y^{2}$$

Step-by-Step Solution

Answer

The completely factored form of the polynomial \(20y^4 - 45y^2\) is \(5y^2(2y - 3)(2y + 3)\).

Step 1: Identify the greatest common factor (GCF)

The greatest common factor (GCF) of \(20y^4\) and \(45y^2\) is \(5y^2\). This is the highest number and variable that divides both terms without leaving a remainder.

Step 2: Factor out the GCF

Now factor out the GCF from both terms. Factoring out \(5y^2\) from \(20y^4 - 45y^2\) gives \(5y^2(4y^2 - 9)\).

Step 3: Identify if the resulting polynomial within the parentheses can be factored further

The polynomial within the parentheses, \(4y^2 - 9\), is a difference of squares. It can be factored further as \((2y)^2 - (3)^2\).

Step 4: Apply the difference of squares formula

Recall that the formula to factor a difference of squares \(a^2 - b^2\) is \((a-b)(a+b)\). Therefore, the polynomial \(4y^2 - 9\) can be factored into \((2y - 3)(2y + 3)\).

Step 5: Write the completely factored form of the polynomial

Putting it all together, the completely factored form of \(20y^4 - 45y^2\) is \(5y^2(2y - 3)(2y + 3)\)