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Chapter 0: Problem 83

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

Short Answer

Expert verified
The fully factored form of the polynomial \(20 y^{4}-45 y^{2}\) is \(5y^{2}(2y - 3)(2y + 3)\).

Step by step solution

01

Identify the GCD

Start by finding the greatest common divisor (GCD) of the coefficients of the terms. The GCD of 20 and 45 is 5. The common power of y that can be factored out is 2 since it is present in both the terms. So, factorize 5 and \(y^2\).
02

Factor out the GCD

After the GCD is identified, factor it out. The expression becomes: \(5 y^{2}(4 y^{2} - 9)\).
03

Apply the Difference of Squares Formula

Now, let's examine the expression inside the bracket: we have \(4y^{2} - 9\), which is a difference of squares (because \(4y^{2}\) is the square of \(2y\) and 9 is the square of 3). Difference of squares can be factored using the formula \(a^{2} - b^{2} = (a - b)(a + b)\). Apply this formula to the expression, this gives: \(5 y^{2}(2 y - 3)(2 y + 3)\).
04

Report the Final Answer

Therefore, the completely factored form of the given polynomial is \(5y^{2}(2y - 3)(2y + 3)\).

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