Question

question
factor completely.
\\(25y^4 - 1\\)

Explanation:

Step1: Recognize as difference of squares

The expression \(25y^4 - 1\) can be written as \((5y^2)^2 - 1^2\), which is a difference of squares \(a^2 - b^2=(a + b)(a - b)\). So, \(a = 5y^2\), \(b = 1\).
\(25y^4 - 1=(5y^2 + 1)(5y^2 - 1)\)

Step2: Factor \(5y^2 - 1\) as difference of squares

Now, \(5y^2 - 1\) can also be written as \((\sqrt{5}y)^2 - 1^2\), again using the difference of squares formula. So, \(a=\sqrt{5}y\), \(b = 1\).
\(5y^2 - 1=(\sqrt{5}y + 1)(\sqrt{5}y - 1)\)

Step3: Combine the factors

Putting it all together, the completely factored form of \(25y^4 - 1\) is \((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). However, if we consider integer coefficients (sometimes factoring over integers is implied), we note that \(25y^4 - 1=(5y^2)^2 - 1^2=(5y^2 + 1)(5y^2 - 1)\), and \(5y^2 - 1\) doesn't factor over the integers. But if we allow real coefficients, we factor further as above. Wait, actually, another way: \(25y^4 - 1=(5y^2)^2 - 1^2=(5y^2 + 1)(5y^2 - 1)\), and \(5y^2 - 1=(\sqrt{5}y)^2 - 1^2=(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). But maybe the problem expects factoring over integers first. Wait, no, let's check again. Wait, \(25y^4 - 1=(5y^2)^2 - 1^2=(5y^2 + 1)(5y^2 - 1)\), and \(5y^2 - 1\) can be factored as a difference of squares with real numbers. But maybe the initial step is to factor as a difference of squares twice. Wait, \(25y^4 - 1=(5y^2)^2 - 1^2=(5y^2 + 1)(5y^2 - 1)\), and then \(5y^2 - 1=( \sqrt{5}y)^2 - 1^2=( \sqrt{5}y + 1)( \sqrt{5}y - 1)\). So the completely factored form is \((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). But if we consider factoring over the integers, then \(25y^4 - 1=(5y^2 + 1)(5y^2 - 1)\), but \(5y^2 - 1\) is irreducible over the integers. However, maybe the problem allows real coefficients. So the final factored form is \((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). Wait, no, actually, another approach: \(25y^4 - 1=(5y^2)^2 - 1^2=(5y^2 + 1)(5y^2 - 1)\), and \(5y^2 - 1\) can be written as \((\sqrt{5}y + 1)(\sqrt{5}y - 1)\). So combining, we get \((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). But maybe the problem expects factoring as \((5y^2 + 1)(5y^2 - 1)\) if we are factoring over integers, but since \(5y^2 - 1\) can be factored further with real numbers, the complete factorization is \((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). Wait, no, let's do it step by step correctly.

Wait, \(25y^4 - 1=(5y^2)^2 - 1^2=(5y^2 + 1)(5y^2 - 1)\) (first difference of squares). Then, \(5y^2 - 1\) is also a difference of squares: \(5y^2 - 1=(\sqrt{5}y)^2 - 1^2=(\sqrt{5}y + 1)(\sqrt{5}y - 1)\) (second difference of squares). So the completely factored form is \((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). Alternatively, if we consider factoring over the integers, we stop at \((5y^2 + 1)(5y^2 - 1)\), but since \(5y^2 - 1\) can be factored over the reals, the complete factorization is with the three factors.

Wait, maybe I made a mistake. Let's check again. The original expression is \(25y^4 - 1\). Let's write it as \((5y^2)^2 - 1^2=(5y^2 + 1)(5y^2 - 1)\). Now, \(5y^2 - 1\) can be written as \((\sqrt{5}y)^2 - 1^2=(\sqrt{5}y + 1)(\sqrt{5}y - 1)\). So yes, that's correct. So the completely factored form is \((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\).

Answer:

\((5y^2 + 1)(\sqrt{5}y + 1)(\sqrt{5}y - 1)\) (or \((5y^2 + 1)(5y^2 - 1)\) if factoring over integers is intended, but the complete factorization over reals is with the three factors)