Certainly! Let's go through the steps to factorize the given expression [tex]\( a^2 + \frac{1}{a^2} + 2 \)[/tex] step-by-step:
1. Identify the Expression:
The expression we are given is:
[tex]\[
a^2 + \frac{1}{a^2} + 2
\][/tex]
2. Simplify the Expression:
Notice that we can rewrite the expression in a form that makes it easier to factor. Observe that:
[tex]\[
a^2 + \frac{1}{a^2} + 2
\][/tex]
can be grouped by recognizing a common pattern that fits into a perfect square form. Specifically, we see that:
[tex]\[
a^2 + \frac{1}{a^2} + 2 = (a^2 + 1 + \frac{1}{a^2}) - 1 + 3
\][/tex]
3. Reframe the Expression:
Recognize that [tex]\(a^2 + \frac{1}{a^2}\)[/tex] can be associated with a perfect square:
[tex]\[
a^2 + 2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2 + 2 = (a + \frac{1}{a})^2
\][/tex]
4. Express It Compactly:
Therefore, the given expression can be compactly written as:
[tex]\[
a^2 + \frac{1}{a^2} + 2 = \left(a + \frac{1}{a}\right)^2 + 2 - 2 = \left(a + \frac{1}{a}\right)^2
\][/tex]
However, note that the symmetry isn't immediately obvious. There may be an alternative manipulation including a single refined step.
5. Recheck for Reconstructive Factors:
Specifically, from our target, the expression can be neatly simplified:
[tex]\[
a^2 + \frac{1}{a^2} + 2 = \frac{(a^2 + 1)^2}{a^2}
\][/tex]
This recognizes both terms together in the numerator and resolves clearly when distributed.
Therefore, the correct and fully factorized form of the initial expression is:
[tex]\[
(a^2 + 1)^2/a^2
\][/tex]
This result shows that the original expression factorizes neatly as a square over rational dimensions.
