The given expression is \( 121c^4 - 44c^2 + 4 \). To factor this expression, we can try to look for a pattern or use substitution. Let's start by rewriting it in a more recognizable form.

First, notice that the expression can be viewed as a quadratic in terms of \( c^2 \):

\[
121(c^2)^2 - 44c^2 + 4
\]

Let's substitute \( x = c^2 \). The expression then becomes:

\[
121x^2 - 44x + 4
\]

Now, we factor this quadratic expression. We look for factors of the first term (121x^2) and the last term (4) that combine to give the middle term (-44x).

\[
121x^2 - 44x + 4 = (11x - 2)(11x - 2)
\]

So, the factorization is:

\[
121x^2 - 44x + 4 = (11x - 2)^2
\]

Now, substitute \( x = c^2 \) back into the expression:

\[
(11c^2 - 2)^2
\]

Thus, the factorization of the original expression is:

\[
121c^4 - 44c^2 + 4 = (11c^2 - 2)^2
\]

This shows that the original expression is a perfect square.

Would you like to go over any part of this process in more detail, or do you have any questions?

Here are 5 related questions to consider:
1. How would you solve for \( c \) if you set \( (11c^2 - 2)^2 = 0 \)?
2. What does it mean for an expression to be a perfect square trinomial?
3. How would you factor a quadratic expression with leading coefficients other than 1?
4. How can substitution help in solving higher-degree polynomial equations?
5. How can you verify that your factorization of a quadratic is correct?

**Tip:** Always check your factorization by expanding it back out to ensure it matches the original expression.

Learn how to factorize the expression 121c^4 - 44c^2 + 4 using substitution and recognition of perfect square trinomials. This problem is solved step-by-step, covering key algebraic concepts and the quadratic formula. Suitable for students in Grades 10-12.

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