Key Concepts and Formulae

To find the coefficient of a specific power of $x$ in a product of two polynomial expansions, say $P(x) \cdot Q(x)$, we sum the products of coefficients from $P(x)$ and $Q(x)$ such that their powers of $x$ add up to the desired power. For example, if we want the coefficient of $x^k$ in $P(x) \cdot Q(x)$: $\text{Coeff}(x^k, P(x) \cdot Q(x)) = \sum_{i=0}^{k} \left( \text{Coeff}(x^i, P(x)) \times \text{Coeff}(x^{k-i}, Q(x)) \right)$

In this problem, we will extensively use the Binomial Theorem for expanding terms of the form $(a+b)^n$: $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

Problem Breakdown

We need to find the coefficient of $x^2$ in the expansion of the product: $(2 - x^2) \cdot \left( (1 + 2x + 3x^2)^6 + (1 - 4x^2)^6 \right)$ Let $P(x) = (2 - x^2)$ and $Q(x) = (1 + 2x + 3x^2)^6 + (1 - 4x^2)^6$. The terms from $P(x)$ that can contribute to $x^2$ in the final product are:

1. The constant term $2$.
2. The $x^2$ term $-x^2$.

To get $x^2$ in the product $P(x) \cdot Q(x)$, we consider two cases:

• Case 1: Constant term from $P(x)$ multiplied by the coefficient of $x^2$ from $Q(x)$. This contributes $2 \times \text{Coeff}(x^2, Q(x))$.
• Case 2: Coefficient of $x^2$ from $P(x)$ multiplied by the constant term from $Q(x)$. This contributes $(-1) \times \text{Constant term}(Q(x))$.

The total coefficient of $x^2$ will be the sum of contributions from Case 1 and Case 2.

Step-by-Step Solution

1. Determine the Coefficient of $x^2$ in $Q(x)$ First, let's find the coefficient of $x^2$ in $Q(x) = (1 + 2x + 3x^2)^6 + (1 - 4x^2)^6$. We'll evaluate each part separately.

a. Coefficient of $x^2$ in $(1 + 2x + 3x^2)^6$ We can treat $(2x + 3x^2)$ as a single term, say $Y$, and expand $(1+Y)^6$ using the Binomial Theorem: $(1 + Y)^6 = \binom{6}{0} (1)^6 + \binom{6}{1} (1)^5 Y^1 + \binom{6}{2} (1)^4 Y^2 + \dots$ Substituting $Y = (2x + 3x^2)$: $(1 + 2x + 3x^2)^6 = \binom{6}{0} + \binom{6}{1} (2x + 3x^2) + \binom{6}{2} (2x + 3x^2)^2 + \dots$ We are looking for terms with $x^2$.

• From $\binom{6}{1} (2x + 3x^2)$: The term contributing to $x^2$ is $\binom{6}{1} (3x^2) = 6 \times 3x^2 = 18x^2$.
• From $\binom{6}{2} (2x + 3x^2)^2$: We need to expand $(2x + 3x^2)^2 = (2x)^2 + 2(2x)(3x^2) + (3x^2)^2 = 4x^2 + 12x^3 + 9x^4$. The term contributing to $x^2$ is $\binom{6}{2} (4x^2) = 15 \times 4x^2 = 60x^2$.
• Higher power terms in the binomial expansion, like $\binom{6}{3} (2x + 3x^2)^3$, will only produce $x$ terms with power $3$ or higher, so we don't need to consider them for $x^2$.

Summing the coefficients of $x^2$ from these parts: $18 + 60 = 78$. So, $\text{Coeff}(x^2, (1 + 2x + 3x^2)^6) = 78$.

b. Coefficient of $x^2$ in $(1 - 4x^2)^6$ This is a simpler binomial expansion. Let $a=1$ and $b=-4x^2$. Using the Binomial Theorem $(a+b)^n = \binom{n}{r} a^{n-r} b^r$: $(1 - 4x^2)^6 = \binom{6}{0} (1)^6 + \binom{6}{1} (1)^5 (-4x^2)^1 + \binom{6}{2} (1)^4 (-4x^2)^2 + \dots$ The term contributing to $x^2$ is when $r=1$: $\binom{6}{1} (1)^5 (-4x^2)^1 = 6 \times (-4x^2) = -24x^2$ So, $\text{Coeff}(x^2, (1 - 4x^2)^6) = -24$.

c. Total Coefficient of $x^2$ in $Q(x)$ Summing the coefficients from parts (a) and (b): $\text{Coeff}(x^2, Q(x)) = 78 + (-24) = 54$

2. Determine the Constant Term in $Q(x)$ Now, let's find the constant term (coefficient of $x^0$) in $Q(x) = (1 + 2x + 3x^2)^6 + (1 - 4x^2)^6$.

a. Constant Term in $(1 + 2x + 3x^2)^6$ Using the Binomial Theorem $(1+Y)^6$, the constant term occurs when $Y=0$, which corresponds to $\binom{6}{0} (1)^6 = 1$. So, $\text{Constant term}((1 + 2x + 3x^2)^6) = 1$.

b. Constant Term in $(1 - 4x^2)^6$ Similarly, for $(1 - 4x^2)^6$, the constant term is $\binom{6}{0} (1)^6 = 1$. So, $\text{Constant term}((1 - 4x^2)^6) = 1$.

c. Total Constant Term in $Q(x)$ Summing the constant terms from parts (a) and (b): $\text{Constant term}(Q(x)) = 1 + 1 = 2$

3. Final Calculation of the Coefficient of $x^2$ in the Product Recall our strategy for the product $(2 - x^2) \cdot Q(x)$:

• Contribution from Case 1: $2 \times \text{Coeff}(x^2, Q(x)) = 2 \times 54 = 108$.
• Contribution from Case 2: $(-1) \times \text{Constant term}(Q(x)) = (-1) \times 2 = -2$.

Adding these contributions: $\text{Total Coeff}(x^2) = 108 + (-2) = 106$

Tips and Common Mistakes

• Systematic Approach: Break down complex products into simpler coefficient calculations. This reduces errors.
• Binomial Theorem Application: Remember that $\binom{n}{r}$ gives the coefficient of $b^r$ (or $a^r$, depending on convention). Ensure you pick the correct terms that contribute to the desired power of $x$.
• Signs: Be extremely careful with negative signs, especially when terms like $-x^2$ or $-4x^2$ are involved. A common mistake is to overlook the negative sign.
• Constant Terms: Don't forget to calculate the constant terms correctly; they are crucial when multiplying by an $x^2$ term from the other factor.
• Ignoring Higher Powers: Only consider terms that, when multiplied, result in $x^2$. Any term with $x^3$ or higher from one factor, when multiplied by $x^0$ or $x^2$ from the other, will result in powers greater than $x^2$.

Summary

By systematically identifying the components that contribute to the $x^2$ term in the product, using the Binomial Theorem for expansion, and carefully summing the resulting coefficients, we found the coefficient of $x^2$ in the given expression to be $106$. This approach ensures all relevant terms are considered and correctly combined.