and $ q(x) = p(x+1) $. To find the coefficient of $ x^2 $ in $ q(x) $, we compute:

["# Understanding $ q(x) = p(x+1) $: Finding the Coefficient of $ x^2 $", "When working with polynomials, one powerful transformation is shifting the input, often represented as $ q(x) = p(x+1) $. This substitution shifts the graph of $ p(x) $ horizontally to the left by 1 unit, which can reveal important properties about the original polynomial’s coefficients. In particular, determining the coefficient of $ x^2 $ in $ q(x) $ when $ q(x) = p(x+1) $ offers insight into how polynomial coefficients transform under horizontal shifts.", "Let $ p(x) $ be a general polynomial of degree at least 2:\n$$\np(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0\n$$", "We define $ q(x) = p(x+1) $. To compute the coefficient of $ x^2 $ in $ q(x) $, we expand $ p(x+1) $ using the binomial theorem.", "## Expanding $ p(x+1) $", "Substituting $ x+1 $ into $ p(x) $, we have:\n$$\nq(x) = p(x+1) = a_n (x+1)^n + a_{n-1} (x+1)^{n-1} + \cdots + a_2 (x+1)^2 + a_1 (x+1) + a_0\n$$", "To find the $ x^2 $ coefficient, we focus only on terms that contribute to $ x^2 $ in the expansion. The relevant terms come from:", "- $ a_n (x+1)^n $: The $ x^2 $ term arises from $ \binom{n}{2} x^2 \cdot 1^{n-2} = \binom{n}{2} x^2 $\n- $ a_{n-1} (x+1)^{n-1} $: The $ x^2 $ term comes from $ \binom{n-1}{2} x^2 $\n- $ a_2 (x+1)^2 $: This contributes $ a_2 (x^2 + 2x + 1) $, so the $ x^2 $ term is $ a_2 x^2 $", "All higher-degree expansions (e.g., $ x^n, x^{n-1} $) contribute to higher powers or lower-order terms and do not affect the $ x^2 $ coefficient in $ q(x) $ unless expanded.", "However, the full expression for the coefficient of $ x^2 $ in $ q(x) = p(x+1) $ can be obtained by summing contributions:", "### Coefficient of $ x^2 $ in $ q(x) = \sum_{k=2}^n a_k \binom{k}{2} x^2 + a_2 \cdot 1 x^2 $", "More precisely:\n- From $ a_k (x+1)^k $: contributes $ a_k \binom{k}{2} x^2 $\n- From $ a_2 (x+1)^2 $: contributes $ a_2 \cdot 1 \cdot x^2 $\n- Other $ a_j $ with $ j < k $ contribute via lower-degree expansions, but only $ k=2 $ and $ k \geq 3 $ terms (but $ a_2 $ already includes $ (x+1)^2 $)", "Actually, to clarify: writing $ p(x+1) $ variationally, the coefficient of $ x^2 $ is:", "$$\n[\ ext{coeff of } x^2 \ ext{ in } q(x)] = a_2 \cdot \binom{2}{2} + a_3 \cdot \binom{3}{2} + \cdots + a_n \cdot \binom{n}{2}\n$$\nBut wait — this assumes $ p(x) $ starts at $ x^2 $. However, $ p(x) $ may not have an $ x^2 $ term unless explicitly present. To make a general claim:", "### Key Insight:\nThe coefficient of $ x^2 $ in $ q(x) = p(x+1) $ depends on $ p(x) $’s coefficients $ a_k $, and equals\n$$\n\sum_{k=2}^n a_k \binom{k}{2}\n$$\nbut only if $ p(x) $ contains $ x^k $ terms up to $ n $. However, since $ p(x+1) $ expands all terms, the correct approach is:", "The coefficient of $ x^2 $ in $ p(x+1) $ is the sum over all $ k \geq 2 $ of $ a_k \cdot \binom{k}{2} $, but only if $ p(x) $ includes $ x^2 $ — in fact, this formula is valid for any polynomial $ p $, because every term $ a_k (x+1)^k $ expands with $ \binom{k}{2} a_k x^2 $, and lower terms contribute only for $ k > 2 $ but do not affect $ x^2 $ unless from $ k=2 $, $ k=3 $, etc.", "Wait — correction: when expanding $ (x+1)^k $, the coefficient of $ x^2 $ is $ \binom{k}{2} $ times the coefficient of $ x^2 $ in $ (x+1)^k $. But $ (x+1)^k $ contributes $ \binom{k}{2} x^2 $ for $ k \geq 2 $. Therefore, the total contribution to $ x^2 $ in $ q(x) = p(x+1) $ is:", "$$\n\ ext{Coeff}{x^2}(q) = \sum}^n a_k \cdot \binom{k}{2\n$$", "This formula works because:\n- $ (x+1)^1 $ contributes only to $ x^0, x^1 $\n- $ (x+1)^2 $ contributes $ \binom{2}{2} = 1 $ to $ x^2 $\n- $ (x+1)^3 $ contributes $ \binom{3}{2} = 3 $ to $ x^2 $\n- etc.", "Thus, the coefficient of $ x^2 $ in $ q(x) = p(x+1) $ is:", "$$\n\boxed{ \sum_{k=2}^{\deg p} a_k \binom{k}{2} }\n$$", "Where $ a_k $ is the coefficient of $ x^k $ in $ p(x) $.", "### Example for Clarity", "Let $ p(x) = 3x^3 + 2x^2 - 5x + 1 $. Then:", "- $ a_3 = 3 $, $ a_2 = 2 $, $ a_1 = -5 $, $ a_0 = 1 $", "$$\nq(x) = p(x+1) = 3(x+1)^3 + 2(x+1)^2 - 5(x+1) + 1\n$$", "Compute $ x^2 $ terms:", "- From $ 3(x+1)^3 $: $ \binom{3}{2} \cdot 3 = 3 \cdot 3 = 9 $\n- From $ 2(x+1)^2 $: $ \binom{2}{2} \cdot 2 = 1 \cdot 2 = 2 $\n- From $ -5(x+1) $: no $ x^2 $\n- Constant only", "Total coefficient: $ 9 + 2 = 11 $", "Using formula: $ a_3 \binom{3}{2} + a_2 \binom{2}{2} = 3 \cdot 3 + 2 \cdot 1 = 9 + 2 = 11 $", "Correct.", "### Conclusion", "The coefficient of $ x^2 $ in $ q(x) = p(x+1) $ is determined by extracting contributions from each monomial in $ p(x) $via binomial coefficients. This transformation is fundamental in calculus and discrete mathematics—particularly in finite differences and Taylor expansions—where $ p(x+1) $ models forward shifts of polynomial behavior. Understanding this operation equips learners to analyze polynomial transformations and extract key coefficients efficiently.", "Whether proving identity, analyzing symmetry, or solving recurrence relations, recognizing how $ p(x+1) $ reshapes coefficients enhances algebraic intuition and problem-solving precision.", "Keywords: $ p(x+1) $, coefficient of $ x^2 $, polynomial transformation, binomial expansion, finite difference, algebraic coefficient shift\nMeta description: Explore how $ q(x) = p(x+1) $ transforms the coefficients of a polynomial, focusing on extracting the $ x^2 $ coefficient using binomial coefficients. Ideal for students and educators studying algebra and polynomial functions."]