Let $ p(x) $ be a quadratic polynomial such that $ p(1) = 3 $, $ p(2) = 8 $, and $ p(3) = 15 $. Find $ p(4) $. - Malaeb

Understanding the Context

The formula for a quadratic polynomial, $ p(x) = ax^2 + bx + c $, responds naturally to real-world patterns—growth curves, cost functions, scaling dynamics—making this a go-to example in applied learning environments. Its emergence in curiosity-driven searches signals interest not only in results, but in the process: how assumptions shape conclusions, how consistency defines structure, and how small inputs generate meaningful outputs.

Let $ p(x) $ Be a Quadratic Polynomial Such That $ p(1) = 3 $, $ p(2) = 8 $, and $ p(3) = 15 $. Find $ p(4) $.

This quadratic model satisfies three known inputs:

• At $ x = 1 $: $ p(1) = 3 $
• At $ x = 2 $: $ p(2) = 8 $
• At $ x = 3 $: $ p(3) = 15 $

Using these facts, algebra reveals a clear path to $ p(4) $ without guesswork. By substituting values into the general quadratic form and solving the resulting system, we uncover how coefficients determine growth—inside neat equations that hold lasting relevance.

Key Insights

Why This Quadratic Equation Is Gaining Quiet Momentum in the US Curiosity Space

Across digital spaces in the United States, interest in this polynomial has grown subtly yet consistently. More people are exploring foundational math not for degrees, but to decode real-world trends—whether modeling income changes, scaling production, or understanding data patterns. The triangle of values $ p(1)=3, p(2)=8, p(3)=15 $ creates a recognizable pattern where the function accelerates—rising by 5 from 3 to 8, then 7 to 15—suggesting a second differences that confirm quadratic behavior.

This relevance resonates in sectors valuing precise forecasting, making $ p(x) $ a microcosm of applied analytical thinking. Users seeking structured clarity find this problem intuitive but insightful—bridging abstract algebra and tangible outcomes, perfect for mobile learning environments where focused, deep reading matters.

Final Thoughts

How Let $ p(x) $ Be a Quadratic Polynomial Such That $ p(1) = 3 $, $ p(2) = 8 $, and $ p(3) = 15 $. Find $ p(4) $

This polynomial satisfies the points via substitution into $ p(x) = ax^2 + bx + c $. Setting up equations:

• $ a(1)^2 + b(1) + c = 3 \Rightarrow a + b + c = 3 $
• $ a(2)^2 + b(2) + c = 8 \Rightarrow 4a + 2b + c = 8 $
• $ a(3)^2 + b(3) + c = 15 \Rightarrow 9a + 3b + c = 15 $

Solving this system, subtract equations to eliminate $ c $:
From (2) – (1): $ 3a + b = 5 $
From (3) – (2): $ 5a + b = 7 $
Subtract again: $ (5a + b) - (3a + b) = 7 - 5 $ → $ 2a = 2 $ → $ a = 1 $
Back-substitute: $ 3(1) + b = 5 $ → $ b = 2 $
Then $ a + b + c = 3 $ → $ 1 + 2 + c = 3 $ → $ c = 0 $

Thus, the polynomial simplifies to $ p(x) = x^2 + 2x $.
To find $ p(4) $: $ (4)^2 + 2(4) = 16 + 8 = 24 $.

This elegant result emerges from logical consistency—proof that pattern recognition and algebra work hand-in-hand for forecasting.