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But as written: **No integer solution exists. - Upplift

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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Substitute \( a = 1 \) into \( a + b = 5 \):
\[ 1 + b = 5 \]
\[ b = 4 \]
Thus, \( h(x) = x^2 + 4x + 3 \).
Next, find \( h(x^2 + 2) \):
\[ h(x^2 + 2) = (x^2 + 2)^2 + 4(x^2 + 2) + 3 \]
Expand each term:
\[ (x^2 + 2)^2 = x^4 + 4x^2 + 4 \]
\[ 4(x^2 + 2) = 4x^2 + 8 \]
\[ h(x^2 + 2) = x^4 + 4x^2 + 4 + 4x^2 + 8 + 3 \]

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