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Problem 5 - Entrance Test
Let P(x) be a polynomial such that P(x^2 + 1) = x^4 + 4x^2 + 8. Find P(x).
A. x^2 + x + 2
B. x^2 + 2x + 5
C. x^2 - 2x + 5
D. x^2 + 3x + 4
E. x^2 + 5x + 1
Check Answer
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Correct: B
Let y = x^2 + 1. From this substitution, we can express x^2 in terms of y: x^2 = y - 1. Now, substitute y and x^2 into the given equation P(x^2 + 1) = x^4 + 4x^2 + 8. Since x^4 = (x^2)^2, we have x^4 = (y - 1)^2. So, P(y) = (y - 1)^2 + 4(y - 1) + 8. Expand and simplify: P(y) = (y^2 - 2y + 1) + (4y - 4) + 8 P(y) = y^2 - 2y + 1 + 4y - 4 + 8 P(y) = y^2 + (4y - 2y) + (1 - 4 + 8) P(y) = y^2 + 2y + 5. Therefore, P(x) = x^2 + 2x + 5.