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2025/26 AHSMC Part 1 Problem 14

Problem 14 of 16

Let bb and cc be real numbers, and define the polynomial P(x)=x2+bx+cP(x) = x^2 + bx + c. Suppose that P(P(2))=P(P(3))=0P(P(2)) = P(P(3)) = 0, and that P(2)≠P(3)P(2) \ne P(3). The value of P(1)P(1) satisfies

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Solution

The values P(2)P(2) and P(3)P(3) are the two distinct roots of PP. Equating coefficients in x2+bx+c=(x−P(2))(x−P(3))x^2 + bx + c = \bigl(x-P(2)\bigr)\bigl(x-P(3)\bigr) gives −b=P(2)+P(3)=13+6b+2c  ⟹  c=−132−3b, -b = P(2)+P(3) = 13 + 6b + 2c \implies c = -\frac{13}{2} - 3b, and then P(2)=−52−bP(2) = -\dfrac{5}{2} - b, P(3)=52P(3) = \dfrac{5}{2}. The product of the roots yields c=−254−52bc = -\dfrac{25}{4} - \dfrac{5}{2}b. Combining, −254−52b=−132−3b  ⟹  b=−12,c=−5. -\frac{25}{4} - \frac{5}{2}b = -\frac{13}{2} - 3b \implies b = -\frac{1}{2},\quad c = -5. Thus P(x)=x2−12x−5P(x) = x^2 - \dfrac{1}{2}x - 5 and P(1)=−92P(1) = -\dfrac{9}{2}. The answer is (A).

University of Alberta