2007 AIME I Problem 8

Attempt Problem 8 of the 2007 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AIME I solutions, or check the answer key.

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8.

The polynomial P(x)P(x) is cubic. What is the largest value of kk for which the polynomials Q1(x)=x2+(k29)xkQ_1(x) = x^2 + (k - 29)x - k and Q2(x)=2x2+(2k43)x+kQ_2(x) = 2x^2 + (2k - 43)x + k are both factors of P(x)?P(x)?

Answer: 30
Concepts:polynomialquadraticfactoring
Difficulty rating: 2500
Solution:

If Q1Q_1 and Q2Q_2 had no common root, their product — of degree 44 — would divide the cubic P(x),P(x), which is impossible. So they share a root r,r, and 2Q1(r)Q2(r)=0.2Q_1(r) - Q_2(r) = 0. Computing, 2Q1(x)Q2(x)=15x3k,2Q_1(x) - Q_2(x) = -15x - 3k, so r=k5.r = -\frac{k}{5}.

Substituting into Q1(r)=0Q_1(r) = 0 gives k225(k29)k5k=0;\frac{k^2}{25} - (k - 29)\frac{k}{5} - k = 0; multiplying by 2525 and simplifying yields 4k2+120k=0,-4k^2 + 120k = 0, so k=0k = 0 or k=30.k = 30.

For k=30,k = 30, Q1(x)=x2+x30Q_1(x) = x^2 + x - 30 =(x+6)(x5)= (x + 6)(x - 5) and Q2(x)=2x2+17x+30Q_2(x) = 2x^2 + 17x + 30 =(x+6)(2x+5),= (x + 6)(2x + 5), and both divide P(x)=(x+6)(x5)(2x+5).P(x) = (x + 6)(x - 5)(2x + 5). The largest value is 30.30.

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