2007 AIME I Problem 8

Question

The polynomial

P(x)

is cubic. What is the largest value of

k

for which the polynomials

Q_1(x) = x^2 + (k - 29)x - k

and

Q_2(x) = 2x^2 + (2k - 43)x + k

are both factors of

P(x)?

Answer

Two quadratic factors of a cubic must share a common root

At a common root

r,

2Q_1(r) - Q_2(r) = 0

is a linear equation giving

r = -\frac{k}{5};

substitute back into

Q_1

If

Q_1

and

Q_2

had no common root, their product — of degree

4

— would divide the cubic

P(x),

which is impossible. So they share a root

r,

and

2Q_1(r) - Q_2(r) = 0.

Computing,

2Q_1(x) - Q_2(x) = -15x - 3k,

so

r = -\frac{k}{5}.

Substituting into

Q_1(r) = 0

gives

\frac{k^2}{25} - (k - 29)\frac{k}{5} - k = 0;

multiplying by

25

and simplifying yields

-4k^2 + 120k = 0,

so

k = 0

or

k = 30.

For

k = 30,

Q_1(x) = x^2 + x - 30

= (x + 6)(x - 5)

and

Q_2(x) = 2x^2 + 17x + 30

= (x + 6)(2x + 5),

and both divide

P(x) = (x + 6)(x - 5)(2x + 5).

The largest value is

30.

Problem 8 in other years