Bulgarian Winter Tournament

Question

Given the equation 2 + x9 + 62 = x5 - 26 + 6 - 23 + 2. a) Write the root of the equation in the form m - n, where m and n are natural numbers. b) Factor the expression a^3 - 3a^2 - 5a + 7 into two non-constant factors with integer coefficients and calculate the value of this expression if a is the root found in a).

Accepted Answer

9 + 62 = 32 + 22 + 1 = 3(2 + 1) = 6 + 3 5 - 26 = 3 - 26 + 2 = |3 - 2| = 3 - 2. The equation takes the form x(6 + 3 - 3 + 2) = 6 + 2 - 23 - 2 x(6 + 2) = (6 + 2)(1 - 2), whence (given 6 + 2 > 0) finally x = 1 - 2. b) We have a^3 - 3a^2 - 5a + 7 = a^3 - a^2 - 2a^2 + 2a - 7a + 7 = (a-1)(a^2 - 2a - 7). The product of a-1 = -2 and a^2 - 2a - 7 = (a-1)^2 - 8 = 2 - 8 = -6 is 62.