Question

Yuzrah's class was given an assignment to solve for the zeroes of a 5th-degree monic polynomial function. Her attention-seeker partner turned off the projector and thus, she was able to write only P(x) = x⁵ + 20x⁴ + 20x³, forgetting the x², x, and constant terms. However, her professor spoiled that the roots formed an arithmetic progression.

Q1. What are the roots of P(x)?
Q2. What is P(x)?

Answer 1

We know the roots are m, m+d, and m+2d, ... an arithmetic progression. So, we have
P(x) = (x-m)(x-(m+d))(x-(m+2d))(x-(m+3d))(x-(m+4d))
= x^5 + (-5m-10d)x^4 + (10m^2+40md+35d^2)x^3 + ...
So now we know that
-5m-10d = 20
10m^2+40md+35d^2 = 20
Unfortunately, that yields an irrational solution, as
m = 4(√7-1) ≈ 6.5830
d = -2√7 ≈ -5.2915
Plugging that in, we get
P(x) = x^5 + 20x^4 + 20x^3 - 1040x^2 - 2304x + 4608
I suspect a typo somewhere; that's just too much work!