The polynomial x^4+px^3+qx^2+rx+6 is exactly divisible by each of (x-1), (x-2) and (x-3). Find the values of p, q and r.

The remainder theorem says that a polynomial f(x) has a remainder of f(c) upon getting divided by x - c. So if

f(x) = x⁴ + p x³ + q x² + r x + 6

is divisible by x - 1, x - 2, and x - 3, then we have

f (1) = 7 + p + q + r = 0

f (2) = 22 + 8p + 4q + 2r = 0

f (3) = 87 + 27p + 9q + 3r = 0

Solve the system for p, q, r. You can eliminate r by substituting the first equation into the other two:

p + q + r = -7

8p + 4q + 2r = -22   →   6p + 2q - 14 = -22   →   3p + q = -4

27p + 9q + 3r = -87   →   24p + 6q - 21 = -87   →   4p + q = -11

Eliminate q by adding (-1) times the first equation to the second one, then solve for p :

- (3p + q) + (4p + q) = - (-4) + (-11)

-3p - q + 4p + q = 4 - 11

p = -7

Solve for q :

3p + q = -4   →   3 (-7) + q = -4   →   q = 17

Solve for r :

p + q + r = -7   →   -7 + 17 + r = -7   →   r = -17

step-by-step explanation: