MATH 433
May 4, 2015
Quiz 12: Solutions
Problem 1. Find the remainder under division of a real polynomial f (x) = x5 − 2x4 + 2 by a
polynomial g(x) = (x − 1)(x − 2).
Solution: x.
Let q be the quotient and r be the remainder under division of f by g. Then
f (x) = (x − 1)(x − 2)q(x) + r(x).
Note that we only need to find the remainder r. Evaluating both sides of the above equality at x = 1 and
x = 2, we obtain that f (1) = r(1) and f (2) = r(2). Since deg(r) < deg(g) = 2, we have r(x) = ax + b,
where a, b ∈ R. The relations f (1) = r(1) and f (2) = r(2) give rise to a system of linear equations
a + b = 1,
2a + b = 2,
which has a unique solution a = 1, b = 0.
Problem 2.
field Z11 .
Factorise into irreducible factors a polynomial p(x) = x3 − x2 + 3x + 1 over the
Solution: p(x) = (x − 2)2 (x + 3).
A quadratic or cubic polynomial is irreducible if and only if it has no roots. Let us look for the roots of
the polynomial p among elements of the field Z11 : p(0) = 1, p(1) = 4, p(2) = 0. Hence p(x) is divisible by
x − 2. We obtain that x3 − x2 + 3x + 1 = (x − 2)(x2 + x + 5).
Now we look for the roots of the polynomial q(x) = x2 + x + 5 within the field Z11 . Note that values
0 and 1 can be skipped this time. Then q(2) = 0. Hence q(x) is also divisible by x − 2. We obtain that
x2 + x + 5 = (x − 2)(x + 3).
Finally, x3 − x2 + 3x + 1 = (x − 2)2 (x + 3) over the field Z11 .