Polynomial theorems worksheet solutions
1. What is the remainder when f(x) = x2005
−3x1980
−2x42 +x−1is divided by x+ 1?
By the Remainder Theorem, we need only calculate f(−1) = −1−3−2−1−1 = −8.
2. For each of the following, find an nth degree polynomial with the given roots and no other
roots.
(a) n= 2; roots: 4,−1.
Apply the Factor Theorem. Example: (x−4)(x+ 1).
(b) n= 3; roots: −2,1,3
Example: (x+ 2)(x−1)(x−3).
(c) n= 3; roots: 1,π.
Example: (x−1)(x−π)2. Note that in order to ensure that the degree is 3, one of the
factors must be squared.
(d) n= 3; roots: 4.
Example: (x−4)3.
3. For each of the problems below, I will give you a polynomial and one or more roots. Use this
information to obtain a complete factorization. (Hint: you can use polynomial division and
the factor theorem to reduce the unfactored part to a quadratic.)
(a) f(x) = x4+ 2x3
−2x2+ 2x−3,f(−3) = 0,f(1) = 0.
We know that x+ 3 and x−1must be factors, so divide x4+ 2x3
−2x2+ 2x−3by
x2+ 2x−3to get x2+ 1. So, we have
x4+ 2x3
−2x2+ 2x−3 = (x+ 3)(x−1)(x2+ 1).
Note that as the discriminant of x2+ 1 is negative, the polynomial x2+ 1 has no roots,
and thus no factors. So, the factorization above is complete.
(b) f(x) = x3
−2x2
−8x,f(0) = 0.
We know that xmust be a factor, so divide x3
−2x2
−8xby xto get x3
−2x2
−8x=
(x2
−2x−8)x. We can factor x2
−2x−8=(x−4)(x+ 2) (either mentally, or using the
quadratic formula) so our factorization is
x3
−2x2
−8x= (x−4)(x+ 2)x.
(c) g(x) = x3
−3x+ 2,g(1) = 0.
We know that x−1must be a factor, so divide x3
−3x+ 2 by x−1to get x3
−3x+ 2 =
(x−1)(x2+x−2). We can factor x2+x−2 = (x+ 2)(x−1), so our factorization is
x3
−3x+ 2 = (x+ 2)(x−1)2.
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