MATH 1050 SECTION 2 QUIZ 7
FALL 2009
1. 5 points. According to the rational root theorem, what are the possible rational roots of
f (x) = 2x3 − 5x2 − x + 6?
solution. Possible values for the numerator: p = 1, 2, 3, 6. Possible values for the denominator:
q = 1, 2. Possible rational roots: ±1, ±2, ±3, ±6, ±1/2, ±3/2.
2. 5 points. Find all the roots of f .
solution. We find roots of f by testing the possible rational roots with synthetic division. We
eventually find that −1 is a root and that division by x + 1 gives 2x2 − 7x + 6. This factors as
(2x − 3)(x − 2), so we have roots −1, 3/2 and 2.
3. 5 points. g(x) = x4 − x3 − x2 − x − 2 and g(i) = 0. Find all the roots of g.
solution. The fact that g(i) = 0 implies that x − i is a factor of g. By conjugate pairing, x + i must
also be a factor. Thus, (x + i)(x − i) = x2 + 1 is a factor. By long division,
g(x) ÷ (x2 + 1) = x2 − x − 2.
This factors as (x − 2)(x + 1). Thus, the roots are i, −i, 2 and −1.
4. 5 points. Find the vertical and horizontal asymptotes of
h(x) =
2x2 + 3x − 4
.
x2 + x − 6
solution. For the horizontal asymptotes, the numerator and denominator have the same degree so
we look at the quotient of the leading coefficients. The horizontal asymptote is y = 2. For vertical
asymptotes, the denominator factors as (x + 3)(x − 2), so the vertical asymptotes are at x = −3 and
x = 2. (You should check that these are not roots of the numerator, but this wasn’t required to get
full credit.)
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