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Worksheet 2.5A, More zeroes of polynomials
MATH 1410
(SOLUTIONS)
1. Find the set of all possible rational zeroes of the given function, as given by the Rational Root Test.
(a) f (x) = 2x3 + 5x2 − 4x − 3
(b) f (x) = 3x3 − 4x2 + 5.
(c) f (x) = 2x4 − 5x2 − 2x + 1.
(d) f (x) = 4x4 − 9x2 + x + 6.
(e) f (x) = 6x6 + 5x2 + x − 35.
Solutions.
(a) The set of rational zeroes of f (x) = 2x3 + 5x2 − 4x − 3 is limited to
1 3
{±1, ±3, ± , ± }.
2 2
(b) The set of rational zeroes of f (x) = 3x3 − 4x2 + 5 is limited to
1 5
{±1, ±5, ± , ± }.
3 3
(c) The set of rational zeroes off (x) = 2x4 − 5x2 − 2x + 1 is limited to
1
{±1, ± }.
2
(d) The set of rational zeroes off (x) = 4x4 − 9x2 + x + 6 is limited to
1 3 1 3
{±1 ± 2 ± 3 ± 6, ± , ± , ± , ± }.
2 2 4 4
(e) The set of rational zeroes off (x) = 6x6 + 5x2 + x − 35 is limited to
1 5 7 35 1 5 7 35 1 5 7 35
{±1, ±5, ±7, ±35, ± , ± , ± , ± , ± , ± , ± , ± , ± , ± , ± , ± }.
2 2 2
2
3 3 3
3
6 6 6
6
2. Use Descartes’s Rule of Signs to determine possible number of positive and negative zeroes of the
given function.
(a) a(x) = x5 − 3x3 − 5x2 + 9x − 7
(b) b(x) = 5x3 − 2x2 − 3x + 4
(c) c(x) = 3x3 + x2 − 9x − 3
(d) d(x) = 2x3 + 5x2 − x + 2
(e) e(x) = 3x4 + 8x3 − 5x2 + 2x − 3
(f) f (x) = 2x5 − 5x3 + 3x2 + 2x − 1
(g) g(x) = 5x6 − 7x4 + 2x3 − 1
Solutions.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
a(x) = x5 − 3x3 − 5x2 + 9x − 7 has 1 or 3 positive real roots and 0 or 2 negative real roots.
b(x) = 5x3 − 2x2 − 3x + 4 has 0 or 2 positive real roots and 1 negative real roots.
c(x) = 3x3 + x2 − 9x − 3 has 1 positive real roots and 0 or 2 negative real roots.
d(x) = 2x3 + 5x2 − x + 2 has 0 or 2 positive real roots and 1 negative real roots.
e(x) = 3x4 + 8x3 − 5x2 + 2x − 3 has 1 or 3 positive real roots and 1 negative real roots.
f (x) = 2x5 − 5x3 + 3x2 + 2x − 1 has 1 or 3 positive real roots and 0 or 2 negative real roots.
g(x) = 5x6 − 7x4 + 2x3 − 1 has 1 or 3 positive real roots and 1 negative real roots.
3. Determine the upper and lower bounds on the zeroes of the given function.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
a(x) = x4 − x3 − 5x2 − x − 6
b(x) = 3x3 − x2 + 9x − 3
c(x) = 2x3 − 3x2 − 14x + 21
d(x) = 3x3 + 2x2 + 5x + 7
e(x) = x3 + 3x2 + x − 4
f (x) = x4 + 3x3 − 15x2 − 9x + 31
g(x) = 3x4 − 20x3 + 28x2 + 19x − 13
Solutions.
(a) a(x) = x4 − x3 − 5x2 − x − 6 = (x − 3)(x + 2)(x2 + 1) An upper bound is x = 3 iand x = −2
is a lower bound. Here is a graph showing the roots:
(b) b(x) = 3x3 − x2 + 9x − 3 = (3x − 1)(x2 + 3) x = 1 is an upper bound and x = 0 is a lower
bound (there are no negative roots.) Here is a graph showing the roots:
(c) c(x) = 2x3 − 3x2 − 14x + 21 = (2x − 3)(x2 − 7) x = 4 (or x = 7) is an upper bound. A lower
bound is around x = −2 or x = −3. Here is a graph showing the roots:
(d) d(x) = 3x3 + 2x2 + 5x + 7 is a cubic with no rational roots (there is a real root ≈ −1.09362
and two complex roots.) There are no positive roots so x = 0 is an upper bound for real roots
and x = −2 or x = −7/3 is a lower bound. Here is a graph showing the roots:
(e) e(x) = x3 + 3x2 + x − 4 is a cubic with no rational roots (there is a real root ≈ 0.893289 and
two complex roots.) An upper bound for real roots is x = 1; x = −3 or x = −4 is a lower
bound. Here is a graph showing the roots:
(f) f (x) = x4 + 3x3 − 15x2 − 9x + 31 has four real roots, two negative, two positive, none rational.
The roots occur between -6 and 3; x = 3 is an upper bound and x = −6 is a lower bound.
Here is a graph showing the roots:
(g) g(x) = 3x4 − 20x3 + 28x2 + 19x − 13 has four real roots, one negative, three positive, none
rational. The roots occur between -1 and 5. Here is a graph showing the roots:
4. Let f (x) = 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30.
(a) Use Descartes’s Rule of Signs to determine possible number of positive and negative zeroes of
f (x).
(b) Write out the Test Set of rational numbers to try.
(c) Use synthetic division to compute f (1), f (2), f ( 32 ) and f (−1).
(d) Factor f (x) completely.
Solution. See the class notes.
5. Let f (x) = x4 − 2x3 − 2x2 + 2
(a) Write out the Test Set of rational numbers to try.
(b) Use Descartes’s Rule of Signs to determine possible number of positive and negative zeroes of
f (x).
(c) Use synthetic division to compute f (−1), f (2) and f (3). Then find the quotientw and remainders when f (x) is divided by x + 1, x − 2 and x − 3.
(d) Is x = 3 an upper bound for the roots of f (x)?
(e) Is x = −1 an upper bound for the roots of f (x)?
(f) Find all rational roots of f (x).
Solutions.
(a) The list of all possible rational roots of f (x) is {1, −1, 2, −2}.
(b) There are 0 or 2 positive roots and 0 or 2 negative roots.
(c) Using synthetic division we divide f (x) by x + 1:
1
−1
1
−2
−2
0
2
−1
3
−1
1
−3
1
−1
3
3
so f (−1) = 3 and f (x) = (x + 1)(x − 3x2 + x − 1) + 3. Thus the quotient when f (x) is divided
by x + 1 is q(x) = x3 − 3x2 + x − 1 and the remainder is r(x) = 3.
Using synthetic division we divide f (x) by x − 2:
1
2
1
−2
−2
0
2
2
0
−4
−8
0
−2
−4
−6
so f (2) = −6 and f (x) = (x − 2)(x3 − 2x − 4) − 6 Thus the quotient when f (x) is divided by
x − 2 is q(x) = x3 − 2x − 4 and the remainder is r(x) = −6.
Using synthetic division we divide f (x) by x − 3:
1
3
1
−2
−2
0
2
3
3
3
9
1
1
3
11
so f (3) = 11 and f (x) = (x − 3)(x3 + x2 + x + 3) + 11. Thus the quotient when f (x) is divided
by x − 3 is q(x) = x3 + x2 + x + 3 and the remainder is r(x) = 11.
(d) When we divided f (x) by x − 3 the bottom row in our synthetic division was all non-negative.
So, YES, x = 3 is an upper bound for the roots of f (x).
(e) When we divided f (x) by x + 1 the bottom row in our synthetic division was all non-positive.
So, YES, x = −1 is a lower bound for the roots of f (x).
(f) There are no rational roots of f (x). We have already plugged in −1 and 2 as possible roots.
If we try x = 1 we have
1
1
1
−2
−2
0
2
1
−1
−3
−3
−1
−3
−3
−1
and so f (1) = −1. There is no reason to try x = −2 since we already know that x = −1 is a
lower bound for the roots.
Here is a graph showing the roots: