The signs of the coefficients, and the bottom row of a synthetic division, can rule out whole ranges of candidate roots before any are tested.
Problem 10.10.1. (a) Let .
(b) Let . Without finding any roots, how can we tell that ?
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(a) At a root we know , so since all ,..
(b) must be (it has to equal ), but look!
cannot be negative.
Problem 10.10.2. Let .
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If then . Now if all , then must be negative; but for positive the sum is positive. The same is true if all : if , then , but since is negative, is positive , which is a contradiction.
Now if for all and for all , note that must then be a negative number, since is even. So assume is even; the number of terms is odd.
(b) If for all odd ,, and for all even , then for odd : each term for odd must be negative. Similarly, if for all even , then , since is positive here and .
Thus all roots must be non-negative (to cancel out the term, which is even and thus negative).
(c) Let for all odd and for all even . If , then and , there cannot exist negative roots.
Note. So if the coefficients are all the same sign, there are no positive roots; if they alternate in sign, there are no negative roots.
Problem 10.10.3. Let .
(a) Use synthetic division to divide by .
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(a)
(b) If then is positive and nonzero. is also positive and nonzero. for is nonzero and positive. So is an upper bound for the roots of .
Problem 10.10.4. Let .
(a) If and all coefficients of are greater than zero, why must have no roots greater than ?
(b) Why must be positive for this statement to hold?
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(a) Let . Then is positive and each term of is positive, so for all , is positive and nonzero. Thus no can be a zero, so there aren’t any roots .
(b) Take 2: if , write . Then , so for all ; but can be negative, so can be negative, while will be positive. can be negative, positive, or zero (and thus a root). TL;DR: this doesn’t give us much useful info.
Exercises
Exercise 10.10.5. (a)
(b)
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(a):;:. Non-integer candidates:
which rules out .
so no root is bigger than either. Left: ,,,.
So , and in fact :
For the candidates are over . Try :, which cancels the :
Check:
(b). Here and , so the candidates are
(not negative: OOPS, there are supposed to be no negative roots!). Since , if any of them has , then .
Dividing :
Exercise 10.10.6.Show solution
,,,. Since is a root, with .
After synthetic division by , the bottom row starts , and the next product is .
Consider:
Exercise 10.10.7.Show solution
Suppose all , but the number of is less than , up to (i.e., there is at least one ). Let be positive. Then each term is either or , and at least one of them is guaranteed to be positive. is always positive for , positive roots. Similarly for .
Exercise 10.10.8. Consider the set of all equations , where for all and for all . Let be the largest positive number () that satisfies at least one of these equations. Which inequality does satisfy?