For a polynomial with integer coefficients, an integer root has to divide the constant term. This page finds that fact, uses it with synthetic division to factor polynomials, and adds the sign-change test for locating roots.
Problem 10.8.1. Let .
(a) Consider . How can we determine without evaluating or doing calculations?
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is odd; all other terms are multiples of and therefore even, so they cannot sum to . Therefore .
(b) Consider . Which terms are divisible by ? Which aren’t? Is a root?
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No, is not a root, because if it were, the term would be divisible by , i.e. end in or .
(c) Suppose with . Show that is a divisor of .
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and since ,.
(d) Find the roots of .
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. Let’s check if is a root (). (Possible roots, if in :.)
So is a factor ( is a root), and the quotient is .
Therefore the roots are , or: with a multiplicity of , and ; or .
Note (Extra). cannot be factored with integer coefficients (as a product of two polynomials). Compute : what is interesting?
The roots:
and are prime; prime. What about these?
Also, what if ? Then . Ah! is divisible by . Wait, if not prime. So this happens earlier. not prime. Let’s try prime. Hmmm.
no new info. (restrict to ).
Let’s try prime; prime; prime; not prime.
So produces primes for .
Let’s study : always odd. An odd prime up to the prime .
What’s going on? I checked and it seems to be that on the roots, :? For : with ,.
Only up to ? What about ?
Problem 10.8.2. Suppose is a root of , where is a nonzero integer, and all the coefficients of are integers. Why must ?
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Let , with ,. Then
Ah, but that is
in which the right side must be an integer, as a finite sum of integers. Thus , so .
Remark. For a polynomial with all : if a root , then .
Problem 10.8.3. Let .
(a) Graph .
(b) Notice that and . Show there exists a root of in .
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(, and too (duh).)
Since and , the -axis was crossed by . Thus there exists a root; let’s check:
So the factors are .
Note (Trick!). If is a polynomial and such that , then there exists a root of between and .
Problem 10.8.4. Factor and find its roots.
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So are possible. Given that we have order , either or out of of the roots are negative (since ).
Start with :, and . Rats. Ah! But and , so if this trend of upwardness continues we might have a root very soon! , a root! So now we can factor quickly and continue.
Now we’ve narrowed down to ! Let’s try and then to see if that tells us anything. , woah! So the root has multiplicity !
So .
Thus the roots are , and the factored form is .
Note (Trick!). If there is no in our polynomial, then is a root of said polynomial.
Problem 10.8.5. Factor and find the roots of .
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At a root,
. We know is a root, and all other roots must be negative:
What’s possible: . Let’s quickly try :
so is a root, and so now we have:
At :,, and , so . So works again!
So so far.
Now does not work anymore, so what else: . But and , so there exists a root between them. , yay!
So fully factored: , and are the roots.
Exercises
Exercise 10.8.6. Find all roots of the following:
(a)
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By the RRT: (and ), so . Since , divides :
is , so is our factored form. Therefore the roots are .
(b)
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So , and since
, so .
So:
and now notice that this cubic vanishes at , so:
So the factored form is . Therefore the roots are .
(c)
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, and :
with divide by again, so:
(d)
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so there exists a root .
So :
so .
Exercise 10.8.7.. Note . We can then check all the other divisors of besides and see that they aren’t roots of . Can we conclude that the other roots are not integers?
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roots: , so or . We know , not , so maybe. Wait, no: the root can repeat, so we cannot conclude that. In fact , and the other roots are all .
Exercise 10.8.8. Suppose with all , and . Show that has no integer roots.
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So there exists . Hm. So
.
Exercise 10.8.9. Let . For what is ?
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Dividing out the root :
which leaves . If the other roots are integers, . Plugging in :, so :
which leaves . So for .