Section 14B, page 301.
15. Following the Euclidean algorithm, we find the following instances of the division theorem in

Thus one gcd is 2x + 2. The monic gcd is x + 1.
Now that we know the answer, let's look at the problem again. We are working over . The two
given polynomials can be factored :

and x2 + x + 2 is irreducible in . From this factorization, it is clear that the gcd is x + 1.

19. First, following the Euclidean algorithm, the have the following (obvious) instances of the
division theorem :

Thus we see that the gcd is indeed 1, so solutions exist .

Secondly, we can unwind this to obtain the Bezout polynomials.


which together give

which is an equation in the desired form.
Section 14C, page 305.

38. We are given that q = pr for some r. Since q is irreducible and p is not a unit, we conclude
that r is a unit. Both p and q are monic, so r = 1.

42. Having read the hint, we observe that . So the two given polynomials


The gcd is

45. Counterexample: In the field of polynomials over the reals, let , with n = 2. Then
f is reducible , but it has no irreducible factor of degree < 1/1.

The author must have meant "< n/2" to be " ≤n/2." Now the result is pretty clear . We have
where k≥2, the gi's are irreducible, and So we could not have
both   and That is, either or

46. In , there are of course two linear polynomials (i.e., degree 1), four quadratic polynomials
(degree 2), eight cubic polynomials (degree 3), and sixteen quartic polynomials (degree 4).

We know that both the linear polynomials , x and x + 1, are irreducible. For the quadratics and the
cubics, any polynomial with a root in is reducible, because it will have a linear factor, by the root
theorem. And if it does not have a root, then it will be irreducible, because for 2 or 3 to be the sum of
two smaller
numbers, one of those smaller numbers must be 1.

In this way, we find one quadratic , , and two cubics, and , that are

For the quartics, there are four without roots. But might they be the product of irreducible quadrat-
ics? Because there is only one irreducible quadratic, this consideration eliminates only one quartic, its
square . The other three are irreducible: , and .

That finishes the problem. Nonetheless, we add here a list of all 32 polynomials in with degree
4 or less, with their factorization into irreducible polynomials.

  0 zero 1 unit
x irreducible x + 1 irreducible







Section 15D, page 319.

18. We are given the valuable clue that 3+2i is a root of f. Therefore 3-2i is also a root, and f
is divisible by the quadratic

When we carry out the division, we obtain

Because x2 + 2x + 2 has no real roots, it is irreducible over the reals, and we are done.

21. We seek a multiple of f that has real coefficients. Following the hint, we calculate

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