2). We already observed how the Rational Root Theorem guarantees that r(x) = x4 - 2 has no 1st degree (and therefore, no 3rd degree) factors: the only possible 1st degree factors with integer coefficients are x ± 1 and x ± 2, but x = ±1, ±2 all evaluate to non-zero numbers. With a bit more work, we can show that p does not factor into two 2nd degree polynomials with integer coefficients either. If it did, the factorization would have to look like:
x4
- 2
= (x2 + ax ± 1)(x2 + bx
2).
since the x2 terms should multiply to give x4, while the constant terms must multiply to give -2.
Multiplying out and combining like terms gives the equation:
x4
- 2
= x4 + (a + b)x3 + (ab 1)x2
± (-2a + b)x - 2
Since the coefficients on both sides of the equation must be equal, this in turn implies the three equations:
a +
b
= 0, ab 1
= 0, and -2a + b = 0
Solving the first equation for a says that a = -b. Plugging this into the other two equations gives:
-b2 1
= 0, and 2b + b = 0
This last equation implies that b =
0, so that a = 0. But this gives the equation -02 1
= 0, which is impossible. In other words, even if we choose a and b
so that the product (x2 + ax ± 1)(x2 +
bx 2)
does not have an x3 or x term, we will be left with an x2
term. In particular, it will not multiply to give x4
- 2. Therefore, it is impossible to factor x4
- 2 over the integers.
Go back to Polynomial Division and Factoring
| Table of Contents |  Send questions or
comments to jwicks@northpark.edu |
Glossary |