Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.

Flight instructor forum-16 = -4 2. In other words, -4 2 does not mean negative four times negative four. That would be positive sixteen. Some calculators give this result; so, be careful and make sure that you understand how the calculator that you are using works. The exponent may be negative. Consider this expression: 4-3

3x2 +5x¡2 3 p 7¡3x Remember that we are able to evaluate the cube root of a negative number. However, since the cube root is in the denominator, we are not allowed to let it be zero. As a result, we will ﬂnd where the denominator is zero, and exclude this value from the domain. 7¡3x = 0 ¡3x = ¡7 x = ¡7 ¡3 x = 7 3 Domain: x 6= 7 3 9. h ...