x³ - 4x² + 21x - 34

The possible rational zeros are ± the divisors of 
34 which are ±1,±2,±17,±34

Try 1

1 |1 -4 21 -34
  |   1 -3  18
   1 -3 18  16

No, that gives a 16 remainder, not 0.

Try -1

-1 |1 -4 21 -34
   |  -1  5 -26
    1 -5 26 -60

No, that gives a -60 remainder, not 0.


Try 2

2 |1 -4 21 -34
  |   2 -4  34
   1 -2 17   0

Yay!  That gives a 0 remainder, so 2 is a zero.

So x = 2 is a solution and that is equivalent
to  x - 2 = 0, so x - 2 is a factor, and we have
factored the polynomial as

(x - 2)(x² - 2x + 17)

So we set each factor equal to 0:

x - 2 = 0      x² - 2x + 17 = 0
    x = 2        doesn't factor.

So we use the quadratic formula:

Compare 1x² - 2x + 17 = 0 to ax² + bx + c = 0

a = 1, b = -2, c = 17















Make two fractions:





Complex zeros: 1 + 4i, 1 - 4i

Edwin