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
