If there are any rational zeros they have to be ± a divisors of 35.

So we try 5

5|1 -9  19 -35
 |   5 -20  -5
  1 -4  -1 -40 

Nope, 5 is not a zero.

So try 7

7|1 -9  19 -35
 |   7 -14  35
  1 -2   5   0

Yep that's a zero, so we have now factored h(x) as

h(x)= x³ - 9x² + 19x - 35
h(x)= (x - 7)(x² - 2x + 5)

Set each factor = 0

x - 7 = 0     x² - 2x + 5 = 0
    x = 7

The first gives back the factor we just found.

The second must be solved with the quadratic formula:

x = 

x = 

x = 

x = 

x = 

x = 

x = 

x = 

x = 1 ± 2i

The 3 zeros are 7, 1 + 2i, and 1 - 2i 

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Find all zeros of 

f(x)= x³ - 2x² + 25x - 50

That one can be factored by grouping:

Factor x² out of the first two terms and 25 out of the last two:

f(x) = x²(x - 2) + 25(x - 2)

Factor (x - 2) out of both terms:

f(x) = (x - 2)(x² + 25)

Set each = 0:

x - 2 = 0
    x = 2

x² + 25 = 0
     x² = -25
      x = 
      x = 
      x = ±5i

The three zeros are 2, 6i, and -5i

Edwin