SOLUTION: How to find the values of constants a and b if 2+3i is a zero of the polynomial P(x) = ax^3 - 9x^2 + 30x - b?

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Question 1056305: How to find the values of constants a and b if 2+3i is a
zero of the polynomial P(x) = ax^3 - 9x^2 + 30x - b?
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
How to find the values of constants a and b if 2+3i is a
zero of the polynomial P(x) = ax^3 - 9x^2 + 30x - b?


Since we are given that 2+3i is a zero, we know that if 
we substitute x = 2+3i in %22P%28x%29%22=ax%5E3+-+9x%5E2+%2B+30x+-+b, we 
must get 0.  Substituting:

a%282%2B3i%29%5E3+-+9%282%2B3i%29%5E2+%2B+30%282%2B3i%29+-+b+=+0
a%282%2B3i%29%282%2B3i%29%282%2B3i%29+-+9%282%2B3i%29%282%2B3i%29+%2B+30%282%2B3i%29+-+b+=+0
a%282%2B3i%29%284%2B12i%2B9i%5E2%29-9%284%2B12i%2B9i%5E2%29%2B60%2B90i-b=0
Since iČ=-1


a%282%2B3i%29%284%2B12i-9%29-9%284%2B12i-9%29%2B60%2B90i-b=0
a%282%2B3i%29%28-5%2B12i%29-9%28-5%2B12i%29%2B60%2B90i-b=0
a%28-10%2B9i%2B36i%5E2%29%2B45-108i%2B60%2B90i-b=0
a%28-10%2B9i%2B36%28-1%29%5E%22%22%29%2B105-18i-b=0
a%28-10%2B9i-36%29%2B105-18i-b=0
a%28-46%2B9i%29%2B105-18i-b=0
-46a%2B9ai%2B105-18i-b=0
%28-46a%2B105-b%29%2B%289ai-18i%29=0
%28-46a%2B105-b%29%2B%289a-18%29i=0
Then the real and the imaginary parts must both = 0

Setting the imaginary part = 0,

9a-18 = 0
   9a = 18
    a = 2

Setting the real part = 0

  -46a+105-b = 0
-46(2)+105-b = 0
   -92+105-b = 0
        13-b = 0
          13 = b

So

%22P%28x%29%22=ax%5E3+-+9x%5E2+%2B+30x+-+b

becomes

%22P%28x%29%22=2x%5E3+-+9x%5E2+%2B+30x+-+13

Edwin