Question 1172086: Find the equation of a degree 4 polynomial with zeros x=5, x=−6, and x=1+i. It passes through the point (1,84). Write the equation in general form.
I'm just not sure about how to use the complex roots to find the equation, since there would have to be a conjugate and the two conjugates multiplied together just make 2 (1+i)(1-i)=-i^2+1=+1+1=2
Thanks for your help!
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
Watch attentively every my step.
The polynomial is ASSUMED to be with real coefficients - it is the condition you missed in your post (!)
Then the roots are 5, -6, 1+i and 1-i (together with 1+i, its conjugate 1-i is the root, too - for such a polynomial).
Then the associate linear binomials are
(x-5), (x+6), (x-(1+i)) and (x-)1-i)).
We seek for the product of these binomials with the unknown real coefficient "a"
f(x) = a*(x-5)*(x+6)*(x-(1+i))*(x-(1-i)).
The product of the last two binomials is
(x-(1+i))*(x-(1-i)) = x^2 - 2x + (1+i)*(1-i) = x^2 - 2x +2.
Therefore, the polynomial f(x) is
f(x) = a*(x-5)*(x+6)*(x^2 - 2x + 2)
To find "a", use the fact that f(1) = 84. It gives
84 = a*(1-5)*(1+6)*(1 - 2 + 2) = a*(-4)*7*1 = -28a.
Hence, a= 84/(-28) = -4.
So, your polynomial is
f(x) = -4*(x-5)*(x+6)*(x^2 - 2x + 2).
It is your ANSWER.
You may transform it further, if you want / (if you need).
Solved and explained.
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