Question 239705
5x^4 - 3x^3 + ax^2 + bx - 7 = 0
f(i)=5-3(-i)-a+bi-7=0 
-3(-i)-a+bi=2 because 5-7=-2
-3(-i)+bi=2+a
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f(-i)=5-3i-a+b(-i)-7=0 If one zero is i then another must be -i.
-3i-a+b(-i)=2
-3i+b(-i)=2+a
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-3(-i)+bi=-3i+b(-i) Since they both equal 2+a.
-3-b=3+b multiply each side by i.
2b=-6 
b=-3
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a=-3(-i)-3i-2
a=-2
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It checks.
This WAS tough!
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Ed