Find all zeros of the function `x^4-x^3+7x^2-9x-18 ` :
The only possible rational roots are factors of 18. We find that x=-1 and x=2 are factors. We can use long division or synthetic division to find the remaining quadratic factor:`x^4-x^3+7x^2-9x-18=(x+1)(x-2)(x^2+9) `
We can factor further in the complex numbers:
`(x+1)(x-2)(x+3i)(x-3i)=0 `
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The real solutions are -1 and 2. The imaginary solutions are 3i and -3i.
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** The problem as stated has no rational roots, so I assume there was a sign error.**
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