Find all zeros of f(x) = x^4 - 6x^3+5x^2 + 22x-40
given (2-i) is a zero

Question

Find all zeros of f(x) = x^4 - 6x^3+5x^2 + 22x-40 
given (2-i) is a zero

phi · Answer

if 2-i is a zero then you know 2+i is also a root.
complex roots come in complex conjugate pairs.

so you could multiply out (x-(2-i))(x-(2+i)) to get a quadratic and divide that into the original expression to get another quadratic.  factor this second quadratic for the remaining 2 zeros

curry · Answer

ye but i keep getting a error in my calculations

anonymous · Answer

you could multiply out (x-(2-i))(x-(2+i)) to get a quadratic = x^2-4x+5

curry · Answer

ye and then itrs the factoreing to 6th degreee that messes me up

phi · Answer

you divide x^2-4x+5 into the quartic.  It will go in evenly, a give you a quadratic that can be factored.

anonymous · Answer

divide: 
x^2-4x+5    |     x^4 - 6x^3+5x^2 + 22x-40

goes in on the first time through x^2 times...

anonymous · Answer

X^2 times x^2-4x+5 =...
then subtract, like in long division...

anonymous · Answer

$$\frac{x^4-6 x^3+5 x^2+22 x-40}{x^2-4 x+5}=x^2-2 x-8  $$