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How do you find a cubic equation with integral coefficients that has the roots 2 and 4 plus i?
Wiki User
∙ 14y ago
If it has integral coefficients and 4+i is a root then its conjugate, 4-i must also be a root.
So the equation is f(x) = (x-2)*(x-4-i)*(x-4+i) where each factor is x minus a root. Then multiply these out.
= (x-2)*(x2 - 8x + 17) = x3 - 2x2 - 8x2 + 16x + 17x - 34 = x3 - 10x2 + 33x - 34
Wiki User
∙ 14y ago
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How many real solutions does a cubic equation have?
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There is a formula, but it is very difficult. I will give you a link to it. http://en.wikipedia.org/wiki/Cubic_equation
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