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How do you find a cubic equation with integral coefficients that has the roots -3 and 5 minus i?
Wiki User
∙ 10y ago
If the equation has real coefficients, and 5 - i is a root, then its conjugate, 5 + i must be a root.
Since 5 - i and 5 + i are roots, then (x - 5 + i) and (x - 5 - i) are factors.
That means x2 - 10x + 26 is a factor.
The other root is x = -3 so x + 3 is the other factor.
So the cubic is (x + 3)*(x2 - 10x + 26) = 0
That is x3 - 7x2 - 4x + 78 = 0
Wiki User
∙ 10y ago
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