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Find a cubic function that has the roots 5 and 3-2i

User Joakimbeng
by
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1 Answer

4 votes

Answer:


P(x)=x^3-11x^2+43x-65

Explanation:

If the complex number
3-2i is a root of a cubic function, then the complex number
3+2i is a root too. Thus, the cubic function has three known roots
5,\ 3-2i,\ 3+2i and can be written as


P(x)=(x-5)(x-(3-2i))(x-(3+2i)),\\ \\P(x)=(x-5)(x^2-x(3-2i+3+2i)+(3-2i)(3+2i)),\\ \\P(x)=(x-5)(x^2-6x+9-4i^2),\\ \\P(x)=(x-5)(x^2-6x+9+4),\\ \\P(x)=(x-5)(x^2-6x+13),\\ \\P(x)=x^3-11x^2+43x-65.


User Austin Brunkhorst
by
8.8k points

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