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alexdok [17]
3 years ago
11

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

Mathematics
1 answer:
Zolol [24]3 years ago
4 0

Answer:

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

Step-by-step 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.


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