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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