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MArishka [77]
3 years ago
14

Suppose $p(x)$ is a monic cubic polynomial with real coefficients such that $p(3-2i)=0$ and $p(0)=-52$. determine $p(x)$ (in exp

anded form).
Mathematics
1 answer:
Maslowich3 years ago
3 0

If p(3-2i)=0, then complex number 3-2i is a root of cubic polynomial.

If polynomial has real coefficients, then conjugated 3+2i is also a root of polynomial.

Then the polynomial will be of a form

p(x)=(x-(3-2i))(x-(3+2i))(x-a).

Since p(0)=-52, then

-52=(0-(3-2i))(0-(3+2i))(0-a),\\ \\-52=(2i-3)(3+2i)a,\\ \\-52=(4i^2-9)a,\\ \\-52=(-4-9)a,\\ \\-13a=-52,\\ \\a=4.

Therefore,

p(x)=(x-(3-2i))(x-(3+2i))(x-4),\\ \\p(x)=(x^2-6x+13)(x-4),\\ \\p(x)=x^3-10x^2+37x-52.

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