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Fofino [41]
3 years ago
8

Note the degree x^3+x+x^2-2-x^3

Mathematics
1 answer:
iren [92.7K]3 years ago
8 0

Answer:

2nd degree

Step-by-step explanation:

can't be x^3 because that cancels out, so the next highest one is x^2

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Find a polynomial with integer coefficients that satisfies the given conditions. R has degree 4 and zeros 3 − 3i and 2, with 2 a
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The required polynomial is P(x)=x^4-10x^3+46x^2-96x+72.

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If a polynomial has degree n and c_1,c_2,...,c_n are zeroes of the polynomial, then the polynomial is defined as

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According to complex conjugate theorem, if a+ib is zero of a polynomial, then its conjugate a-ib is also a zero of that polynomial.

Since 3-3i is zero, therefore 3+3i is also a zero.

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R(x)=(x-3+3i)(x-3-3i)(x-2)(x-2)

R(x)=((x-3)+3i)((x-3)-3i)(x-2)^2

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R(x)=(x^2-6x+18)(x^2-4x+4)

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Therefore the required polynomial is P(x)=x^4-10x^3+46x^2-96x+72.

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