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Tresset [83]
3 years ago
7

Write a polynomial function of least degree with integral coefficients that has the given zeros. –2, –3,3 – 6i

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
5 0

(x-(-2))(x-(-3))(x-(3-6i)(x-(3+6i))=\\(x+2)(x+3)(x-3+6i)(x-3-6i)=\\(x^2+3x+2x+6)((x-3)^2+36)=\\(x^2+5x+6)(x^2-6x+9+36)=\\(x^2+5x+6)(x^2-6x+45)=\\x^4-6x^3+45x^2+5x^3-30x^2+225x+6x^2-36x+270=\\x^4-x^3+21x^2+189x+270

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Answer:

x = \frac{7}{2} ± \frac{\sqrt{13} }{2}

Step-by-step explanation:

Given

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add ( half the coefficient of the x- term)² to both sides

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x - \frac{7}{2} = ± \sqrt{\frac{13}{4} } = ± \frac{\sqrt{13} }{2} ( add

x = \frac{7}{2} ± \frac{\sqrt{13} }{2} = \frac{7+/-\sqrt{13} }{2}

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