How do you write a polynomial with zeros 4, 4, 2+i and leading coefficient 1?

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How do you write a polynomial with zeros 4, 4, 2+i and leading coefficient 1?

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Answer 1 · 2017-02-18T22:34:48

$P (x) = x^{4} - 12 x^{3} + 53 x^{2} - 104 x + 80$ $P(x)=x^4-12x^3+53x^2-104x+80$

Explanation:

$4, 4, 2 + i$ $4,4,2+i$
Root $4$ $4$ is multiplicity $2$ $2$
$2 + i$ $2+i$ Complex Conjugate Zero is $2 - i$ $2-i$ , we have:
$P (x) = {(x - 4)}^{2} (x - 2 - i) (x - 2 + i)$ $P(x)=(x-4)^2(x-2-i)(x-2+i)$
$P (x) = (x^{2} - 8 x + 16) [{(x - 2)}^{2} - i^{2}]$ $P(x)=(x^2-8x+16)[(x-2)^2-i^2]$
$P (x) = (x^{2} - 8 x + 16) (x^{2} - 4 x + 4 - (- 1)]$ $P(x)=(x^2-8x+16)(x^2-4x+4-(-1)]$
$P (x) = (x^{2} - 8 x + 16) (x^{2} - 4 x + 5)$ $P(x)=(x^2-8x+16)(x^2-4x+5)$
$P (x) = x^{4} - 4 x^{3} + 5 x^{2} - 8 x^{3} + 32 x^{2} - 40 x + 16 x^{2} - 64 x + 80$ $P(x)=x^4-4x^3+5x^2-8x^3+32x^2-40x+16x^2-64x+80$
$P (x) = x^{4} - 12 x^{3} + 53 x^{2} - 104 x + 80$ $P(x)=x^4-12x^3+53x^2-104x+80$

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How do you write a polynomial with zeros 4, 4, 2+i and leading coefficient 1?

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