BBB83
form a third degree polynomial function with real coefficients such that 8+i and 7 are zeros
f(x) = ???



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rivermaker
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Since coefficients are real, the two complex roots have to be conjugates. So the simplest polynomial is
\[(x7)(x 8 i)(x8+i)\]

BBB83
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The question wants the answer in a X^3 +bx^2+cX+# ?? so im confused.....

dumbcow
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x = 8 + i
x8 = +i
(x8)^2 = 1
(x8)^2 +1 = 0
x^216x +65 = 0
> (x7)(x^216x+65)
Now distribute

jerwyn_gayo
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@dumbcow, why i if the given is positive(+)?

dumbcow
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because complex zeros always come in pairs

jerwyn_gayo
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ahh, yeah. . (+) and ()

BBB83
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x^323x^257x455? is that right?

dumbcow
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everything except the "57x" term

BBB83
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177x?

dumbcow
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yep:)

BBB83
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thanks for the direction much appreciated