Yahoo Answers: Answers and Comments for Math problem 3? [Mathematics]
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From Anonymous
enUS
Thu, 23 Jan 2020 06:46:12 +0000
3
Yahoo Answers: Answers and Comments for Math problem 3? [Mathematics]
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From Philip: P(x) = A(x2)[(x1)^2 +1]. In a function with ...
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Thu, 23 Jan 2020 08:18:30 +0000
P(x) = A(x2)[(x1)^2 +1]. In a function with real
coefficients all complex roots will occur in
conjugate pairs.
P(1) = A(1)1 =A.
P(3) = A(5).
P(1) + P(3) = 4A = 8. Then A = 2.
Then P(1) = 2(3)[(2)^2 + 1] = 30.

From MyRank: 1+i , 1i and 2 roots
p(x) = (xα) (xβ) (x...
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Thu, 23 Jan 2020 11:06:04 +0000
1+i , 1i and 2 roots
p(x) = (xα) (xβ) (xγ)
p(x) = (x(1+i))(x(1i)) (x2)
put x = 1
p(1) = (1(1+i))(1(1i)) (12)
p(x) = (x(1+i))(x(1i)) (x2)
put x = 3
p(3) = (3(1+i))(3(1i)) (32)
put x = 1
p(3) = (1(1+i))(1(1i)) (12)

From atsuo: I think a polynomial has no root , it has zero...
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Thu, 23 Jan 2020 07:29:25 +0000
I think a polynomial has no root , it has zero(s) .
The third zero is the complex conjugate of 1+i , so it is 1i .
Therefore the polynomial is
A * (x2) * (x(1+i)) * (x(1i))
= A * (x2) * ((x1)+i)) * ((x1)i)
= A * (x2) * ((x1)^2i^2)
= A * (x2) * (x^22x+1+1)
= A * (x2) * (x^22x+2)
= A * (x^34x^2+6x4)
P(1) = A * (14+64) = A
P(3) = A * (2736+184) = 5A
P(1)+P(3) = 4A = 8 , so A = 2 .
The polynomial is
2 * (x^34x^2+6x4) = 2x^38x^2+12x8