# math problem 3?

Two of the roots of a cubic polynomial with real coefficients are 1+i and 2. If P(1)+P(3)=8, determine the value of P(−1).

### 3 Answers

- MyRankLv 63 months ago
1+i , 1-i and 2 -roots

p(x) = (x-α) (x-β) (x-γ)

p(x) = (x-(1+i))(x-(1-i)) (x-2)

put x = 1

p(1) = (1-(1+i))(1-(1-i)) (1-2)

p(x) = (x-(1+i))(x-(1-i)) (x-2)

put x = 3

p(3) = (3-(1+i))(3-(1-i)) (3-2)

put x = -1

p(3) = (-1-(1+i))(-1-(1-i)) (-1-2)

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- PhilipLv 63 months ago
P(x) = A(x-2)[(x-1)^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.

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- atsuoLv 63 months ago
I think a polynomial has no root , it has zero(s) .

The third zero is the complex conjugate of 1+i , so it is 1-i .

Therefore the polynomial is

A * (x-2) * (x-(1+i)) * (x-(1-i))

= A * (x-2) * ((x-1)+i)) * ((x-1)-i)

= A * (x-2) * ((x-1)^2-i^2)

= A * (x-2) * (x^2-2x+1+1)

= A * (x-2) * (x^2-2x+2)

= A * (x^3-4x^2+6x-4)

P(1) = A * (1-4+6-4) = -A

P(3) = A * (27-36+18-4) = 5A

P(1)+P(3) = 4A = 8 , so A = 2 .

The polynomial is

2 * (x^3-4x^2+6x-4) = 2x^3-8x^2+12x-8

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- atsuoLv 63 months agoReport
Sorry , it is your task . Substitute x = -1 into the polynomial .

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