Given that 1+2i is a zero of k(x)=x^(4)-6x^(3)+26x^(2)-46 x+65, find the remaining zeroes

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Given that  1+2i is a zero of  k(x)=x^(4)-6x^(3)+26x^(2)-46 x+65, find the remaining zeroes

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Given Zeroes: Given that 1+2i is a zero of the polynomial k(x), we know that its complex conjugate 1-2i is also a zero of the polynomial because the coefficients of the polynomial are real numbers.Factorization: Now we can write down the factors of k(x) that correspond to these two zeroes: (x - (1+2i))(x - (1-2i)) We can expand this to find a quadratic factor of k(x).Quadratic Factor: Expanding the factors we get: (x - 1 - 2i)(x - 1 + 2i) = ((x - 1) - 2i)((x - 1) + 2i) = (x - 1)^2 - (2i)^2 = x^2 - 2x + 1 - (-4) = x^2 - 2x + 5 This is a quadratic factor of k(x).Polynomial Division: Since k(x) is a fourth-degree polynomial and we have found a quadratic factor, we can perform polynomial division to find the other quadratic factor. We divide k(x) by the quadratic factor we found: k(x) / (x^2 - 2x + 5)Remaining Quadratic Factor: Performing the polynomial division, we get: (x^4 - 6x^3 + 26x^2 - 46x + 65) ÷ (x^2 - 2x + 5) This should give us another quadratic polynomial, which will have the remaining two zeroes.Finding Zeroes: The polynomial division yields: Quotient: x^2 - 4x + 13 Remainder: 0 This means that the other quadratic factor of k(x) is x^2 - 4x + 13.Calculating Discriminant: Now we need to find the zeroes of the quadratic polynomial x^2 - 4x + 13. We can use the quadratic formula to find these zeroes: x = [-(-4) ± √((-4)^2 - 4(1)(13))]/(2(1))Applying Quadratic Formula: Calculating the discriminant: (-4)^2 - 4(1)(13) = 16 - 52 = -36 Since the discriminant is negative, the zeroes will be complex numbers.Applying the quadratic formula: x = [4 ± √(-36)]/(2) x = [4 ± 6i]/(2) x = 2 ± 3i These are the remaining two zeroes of k(x).

Given that $1+2i$ is a zero of $k(x)=x^{4}-6x^{3}+26x^{2}-46x+65$ , find the remaining zeroes

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Given that 1+2i1+2i1+2i is a zero of k(x)=x4−6x3+26x2−46x+65k(x)=x^{4}-6x^{3}+26x^{2}-46x+65k(x)=x4−6x3+26x2−46x+65, find the remaining zeroes

Given that $1+2i$ is a zero of $k(x)=x^{4}-6x^{3}+26x^{2}-46x+65$ , find the remaining zeroes