Given that 1+2i is a zero of k(x)=x^(4)-6x^(3)+26x^(2)-46 x+65, find the remaining zeroe {:[1+2i=0quad1-2i=0],[(x-(1+2i))(x-(7-2i))]:}

Question

Given that  1+2i is a zero of  k(x)=x^(4)-6x^(3)+26x^(2)-46 x+65, find the remaining zeroe   {:[1+2i=0quad1-2i=0],[(x-(1+2i))(x-(7-2i))]:}

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Use Conjugate Zeros: Use the fact that complex zeros of polynomials with real coefficients come in conjugate pairs. Since 1+2i is a zero of k(x), its complex conjugate 1-2i is also a zero of k(x).Multiply Factors: Multiply the factors corresponding to the known zeros to get a quadratic factor of k(x). The factors are (x - (1+2i)) and (x - (1-2i)). Multiply these factors: (x - (1+2i))(x - (1-2i)) = x^2 - (1-2i)x - (1+2i)x + (1+2i)(1-2i) = x^2 - x + 2ix - x - 2ix + 1 - 4i^2 = x^2 - 2x + 1 + 4 = x^2 - 2x + 5Divide by Quadratic Factor: Divide the original polynomial k(x) by the quadratic factor to find the other quadratic factor. k(x) = x^4 - 6x^3 + 26x^2 - 46x + 65 Divide k(x) by x^2 - 2x + 5 using polynomial long division or synthetic division.Perform Polynomial Division: Perform the polynomial division. Since the division process is lengthy and not shown here, we will assume it is done correctly and check the result by confirming that the remainder is zero. After division, we get a quotient of x^2 - 4x + 13 with a remainder of 0.Factor Quadratic Quotient: Factor the quadratic quotient if possible to find the remaining zeros. The quadratic quotient is x^2 - 4x + 13. We need to check if this can be factored or if we need to use the quadratic formula to find the zeros. The discriminant is b^2 - 4ac = (-4)^2 - 4(1)(13) = 16 - 52 = -36, which is negative, so the zeros are complex.Use Quadratic Formula: Use the quadratic formula to find the zeros of the quadratic quotient. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a). For x^2 - 4x + 13, a = 1, b = -4, and c = 13. x = (4 ± √(-36)) / 2 x = (4 ± 6i) / 2 x = 2 ± 3i These are the remaining zeros 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 $\newline$ \begin{align} 1+2i &= 0 & 1-2i &= 0 \ (x-(1+2i))(x-(7-2i)) & \end{align}

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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\newline\begin{align*} 1+2i &= 0 & 1-2i &= 0 \ (x-(1+2i))(x-(7-2i)) & \end{align*}

Given that $1+2i$ is a zero of $k(x)=x^{4}-6x^{3}+26x^{2}-46x+65$ , find the remaining zeroes $\newline$ \begin{align} 1+2i &= 0 & 1-2i &= 0 \ (x-(1+2i))(x-(7-2i)) & \end{align}