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A polynomial \( f(x) \) and one or more of its zeros are given. \[ f(x)=x^{4}-4 x^{3}+22 x^{2}+28 x-203 \] \[ 2-5 i \text { is a zero } \] Part: \( \mathbf{0} / 3 \) Part 1 of 3 (a) Find all the zeros. Write the answer in exact form. If there is more than one answer, separate them with a comma. Select "None" if applicable. The zeros of \( f(x): \square \).

Ask by Lowe Harris. in the United States
Mar 15,2025

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The zeros of \( f(x) \) are: \( 2-5i,\ 2+5i,\ \sqrt{7},\ -\sqrt{7} \).

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To find all the zeros of the polynomial \( f(x) = x^{4} - 4x^{3} + 22x^{2} + 28x - 203 \), we start with the fact that one of its zeros is \( 2 - 5i \). Since the coefficients of the polynomial are real, the complex conjugate \( 2 + 5i \) is also a zero. Next, let's use these zeros to factor the polynomial. We can build the quadratic factor corresponding to the roots \( 2 - 5i \) and \( 2 + 5i \): \[ (x - (2 - 5i))(x - (2 + 5i)) = (x - 2 + 5i)(x - 2 - 5i) = ((x - 2) + 5i)((x - 2) - 5i) = (x - 2)^2 + 25 = x^2 - 4x + 29. \] Now, we divide \( f(x) \) by \( x^2 - 4x + 29 \). Performing polynomial long division gives us: \[ f(x) = (x^2 - 4x + 29)(x^2 + ax + b). \] After division, we find that the other quadratic factor is \( x^2 + 4x - 7 \). Now we need to find the zeros of this quadratic: Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-7)}}{2 \times 1} = \frac{-4 \pm \sqrt{16 + 28}}{2} = \frac{-4 \pm \sqrt{44}}{2} = \frac{-4 \pm 2\sqrt{11}}{2} = -2 \pm \sqrt{11}. \] Thus, the complete set of zeros for \( f(x) \) is: \( 2 - 5i, 2 + 5i, -2 + \sqrt{11}, -2 - \sqrt{11} \). The zeros of \( f(x): 2 - 5i, 2 + 5i, -2 + \sqrt{11}, -2 - \sqrt{11} \).

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