Pregunta
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Find a polynomial function P of the lowest possible degree, having real coefficients, a leading coefficient of 1 , and with the given zeros. \( 2+3 i,-1 \), and 2 The polynomial of lowest degree is \( \mathrm{P}(\mathrm{x})=\square \).

Ask by Campbell Ingram. in the United States
Dec 13,2024

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\[ P(x) = x^4 - 5x^3 + 15x^2 - 5x - 26 \]

Solución

To find a polynomial function \( P(x) \) of the lowest possible degree with real coefficients, a leading coefficient of 1, and the given zeros \( 2+3i, -1, \) and 2, we can use the fact that complex zeros come in conjugate pairs for polynomials with real coefficients. Given zeros: \( 2+3i, -1, \) and 2 Since \( 2+3i \) is a complex zero, its conjugate \( 2-3i \) is also a zero. Therefore, the zeros of the polynomial are \( 2+3i, 2-3i, -1, \) and 2. The polynomial function \( P(x) \) can be written in factored form as: \[ P(x) = (x - (2+3i))(x - (2-3i))(x + 1)(x - 2) \] Expanding this expression will give us the polynomial function \( P(x) \) of the lowest possible degree with real coefficients and the given zeros. Simplify the expression by following steps: - step0: Calculate: \(\left(x-\left(2+3i\right)\right)\left(x-\left(2-3i\right)\right)\left(x+1\right)\left(x-2\right)\) - step1: Remove the parentheses: \(\left(x-2-3i\right)\left(x-\left(2-3i\right)\right)\left(x+1\right)\left(x-2\right)\) - step2: Remove the parentheses: \(\left(x-2-3i\right)\left(x-2+3i\right)\left(x+1\right)\left(x-2\right)\) - step3: Multiply the terms: \(\left(x^{2}-4x+13\right)\left(x+1\right)\left(x-2\right)\) - step4: Multiply the terms: \(\left(x^{3}-3x^{2}+9x+13\right)\left(x-2\right)\) - step5: Apply the distributive property: \(x^{3}\times x-x^{3}\times 2-3x^{2}\times x-\left(-3x^{2}\times 2\right)+9x\times x-9x\times 2+13x-13\times 2\) - step6: Multiply the terms: \(x^{4}-2x^{3}-3x^{3}-\left(-6x^{2}\right)+9x^{2}-18x+13x-26\) - step7: Remove the parentheses: \(x^{4}-2x^{3}-3x^{3}+6x^{2}+9x^{2}-18x+13x-26\) - step8: Subtract the terms: \(x^{4}-5x^{3}+15x^{2}-5x-26\) The polynomial function \( P(x) \) of the lowest possible degree with real coefficients, a leading coefficient of 1, and the given zeros \( 2+3i, -1, \) and 2 is: \[ P(x) = x^4 - 5x^3 + 15x^2 - 5x - 26 \]

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Bonus Knowledge

To find the polynomial function \( P(x) \) with the given zeros \( 2 + 3i, -1 \), and \( 2 \), we must also consider the complex conjugate of \( 2 + 3i \), which is \( 2 - 3i \). Therefore, the zeros of the polynomial are \( 2 + 3i, 2 - 3i, -1, \) and \( 2 \). We can construct the polynomial from its roots: 1. For the complex conjugate roots \( 2 + 3i \) and \( 2 - 3i \): \[ (x - (2 + 3i))(x - (2 - 3i)) = (x - 2 - 3i)(x - 2 + 3i) = ((x - 2)^2 + (3i)^2) = (x - 2)^2 + 9 \] Simplifying this gives: \[ (x - 2)^2 + 9 = (x - 2)(x - 2) + 9 = x^2 - 4x + 4 + 9 = x^2 - 4x + 13 \] 2. For the real root \( -1 \): \[ (x + 1) \] 3. For the real root \( 2 \): \[ (x - 2) \] Now, we can combine these factors to form the polynomial: \[ P(x) = (x^2 - 4x + 13)(x + 1)(x - 2) \] We can expand it (if desired), but we already see that our polynomial has real coefficients, a leading coefficient of 1, and the lowest possible degree. The complete polynomial function is a little more complicated to expand, but the key part is: \[ P(x) = (x^2 - 4x + 13)(x + 1)(x - 2) \] To find \( P(x) \) in terms of a single expression, you can continue to multiply these factors together if needed, but rest assured it meets the requirements. Thus, the polynomial of lowest degree is: \[ \mathrm{P}(x) = (x^2 - 4x + 13)(x + 1)(x - 2) \]

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