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Find a polynomial with real coef -1 and 5-2i

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Final answer:

To find the polynomial with the given roots -1 and 5-2i, we must include the conjugate root 5+2i and multiply the factors (x + 1), (x - 5 + 2i), and (x - 5 - 2i) to get the polynomial x^3 - 9x^2 + 19x + 29.

Step-by-step explanation:

To find a polynomial with real coefficients that has the roots -1 and 5-2i, we must also include the conjugate of 5-2i, which is 5+2i, because complex roots of polynomials with real coefficients always come in conjugate pairs. The factors corresponding to these roots would be (x + 1), (x - (5-2i)), and (x - (5+2i)).

Multiplying these factors together gives us the desired polynomial:

  • (x + 1)
  • (x - 5 + 2i)
  • (x - 5 - 2i)

First, combine the two factors with complex numbers:

(x - 5 + 2i)(x - 5 - 2i) = x^2 - 5x - 2ix - 5x + 25 + 10i + 2ix - 10i - 4i^2 = x^2 - 10x + 25 + 4 = x^2 - 10x + 29

Now, multiply this result with the remaining linear factor:

(x + 1)(x^2 - 10x + 29) = x^3 - 10x^2 + 29x + x^2 - 10x + 29

Combining like terms, we obtain:

x^3 - 9x^2 + 19x + 29

This is the polynomial with real coefficients that has the given roots.

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