How do you write a polynomial in standard form given the zeros x=-9 and -i?

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How do you write a polynomial in standard form given the zeros x=-9 and -i?

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Answer 1 · 2016-03-13T17:14:20

$P_{2} = {(x + 1)}^{2} = x^{2} + 2 x + 1$ $P_2=(x+1)^2 = x^2+2x+1$
$P_{3} = P_{2} (x + 9) = x^{3} + 11 x^{2} + 19 x + 9$ $P_3=P_2 (x+9) = x^3+11x^2+19x+9$

Explanation:

Clearly the $x = - i$ $x=-i$ it satisfies the roots $x = \pm i$ $x=+-i$
Thus we have ${(x + 1)}^{2}$ $(x+1)^2$ . Now using the other root (x+9) we can write a 3rd order polynomial...
$P_{2} = {(x + 1)}^{2} = x^{2} + 2 x + 1$ $P_2=(x+1)^2 = x^2+2x+1$
$P_{3} = P_{2} (x + 9) = x^{3} + 11 x^{2} + 19 x + 9$ $P_3=P_2 (x+9) = x^3+11x^2+19x+9$

Remark: Notice that I did not use the root given in the question $x = - I$ $x=-I$ . The reason is that the complex number in general comes in pair out of even powered polynomial. For example out of the binomial. Recall the quadratic formula:
$x_{1, 2} = \frac{- b \pm \sqrt{b - 4 a c}}{2 a}$ $x_(1,2) = (-b+- sqrt(b-4ac))/(2a)$ Note that the complex number emerges for negative value of $(b - 4 a c) i . e . \Rightarrow b < 4 a c$ $(b-4ac) i.e. => b< 4ac$ and note the true set is a pair of $\pm complex numbers$ $+- " complex numbers"$

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How do you write a polynomial in standard form given the zeros x=-9 and -i?

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