# Question #65294

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mason m Share
Apr 1, 2017

I'm assuming you're asking for the equation of the polynomial with least degree and real coefficients that has roots at $x = 4$ and $x = 3 i$.

One property of polynomials with rational and real coefficients is that if there is one complex zero, it has to be accompanied by another complex zero that's its complex conjugate.

So, if $x = 3 i$ is a zero, we know that in order to have real coefficients $x = - 3 i$ must also be a zero.

So we will have a cubic with zeros at $x = 4$, $x = 3 i$ and $x = - 3 i$. This is given by:

$p \left(x\right) = \left(x - 4\right) \left(x - 3 i\right) \left(x + 3 i\right)$

Expanding the last two as a difference of squares:

$p \left(x\right) = \left(x - 4\right) \left({x}^{2} - {\left(3 i\right)}^{2}\right)$

Note that ${\left(3 i\right)}^{2} = 9 {i}^{2} = - 9$, since ${i}^{2} = - 1$:

$p \left(x\right) = \left(x - 4\right) \left({x}^{2} + 9\right)$

Expanding further:

$p \left(x\right) = {x}^{3} - 4 {x}^{2} + 9 x - 36$

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