If $x + 2 + \sqrt{3}i=0$ then find the value of $2x^4+3x^3-x^2-15x+36$
If you try to find the values of $x^2, x^3 $and $ x^4$, and then put the values in, we can find the solution to be 1. But is there a better way to solve this problem? Where you can probably break the second polynomial into a simple one and then solve it, saving a lot of time and making the calculations less prone to errors?
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Well, notice that:
$$-2-\sqrt{3}i=\left|-2-\sqrt{3}i\right|\exp\left(\arg\left(-2-\sqrt{3}i\right)i\right)=$$ $$\sqrt{\left(-2\right)^2+\left(-\sqrt{3}\right)^2}\exp\left(\left(\pi+\arctan\left(\frac{\sqrt{3}}{2}\right)\right)i\right)=$$ $$\sqrt{7}\exp\left(\left(\pi+\arctan\left(\frac{\sqrt{3}}{2}\right)\right)i\right)\tag1$$
Real coefficients in the quartic inspired me to multiply $x + 2 + \sqrt{3}i\;$ by the conjugate, which gives
$$(x + 2 + \sqrt{3}i)(x + 2 - \sqrt{3}i)=x^2+4x+7.$$
Dividing the given expression by $x^2+4x+7$ (just to see if we get a simple form) leads to $$2x^4+3x^3-x^2-15x+36=\underbrace{(x^2+4x+7)}_{0}\cdot(2x^2-5x+5)+1$$ Yes, if $x + 2 + \sqrt{3}i=0\;$ then $2x^4+3x^3-x^2-15x+36=1.$