# Find the polynomial equation when I know the roots

A polynomial of minimum degree has rational coefficients and has the roots: $$x_1=-1-\sqrt5;x_2=1+2i$$ so there are $$x_3=-1+\sqrt5$$ and $$x_4=1-2i$$. I need to find the polynomial equation.

I tried to use $$(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$ but the calculations are too "heavy" and too long.There is an easier method to solve this? Right answer: $$x^4-3x^2+18x-20$$

• You have the right method, just keep going. Many terms should cancel out when you multiply with conjugates. – Landuros Jul 9 '19 at 10:09
• The "right" answer runs afoul of Vieta: $x_1+x_2+x_3+x_4=-2+1-2+1=-2\not=0$. – Barry Cipra Jul 9 '19 at 10:16
• My mistake, $x_1=-1-\sqrt5$ – DaniVaja Jul 9 '19 at 10:18
• You can also check the result at Wolphram Alpha ..... expand((x+1-sqrt(5))(x+1+sqrt(5))(x-1+2i)(x-1-2i)) – Widawensen Jul 9 '19 at 10:22

Just to give a different approach, from $$x_1+x_2+x_3+x_4=0$$ and $$x_1x_2x_3x_4=-20$$, we know the answer has the form

$$P(x)=x^4+ax^2+bx-20$$

Now

$$P(1)=(2-\sqrt5)(2+\sqrt5)(-2i)(2i)=(4-5)(4)=-4$$

and

$$P(-1)=(-\sqrt5)(\sqrt5)(-2-2i)(-2+2i)=(-5)(4+4)=-40$$

so $$1+a+b-20=-4$$ and $$1+a-b-20=-40$$, or

\begin{align} a+b&=15\\ a-b&=-21 \end{align}

from which we find $$2a=-6$$ and $$2b=36$$, so $$a=-3$$ and $$b=18$$.

Whether this is "easier" than computing the two quadratic factors of $$P(x)$$ and simply multiplying them together is unclear.

• Nice method.Thank you! :) – DaniVaja Jul 9 '19 at 11:16

$$(x+1+\sqrt 5) (x+1+\sqrt 5)=(x+1)^{2}-5$$ and $$(x-1+2i)(x-1-2i)=(x-1)^{2} +4$$. Now multiply $$(x+1)^{2}-5$$ and $$(x-1)^{2} +4$$.