<s>[INST] Value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$ 
Find the value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$ 

Now the trivial method is to put $X=5+2\sqrt{-4}$  in the polynomial  and calculate but this is for $2$ marks only and that takes a hell lot of time for $2$! So I was thinking may be there is some trick or other technique   to get the result quicker . Can anybody help please $?$ 
Thank you .
 [/INST] It's not the that much work.  
1) Just calculate $x, x^2, x^4 = (x^2)^2, x^3 = x^2 * x$ to the side first.  and 
2) consider every number as having two "parts"; a "normal" number part and a "square root of negative one part"
$x = -5 + \sqrt{-4} = -5 + 2\sqrt{-1}$
$x^2 = 25 + 4(-1) + 20\sqrt{-1} = 21 + 20\sqrt{-1}$
$x^4 = 21^2 + 20^2(-1) + 2*20*21\sqrt{-1} = 441 - 400 + 840\sqrt{-1}= 41 + 840\sqrt{-1}$
$x^3 = (21 + 20\sqrt{-1})(-5 + 2\sqrt{-1}) = -110 + 40(-1) + (-100 + 42)\sqrt{-1} = -150 - 58\sqrt{-1}$
So $X^4 + 9x^3+35X^2-X+4 = (41 + 9(-150) + 35(21) -(-5) + 4) + \sqrt{-1}(840 + 9(-58) +  35(2) - 2) = -565 + 386\sqrt{-1}$
This is a hint to complex numbers.  We call $\sqrt{-1} = i$ and all numbers are of the form $(a, b*i)$.  $(a,bi) + (c,di) = (a+c, (b+d)i)$ and $(a,bi)(c,di) = (ac - bd, (ad + bc)i)$.
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