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 .
 A: You could consider carrying out polynomial long division, dividing the given polynomial by the minimal polynomial of $-5+4i$, which is $x^2+10x+41$. Viz:
$x^4+9x^3+35x^2-x+4=(x^2-x-4)(x^2+10x+41)+160$
We know that if $x=-5+4i$, then $x^2+10x+41$ vanishes, so the given polynomial evaluates to $160$ there.
A: When evaluating polynomials, usually the first method taught is to simply replace each $x$ you see with the prescribed value.  While this works in general, it can feel time consuming.
Depending on specific scenarios, one can find tricks to make things easier (e.g. factoring or polynomial long division as in @$\pi$r8's answer above).
There is however a general method that works in every case that is more efficient than the grade-school method of replacing the $x$'s, calculating each of the powers of $x$ separately, multiplying and then adding.
It is known as Horner's Method.
The gist of it is that you can instead write the polynomial in the form $a_0+x(a_1+x(a_2+\dots+(x(a_{n-1}+a_nx))\dots )$
For your specific polynomial, you have:
$x^4+9x^3+35x^2-x+4=4+x(-1+x(35+x(9+x)))$
Now, evaluate from the innermost parenthesis outwards.  This is still tedious for your desired $x$ value so I will not do it here, but the number of arithmetic operations involved decreases from being on the order of $O(n^2)$ to instead being on the order of $O(n)$.  (specifically evaluating a polynomial of degree $n$ can take at most $\frac{n^2+3n}{2}$ steps using the grade-school method and at most $2n$ steps using Horner's method)
A: $X+5=4i$ squaring you get $x^2+10x+41=0$ now just divide the given polynomial with this equation so solving $$\frac{x^4+9x^3+35x^2-x+4}{x^2+10x+41}$$ you get it equal to $(160)$ 
A: 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 $(a,bi)(c,di) = (ac - bd, (ad + bc)i)$.
