I've come across multiple questions like these. $\left(\iota=\sqrt{-1}\right)$

If $f\left(x\right)=x^4-4x^3+4x^2+8x+44$, find the value of $f\left(3+2\iota\right)$.

If $f\left(z\right)=z^4+9z^3+35z^2-z+4$, find $f\left(-5+2\sqrt{-4}\right)$.

Find the value of $2x^4+5x^3+7x^2-x+41$, when $x=-2-\sqrt{3}\iota$.

...and so on. The answers to these are real numbers. And the methods given to solve these include random manipulations of $x=\text{<the given complex number>}$ until we reach the given $f\left(x\right)$ or a term which $f\left(x\right)$ can be rewritten in terms of, so that we get the remainder. But all these approaches are completely random, and I see no logical approach. Can someone explain me exactly how would you solve a question of this format?

EDIT : Here's what I mean by random manipulations. Here's the solution to the third one.

$x+2=-\sqrt{3}\iota\Rightarrow x^2+4x+7=0$
Therefore, $2x^4+5x^3+7x^2-x+41$

Now how are we supposed to observe that the given quartic could be factorized? We have not been taught any method to factorise degree four polynomials, unless one root is known, in which case I can reduce it into a product of a linear and a cubic. Further if a root of the cubic is visible.

  • 1
    $\begingroup$ What do you mean by the word random, could you give an example to illustrate your point? I'd somewhat understand if you're asked to solve a complex-value equation but isn't what you're describing just a polynomial evaluation? $\endgroup$
    – BigbearZzz
    Nov 21, 2015 at 7:24

2 Answers 2


Ill give a guide line we have $x=-2-\sqrt{3}i$ now bring two on lhs and square both sides. So we get $x^2+4x+4=-3$ so $x^2+4x+7=0$ ..(1).now just divide the given expression by this qyadratic so we have . $\frac{2x^4+5x^3+7x^2-x+41}{x^2+4x+7}$ . I hope now you know this type of division . If not look up on net. Sowe get the answer as 6 on division. Now we have $2x^4+5x^3+7x^2-x+41=(x^2+4x+7)(2x^3-3x+5)+6$ but according to eqn(1) we have $x^2+4x+7=0$ . Thus the value of $2x^4+5x^3+7x^-x+41=6$ . Hope now you can get ideas for other problems. Look up on the net for division of polynomials and you will know how did i divide the quartic with a quadratic. For first take value as $x=3+2i$ and for second we have $z=-{5}+2.2i$ so we have $z=-{5}+4i$. Suppose if we have $x=a+bi$ the best trick is to bring the iota independent term together with x here 'a' and squaring both the sides . This removes the iota thus dqn becomes $x^2-a^2=-b^2$ note $i^2=-1$ . Then we divide the given polynomial with this quadratic and get the answer. As a referrence take the example of my answer. Thats what a general solution can be like.

  • $\begingroup$ But this is a special case, where the division of the quartic by the quadratic resulted in a remainder independent of $x$. 1) How am I to observe this? 2) I'm trying to find a general method. This method would come useful in this case of problems, so thank you. $\endgroup$ Nov 21, 2015 at 7:40
  • $\begingroup$ Okay i edit the answer. $\endgroup$ Nov 21, 2015 at 7:51
  • $\begingroup$ We have g(x)=0 where g is quadratic ,Any polynomial function f(z) is equal to h(z)g(z)+az+b for all z, with constants a,b, and polynomial h.The values a, b (and h,if you want it) are found by "synthetic long division". So we get f(x)=ax+b. $\endgroup$ Nov 21, 2015 at 7:56
  • $\begingroup$ Oh I forgot to mention that the answers to these questions are real numbers. I'll add that. That's what makes these questions so tough. In the explanation given above this comment, we have f(x)=ax+b, and as x is imaginary, f(x) is imaginary. All these questions have real answers. $\endgroup$ Nov 21, 2015 at 8:01
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    $\begingroup$ The idea is what @ArchisWelankar explains: in each case, to compute $f(z)$, find some quadratic $q$ such that $q(z)=0$ and perform the Euclidean division $f(x)=h(x)q(x)+r(x)$. Since the degree of $q$ is $2$ the degree of the rest $r$ is at most $1$, that is, $r(x)=ax+b$, hence $f(z)=az+b$. Since $z$ is complex not real and $f(x)$ and $q(x)$ are real, $a$ and $b$ are real, thus $f(z)$ is real if and only if $a=0$. Thus, if you know in advance that $f(z)$ is real, you know in advance that the result of the Euclidean algorithm should be $f(x)=h(x)q(x)+b$ (for some real $b$) and that $f(z)=b$. $\endgroup$
    – Did
    Nov 21, 2015 at 8:32

Unless there is some noticeable relation between f and x(for example if $ f'(x)=0$ ) or some simplifying expression for f (for example if $f(x)=1+x+x^2+x^3 +x^5$ and $x\ne 1$), or some special property of x (for example if $x^3=i$), all you can do is compute the complex arithmetic.There is considerable computer-science theory on efficient ways to compute polynomials.

  • $\begingroup$ In the example given for 'some simplifying expression for f', do you mean rewriting f(x) as $\frac{x^4-1}{x-1}+x^5$? $\endgroup$ Nov 21, 2015 at 7:26
  • $\begingroup$ yes.And what I meant to say about $f'$ was that if know where the zeroes of $f'$ are we can use it. $\endgroup$ Nov 21, 2015 at 7:50
  • $\begingroup$ Oh okay. I haven't studied derivatives in depth yet, so I'll probably understand what you mean some other day. But thanks! It helped. $\endgroup$ Nov 21, 2015 at 7:52

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