Finding an unknown coefficient of a polynomial given a factor Q:Find the value of $a$ given that $x^2+1$ is a factor of $x^4-3x^3+3x^2+ax+2$
No idea where to start, I was going to use the factor theorem but it didn't work out. 
Question from year 10 Cambridge maths textbook
 A: Probably too simplistic.
Write $$x^4-3x^3+3x^2+ax+2=(x^2+1)(x^2+Ax+B)$$ Expand and simplify; this would give $$(2-B)+x (a-A)+(2-B) x^2-(A+3) x^3=0$$ SInce this needs to be true for all $x$, set all coefficients equal to $0$. The first term leads to $B=2$, the last term to $A=-3$ and the second term to $a=A=-3$.
A: Divide your polynomial by $x^{2} + 1$ and set the remainder equal to zero.
Or, since you are only interested in the remainder, set $x^{2} = -1$ in the polynomial to get that the remainder is
$$
(x^2)^{2}-3 x x^2+3 x^2+ax+2
=
(-1)^{2} + 3 x - 3 + a x + 2
=
(a + 3) x.
$$
So $x^{2} + 1$ divides your polynomial iff $a = -3$.
A: $x^2+1$ is a factor of $x^4-3x^3+3x^2+ax+2$ means $x=\pm i$ are the roots of the latter polynomial. Then you know what to do next.
A: \begin{align}
x^4-3x^3+3x^2+ax+2 &= (x^4 + x^2) - 3 \, (x^3 + x) + 2 \, (x^2 + 1) + (a+3) \, x  \\
&= (x^2 + 1) \, \left( x^2 - 3 \, x + 2 + \frac{(a+3) \, x}{x^2 + 1} \right)
\end{align}
In order for this to be factored then it is required that $a = -3$.
A: Don't make this question so hard. It is not necessary to factor x^2 + 1. Simply treat this as a long division problem, the type you did in grammar school, before calculators were invented. As a hint, it might be useful to think of the divisor as x^2 + 0x +1.
Another approach is to let x = 10. Then the problem reduces to 73A2 divided by 101, where A might be a negative integer. The problem then reduces to 72(10 + A)2 divided by 101 so that the second and fourth digits coincide. Eventually you will come to the equation 10 + A - 7 = 0.
