condition for a cubic polynomial to have a real root Let $a,b \in R$ and assume that $x=1$ is a root of the polynomial
  $p(x)= x^4+ax^3+bx^2+ax+1$. Find the ranges of values of $a$ for which $p$ has a complex root which is not real.
Here first I factored out $x-1$ which left me with a cubic polynomial and then i thought of using the discriminant of a cubic polynomial, $\Delta= 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$ $< 0$ to get the condition on a.
But I don't know how this discriminant is derived. And I want to know if there is another method where we don't need to use the discriminant. 
 A: There is a well known method to solve equations of the type
$$x^4+ax^3+bx^2+ax+1=0 \tag{1}$$
If  $x \ne 0$ we can transform it to 
$$x^2(x^2+\frac{1}{x^2})+ax^2(x+\frac{1}{x})+bx^2=0 \tag{2}$$
Now we observe that 
$$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2$$
Using the substitution $y=x+\frac{1}{x}$ from  $(2)$  a quadratic equation of $y$ can be derived that is easier to analyze than $(1)$.
The quadratic equation in $y$ is
$$y^2+ay+b-2=0 \tag{3}$$
$x=1$ is a solution of $(1)$ therefore $b=-2a-2$. 
Substituing this in $(3)$ gives
$$y^2+ay-2a-4=0$$
This equation has two roots $y=2$ and $y=-a-2$. So the much simpler question to solve is when does have
$$x+\frac{1}{x}=-a-2 \tag{4}$$ 
a complex root that is not real.
A simpler method:
If $x=1$ is a solution we get $b=−2a−2$ by substituting $x$ in $(1)$ and therefore 
$$ x^4+ax^3-2(a+1)x^2+ax+1=0$$
Therefore we can divide this polynomial by $x-1$ and get 
$$x^3+\left(a+1\right)\,x^2+\left(-a-1\right)\,x-1$$
$x=-1$ is a solution of this polynomial to and so finally $(1)$ can be transformed to
$$(x-1)^2(x^2+(a+2)x+1)=0$$
So we have to investigate 
$$x^2+(a+2)x+1=0$$
which is equivalent to $(4)$
A: Here is how I got that $b=-2a-2$ before reading miracle173's answer. The method at the end of their post is much quicker than mine, though. Anyway, I always find a good idea to use the symmetric polynomials on the roots to deal with problems about polynomials, so here I go:
Let $r_1,r_2,r_3$ be the other three roots of your polynomial. Using that the coefficients of the polynomial are the symmetric polynomials on the roots, you get that
$$1+r_1+r_2+r_3=-a$$
$$r_1+r_2+r_3+r_1r_2+r_1r_3+r_2r_3=b$$
$$r_1r_2+r_1r_3+r_2r_3+r_1r_2r_3=-a$$
$$r_1r_2r_3=1$$
Combining these equations you easily get that $b=-2a-2$. How to finish from here has already been well told in the body and comments of miracle173's answer.
A: The solution above is probably the best but here is one using the derivative test.
We know that $p(1) = 0$ and so $b + 2a + 2 = 0$.
Now what we actually see is that $x=1$ is a double root since:
$p'(1) = 2b + 4a + 4 = 0$
$p''(1) =  12 + 6a + 2b \neq 0$.
Factoring out $(x-1)^2$ gives:
$p(x) = (x-1)^2 (x^2 + (a+2)x + 1)$
and from there we are left to discover when the quadratic factor has complex conjugate roots. This is easy, we want the discriminant to be negative, i.e. $(a+2)^2 - 4 <0$ giving $-4 < a < 0$.
