Factorizing $f(x) = 2x^4 +x^3 -x^2 +8x-4$ I tried plugging in values of $x$ until I found $f(x)=0$. I got that $(x+2)$ is a factor. Is this the only way to find factors of this polynomial?
Then I used polynomial division to get $f(x) = (x+2)(2x^3 -3x^2+5x-2)$. However I tried different values of x but still couldn't get the cubic to equal zero. Does this mean the cubic has imaginary roots? 
Also what does it mean for a polynomial to have imaginary roots? When you sketch them on the Ragland diagram, how does this bare relevance to the sketch of the function on the Cartesian plane?
Thanks
 A: You might want to take some of the guessing out of finding rational roots by making use of the rational root theorem. From Wikipedia:

In algebra, the rational root theorem (or rational root test) states a constraint on rational solutions (or roots) of the polynomial equation
$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0$$
with integer coefficients.
If $a_0$ and $a_n$ are nonzero, then each rational solution $x$, when written as a fraction $x = p/q$ in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies

*

*p is an integer factor of the constant term $a_0$, and

*q is an integer factor of the leading coefficient $a_n.$

Using the Rational Root Theorem, we can check and confirm first that $-2$ is a root and $\frac 12$ is a rational root, and use polynomial long division to find the remaining quadratic:
$$f(x) = 2x^4 +x^3 -x^2 +8x-4 = (x+2)(2x^3 -3x^2+5x-2)=(x+2)(2x-1)(x^2-x+2)$$
Now, with only a quadratic left, we can find the roots of $x^2 - x + 2$ using the quadratic formula, and since $\Delta = b^2 - 4ac = 1 - 8 = -7 < 0$, those roots will be complex roots: $$\dfrac {1 \pm \sqrt 7 i}{2}$$
So, $f(x) = (x+2)(2x-1)(x^2-x+2)$ has roots $$-2, \frac 12, \dfrac {1 \pm \sqrt 7 i}{2}$$
A: By the Rational root theorem we verify that $-2$ and $\frac 1 2$ are two rational roots for the given polynomial and we have by long division
$$f(x) = 2x^4 +x^3 -x^2 +8x-4=(x+2)(2x-1)(x^2-x+2)$$
A: A long time ago I found this very interesting method of visualing the complex roots of a quadratic equation: Complex Roots Made Visible.
I am not sure if there is a general rule that can be applied to higher degree polynomials but I found it a very interesting and unusual approach.
