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

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 $$-2, \frac 12, \dfrac {1 \pm \sqrt 7 i}{2}$$
