Solving quartic equations Given the following quartic equation:
$$x^4-2x^3-7x^2+8x+12=0$$
Could anyone give some techniques required to solve any quartic equation (apart from this one) if they exist?

As an A level student, I would first look at this with the Remainder Theorem, which states that if we sub in a value, say 1, and we get zero, then (x-1) is a factor.
Let $$p(x)=x^4-2x^3-7x^2+8x+12$$
Sub in -1:
$$p(-1)=(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12 = 0$$
Therefore, (x + 1) must be a factor. We can then use algebraic division to gain a cubic equation. After doing the division, we are left with:
$$p(x)=(x+1)(x^3-3x^2-4x+12)$$
We can then use the remainder theorem to try and factorise the cubic. Subbing in -2, we get another 0. Therefore, (x+2) is a factor of the cubic. Long division gives that p(x) equals:
$$p(x)=(x+1)((x+2)(x^2-5x+6))$$
The quadratic can be factorised to give:
$$p(x)=(x+1)(x+2)(x-3)(x-2)$$
Since this equals 0, then:
$$x+1 = 0 \implies x = -1$$
$$x+2 = 0 \implies x = -2$$
$$x-3 = 0 \implies x = 3$$
$$x-2 = 0 \implies x = 2$$
To confirm, a graph gives:

