Simple AS method of solving cubics, well, the one you are meant to use anyway...
x³ - 2x² - 13x - 10 = 0
Use Trial and Error to find one root, normally between +/-3
try x=0, 0-0-0-10 = -10
x=-1, -1-2+13-10 = 0
so one root is shown to be x=-1, so x+1 is a factor...
factorise using this factor which you ahve now found..
(x+1)(x²+bx-10)
write the original and above equal to each other...
x³-2x²-13x-10=(x+1)(x²+bx-10)
collect the x² terms from both sides...
-2x²=x²+bx² (x*bx = bx², x²*1 = x²)
divide through by x²
-2=1+b
rearrange to find b
b = -2-1
b=-3
put b=-3 back into the partially-factorised cubic, (x+1)(x²+bx-10)
(x+1)(x²+bx-10) => (x+1)(x²-3x-10)
solve the quadratic, ie, (x²-3x-10) either by using the quadratic equation, completing the square or factorising. as you can see, this factorises really easily...
(x²-3x-10) = (x-5)(x+2)
from this we can see, x=5, or x=-2. from before we showed that x=-1
so x=-2, x=-1, x=5
they always say express the cubic it as the product of its factors
so
x³ - 2x² - 13x - 10 = (x+2)(x+1)(x-5)
done, easy as Pi... this may seem long and tedious, but once you know the method the answer pops out like a dream after a couple of minutes calculating...
just wait till p4, when you do this exact same method for complex cubics, ie using imaginary factors
El Stevo