Using rational root theorem: The rational roots are ± (factors of 63 divided by factors of 2)
The options are ±(1, 3, 7, 9, 21, 63 and 1/2, 3/2, 7/2, 9/2, 21/2, 63/2) A bit of a mess.
1,-1 are always first try: for 1, just add coefficients and for -1 add even power coefficients and negate odd coefficients: P(1) = 80, P(-1) = -180 (note there is a change of sign, so try the zero between these. (BTW, Descartes' Law of signs says there are 1 positive real root and 1 or 3 negative roots which can be helpful)
± 1/2 You find that 1/2 works P(1/2) = 0
You can now divide P(x) by 2x-1 to get a new polynomial (you can also use synthetic division.
You get x3 + 7x2 + 9x + 63 = 0 The choices here that are left (negative roots only) -3,-7, -9, -21, -63
-3 is too small, try -7 which works. You can actually see it because the 1st two terms cancel and the last two terms cancel. Factor out x+7 (by division or S.D) and now you have a quadratic with no rational roots (you could use quadratic formula, but you can tell that the solution is imaginary - no factoring)
