Given that x=3 is a solution to f(x)= 2x^3 - 8x^2 + 7x - 3 = 0, solve f(x)=0 completely.
As x=3 is a solution, (x - 3) is a factor of f(x) = 2x3- 8x2+ 7x - 3 = 0 ie. 2x3- 8x2+ 7x - 3 = (x - 3)(Ax2 + Bx + C) where A,B,C are constants to be found. Expanding and then grouping terms by power, we obtain 2x3- 8x2+ 7x - 3 = Ax3+ (B - 3A)x2 + (C - 3B)x - 3C. Observe that A is the only coefficient of the x3 component, and therefore A = 2. By substituting A = 2 into the equation B - 3A = - 8, we have B = 3A - 8 = 3(2) - 8 = - 2 ie. B = -2. Observe that C is the only non-x coefficient and therefore C = 1. Therefore 2x3- 8x2+ 7x - 3 = (x - 3)(2x2 - 2x + 1)Now using the quadratic formula x = (-b +/- sqrt(b2 - 4ac))/2a, factorise 2x2 - 2x + 1. Giving x = (1 +/- i)/2. Therefore we have that x = 3, (1 + i)/2 and (1 - i)/2 are the three solutions of f(x)=0.
