x²-2x-3=(x+1)(x-3) so -1 and 3 are zeroes and must therefore be solutions to:
By substituting the zeroes in the above, we get a system of equations for a and b:
x=-1 gives us -4+b-9a+15=0⇒-9a+b=-11 ①
x=3 gives us 108+9b+27a+15=0⇒27a+9b=-123⇒9a+3b=-41 ②
Add ① and ②:
4b=-52, b=-52/4=-13. So rewriting ① we have 9a=b+11=-2 and a=-2/9.
So now we have the original cubic with values for a and b:
We know that this can have no more than three zeroes and we already have two of them. By looking at the coefficient of x³ and the constant we can deduce that the third factor is 4x-5 because 4x³/x²=4 and 15/-3=-5. The x² term and constant come from the quadratic divisor x²-2x-3.
So the zeroes are -1, 3 and 5/4.