Since each has a common factor of x+2, -2 is a zero
of each polynomial. Therefore if we substitute x=-2
in each we will get 0.
x³+ax²-x+b x³+bx²-5x+3a
(-2)³+a(-2)²-(-2)+b=0 (-2)³+b(-2)²-5(-2)+3a=0
-8+a(4)+2+b=0 -8+b(4)+10+3a=0
-6+4a+b=0 2+4b+3a=0
4a+b=6 3a+4b=-2
Solve the system of equation:
4a+ b= 6
3a+4b=-2
Get a=2, b=-2.
Edwin