All rational roots will be rational factors of P/Q such that:
Q = 3 Factors: 1,3 P = -6 Factors: [+/-] 1,2,3,6
Possible Rational Roots: [1/1,1/3,2/1,2/3,3/1,3/3,6/1,6/3] = [+/-] 1,1/3,2,2/3,3,6
Now you just test them in the equation itself and where the input makes the function equal 0, you have a root.
For this polynomial, no roots are rational, so when you test it you'll find that it must only contain irrational roots and may only be solved by other means.
Steps I’m using elimination so I’m trying to get rid of y To do that I multiply the first equation by 2 and the second by -3. Then solve 4x + 6x = 12 15x + -6x = -12 ————————— 19x + 0 = 0 Divid by 19 to get x to one side. 0/19 is 0 so X= 0 Plug that into an equation so 0 + 3y = 6 Divid by 3 and y = 2 Answer x = 0 y = 2