When the rational number is negative
I'm assuming this is x^2 + 3x - 4 and x(x^2 + 3x - 2)
1.) First distribute x(x^2 + 3x - 2) to get x^3 + 3x^2 - 2x.
2.) Because you are subtracting all the terms from x^3 + 3x^2 - 2x, it's the same thing as distributing -1 to x^2 + 3x - 4 and then adding it to x^3 + 3x^2 - 2x.
3.) -1(x^2 + 3x - 4) = -x^2 - 3x + 4
4.) Add (x^3 + 3x^2 - 2x) + (-x^2 - 3x + 4)
5.) x^3 + 2x^2 - 5x + 4 is your final answer.
Answer:
none of the two graphs in the pic.
it should be one in the II quadrant
Step-by-step explanation:
the plus four means to move 4 to the left
the plus 2 means to move up 2
That would make the vertex in the II quadrant.
