Step-by-step explanation:
:) I have a graphic that illustrates this beautifully
You have two polynomials, one divided by the other. If you re-arrange them in the proper fashion (as shown below), then you can divide them. Just think of the first value of the divisor as being the first digit in an ordinary number; same for the dividend.
(ax^2 + bx + c)/(ax + c)
Illustrated but i dont know how to put it in the middle so so^y
Note that the remainder is actually -1, not -11 as noted in the notes on the left side of the graphic.
The values in the quotient are multiplied by the divisor, and then subtracted in each step.
I came up with x^2 for the first value in the quotient. Multiply that by (3x-2) to get 3x^3 - 2x^2
Subtract this from the dividend in the first step. Remember that you are subtracting -2x^2, which changes it sign to +
thus in the first step you subtract 3x^3 and add 2x^2 from the first two values in the dividend, which are 3x^3 + x^2; the 3x^3 cancels out, and you add 2x^2 to x^2 to get 3x^2. This is carried down into the next step, and 4x drops down from the dividend.
You then draw +x from that and put it in the quotient. Then multiply x by (3x-2) to get 3x^2 + 4x. Subtract this from (3x^2 - 4x). The 3x^2 cancels out, and +2x is added to 4x. The result is 6x, which is brought down to the next step. The last value of the dividend, -5, is also brought down.
You then end up with 6x-5. The final value in the quotient is +2; multiply that by (3x-2) and you end up with 6x-4. Subtract this from 6x-5 and you end up with just -1.
So the quotient is x^2 + x + 2 with a remainder of 1/(3x-2).
You’re essentially trying to find a number (which will be in the quotient) to multiply by the divisor which should cancel out the relevant part of the dividend.
Compare this to regular long division (one number divided by another) and you’ll see how the logic between the two overlaps
