Respuesta :

1.) Set it up like a long division problem.
(2x + 1)/(6x^3 - 1x^2 + 1)

2.) How many times does 2x go into 6x^3? (just look at the first term for long division polynomials). It goes in 3x^2 times.

3.) Do the same thing you would in a normal long division problem. Because you subtract and bring down 3x^2(2x+1), you multiply -3x^2 instead of 3x^2. You will add that.

4.) Distribute -3x^2(2x+1)
-6x^3 - 3x^2

5.) Add -6x^3 - 3x^2 to 6x^3 - 1x^2

6.) You get -4x^2

7.) Repeat. How many times does (2x + 1) go into -4x^2?

8.) -2x times. Multiply NEGATIVE ONE to it as you are subtracting it (even if it starts out negative), add it, repeat until you run out of terms.

9.) Any remainders should be the remainder over (2x + 1)

You should get 3x²-2x+1

Hope this helps, ask for any specifications so I can edit this :)
6x^3-1x^2+1/2x+1

Multiply 1 times x^2 = x^2=6x^3-x^2+1/2x+1

Use the rational root theorem

(2x-1)(3x²-2x+1)/2x+1

Cancel the common factor: 2x+1

3x²-2x+1