Help solve rational expression

Question

I need help solving this rational expression.

Divide $$\frac{4x^4 + 6x^3 + 3x - 1}{2x^2 + 1}$$

How do you solve this problem? Where do I start?

Answer 1

I'd use polynomial division, on $\frac{4x^4 + 6x^3 + 3x - 1}{2x^2+1}$, this results in,

$2x^2+3x-1$

Since, that is quadratic, it should be easy from here.

Answer 2

$$\frac{4x^4 + 6x^3 + 3x - 1}{2x^2 + 1}=\frac{4x^4 -1 +6x^3 + 3x}{2x^2 + 1}=\frac{(2x^2+1)(2x^2-1)+3x(2x^2+1)}{2x^2 + 1}=\frac{(2x^2+1)(2x^2+3x-1)}{2x^2 + 1}=2x^2+3x-1$$

Answer 3

The long division method for polynomials is same as that of numbers, except for the fact that you need to keep a track of the powers of x while performing long divison.

I don't know if there exist a way to show long divison like on pen and paper, but I am putting the exact steps below.

Okay, so the divisor(d) is $(2 x^2+1)$ and dividend(D) is $4x^4 + 6x^3 + 3x - 1$. Let Quotient(Q) =$0$. Whatever you are multiplying to d gets added as a quotient to Q.

1) Multiply d by $2x^2$(add this to Q) you get $4x^4+2x^2$ and subtract this from D, you are left with $6x^3-2x^2+3x-1$ which becomes our new D.

2) Multiply d with $3x$(Q now becomes $2x^2+3x$) you get $6x^3+3x$, subtract this from D, you are left with $-2x^2-1$ and this becomes our new D.

3) Multiply d with $-1$ (Q now becomes $2x^2+3x-1$), you get $-2x^2-1$, subtract this from D, remainder is $0$.

If anyone is down voting, please be merciful and let me know so I can correct mistake.