Divide 2x^3 + x^2 - 4x - 2 by 2x + 1.

Step-by-step solution

1. 1

   Set up the long division. Both polynomials are written in descending powers with no missing terms, so no zero placeholders are needed. Divide leading term by leading term:

   $\frac{2x^{3}}{2x}=x^{2}.$

   Note that synthetic division is awkward here because the divisor is not monic; ordinary long division handles the leading $2$ without any rescaling.

2. 2

   Multiply back and subtract. Multiplying the divisor by the first quotient term:

   $x^{2}(2x+1)=2x^{3}+x^{2},$

   and subtracting leaves nothing from the first two terms:

   $\left(2x^{3}+x^{2}\right)-\left(2x^{3}+x^{2}\right)=0.$

   Bringing down the rest gives the new dividend $-4x-2$.

3. 3

   Repeat on the remainder. Dividing leading terms again:

   $\frac{-4x}{2x}=-2,\qquad -2(2x+1)=-4x-2,$

   and subtracting gives $0$. The division terminates with remainder zero, so $2x+1$ is a genuine factor.

4. 4

   Confirm by grouping. The same result follows without division at all:

   $2x^{3}+x^{2}-4x-2=x^{2}(2x+1)-2(2x+1)=(2x+1)\left(x^{2}-2\right).$

   Spotting that the first two terms and the last two terms each contain $2x+1$ is faster than long division whenever a cubic has this paired structure.

5. 5

   State the answer and check. The quotient is

   $\frac{2x^{3}+x^{2}-4x-2}{2x+1}=x^{2}-2,\qquad x\neq-\tfrac12.$

   As a bonus the full factorisation is $(2x+1)(x-\sqrt2)(x+\sqrt2)$, so the roots are $-\tfrac12$ and $\pm\sqrt2$. Numerical check at $x=2.3$: the quotient gives $3.29$ and the original ratio gives $3.29$ ✓.