Question

polynomial long division​

Answer 1

You're dividing $2x^3-2x^2+3x-4$ (1) by $x+1$ (2).

$2x^3=2x^2\cdot x$, and $2x^2(x+1)=2x^3+2x^2$ (3). Subtract this from (1) to get a remainder of

$(2x^3-2x^2+3x-4)-(2x^3+2x^2)=-2x^2+3x+4$ (4)

$-2x^2=-2x\cdot x$, and $-2x(x+1)=-2x^2-2x$ (5). Subtract this from (4) to get a new remainder of

$(-2x^2+3x+4)-(-2x^2-2x)=5x+4$ (6)

$5x=5\cdot x$, and $5(x+1)=5x+5$ (7). Subtract this from (6) to get a new remainder of

$(5x+4)-(5x+5)=-1$ (8)

$-1$ doesn't contain any factors of $x$, so we're done and we've shown

$\dfrac{2x^3-2x^2+3x-4}{x+1}=2x^2+\dfrac{-2x^2+3x+4}{x+1}$

$\dfrac{2x^3-2x^2+3x-4}{x+1}=2x^2-2x+\dfrac{5x+4}{x+1}$

$\dfrac{2x^3-2x^2+3x-4}{x+1}=2x^2-2x+5-\dfrac1{x+1}$

so that the quotient is $2x^2-2x+5$ (9).