Division of polynomials

Consider the following polynomials:

$$p(x)=-x^3+x$$

$$q(x)=2x^3-x-3$$

$$r(x)=-x+1$$

Do the following operation: $(r(x)+q(x))\cdot p(x)$

First we do the addition,

  r(x) q(x) r(x)+q(x)
degree 0 $1$ $-3$ $-2$
degree 1 $-x$ $-x$ $-2x$
degree 2 $0$ $0$ $0$
degree 3 $0$ $2x^3$ $2x^3$

$$r(x)+q(x)=2x^3-2x-2$$

Then, the product by the monomials of $p(x)$,

$$x\cdot(r(x)+q(x))=x\cdot(2x^3-2x-2)=2x^4-2x^2-2x$$

$$-x^3\cdot(r(x)+q(x))=-x^3\cdot(2x^3-2x-2)=-x^6+2x^4+2x^3$$

We join both polynomials and make groups of similar terms:

$$(r(x)+q(x))\cdot p(x)=(2x^4-2x^2-2x)+(-x^6+2x^4+2x^3)=$$

$$=-x^6+4x^4+2x^3-2x^2-2x$$

Calculate the following polynomial division $\dfrac{x^4-2x+3}{x^2-2}$

We complete and we design the initial table

$x^4$ $0$ $0$ $-2x$ $3$ $x^2-2$

Step 1:

$$\dfrac{x^4}{x^2}=x^2$$

$$x^2\cdot(x^2-2)=x^4-2x^2$$

$x^4$ $0$ $0$ $-2x$ $3$ $x^2-2$
$-x^4$ $0$ $+2x^2$ $0$ $0$ $x^2$
$0$ $0$ $+2x^2$ $-2x$ $3$  

Step 2:

$$\dfrac{2x^2}{x^2}=2$$

$$2\cdot(x^2-2)=2x^2-4$$

$x^4$ $0$ $0$ $-2x$ $3$ $x^2-2$
$-x^4$ $0$ $+2x^2$ $0$ $0$ $x^2+2$
$0$ $0$ $+2x^2$ $-2x$ $3$  
    $-2x^2$ $0$ $+4$  
    $0$ $-2x$ $7$  

The process ends here because:

degree$(-2x+7)=1 < 2=$degree$(x^2-2)$

Quotient: $x^2+2$

Reminder: $-2x+7$

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