To calculate the quotient of two polynomials the procedure used needs many intermediate calculations. A rule that can help us to simplify them is Ruffini's rule. This rule will only be valid when the divisor is a polynomial, such as $$x-a$$, with $$a$$ being a real number.
We will use an example to explain the methodology:
Do the division $$\dfrac{p(x)}{q(x)}$$, where $$p(x)=x^4-3x^2+x+5$$ and $$q(x)=x+2$$.
1) Complete and arrange the dividend polynomial.
Write the dividing polynomial as $$x-a$$, if necessary.
In our case:
$$p(x)=x^4+0x^3-3x^2+x+5$$
$$q(x)=x-(-2)$$
Notice that in this example the value of $$a=-2$$.
2) We write down the elements in a table like the following one.
| $$1$$ | $$0$$ | $$-3$$ | $$1$$ | $$5$$ | |
| $$-2$$ | |||||
In the top row, we write the coefficients of the polynomial (arranged and completed!) $$p(x)$$.
In the left cell, we write the value of $$a$$.
3) We put the first coefficient in, and multiply it by the value of $$a$$. The result of which we write just under the second coefficient:
| $$1$$ | $$0$$ | $$-3$$ | $$1$$ | $$5$$ | |
| $$-2$$ | $$1\cdot(-2)=-2$$ | ||||
| $$1$$ |
4) We add up the second column and put the obtained result in, repeating the process until the last column:
| $$1$$ | $$0$$ | $$-3$$ | $$1$$ | $$5$$ | |
| $$-2$$ | $$1\cdot(-2)=-2$$ | $$(-2)\cdot(-2)=4$$ | $$1\cdot(-2)=-2$$ | $$(-1)\cdot(-2)=2$$ | |
| $$1$$ | $$0+(-2)=-2$$ | $$(-3)+4=1$$ | $$1+(-2)=-1$$ | $$5+2=7$$ |
5) The digit on the bottom-right corner is the remainder. The other digits of the last row are the coefficients, arranged, for the polynomial quotient.
And so, in our case:
quotient: $$x^3-2x^2+x-1$$
remainder: $$7$$
As we can see, the relation od degrees is satisfied:
$$3=$$degree$$(x^3-2x^2+x-1)=$$degree$$(x^4-3x^2+x+5)-$$degree$$(x+2)=4-1=3$$
degree$$(7)=0 < 1 =$$degree$$(x+2)$$
Do the division $$\dfrac{p(x)}{q(x)}$$, where $$p(x)=x^5+2x^4-3x^3+x^2-1$$ and $$q(x)=x-1$$.
1) $$p(x)=x^5+2x^4-3x^3+x^2+0x-1$$
$$q(x)=x-1$$
$$a=1$$.
2)
| $$1$$ | $$2$$ | $$-3$$ | $$1$$ | $$0$$ | $$-1$$ | |
| $$1$$ | ||||||
3)
| $$1$$ | $$2$$ | $$-3$$ | $$1$$ | $$0$$ | $$-1$$ | |
| $$1$$ | $$1$$ | |||||
| $$1$$ | $$3$$ |
4)
| $$1$$ | $$2$$ | $$-3$$ | $$1$$ | $$0$$ | $$-1$$ | |
| $$1$$ | $$1$$ | $$3$$ | $$0$$ | $$1$$ | $$1$$ | |
| $$1$$ | $$3$$ | $$0$$ | $$1$$ | $$1$$ | $$0$$ |
5)
quotient: $$x^4+3x^3+x+1$$
remainder: $$0$$
And it is satisfied that:
$$4=$$degree$$(x^4+3x^3+x+1)=$$degree$$(x^5+2x^4-3x^3+x^2-1)-$$
$$-$$degree$$(x-1)=5-1=4$$
degree$$(0)=0 < 1 =$$degree$$(x-1)$$