Remainder theorem
The remainder of dividing a polynomial $$p(x)$$ by another one of the form $$x-a$$, coincides with the value of $$p(a)$$.
Notice that this kind of division satisfies the hypotheses of the Ruffini's rule.
Calculate the remainder of the division $$\dfrac{p(x)}{q(x)}$$, where $$p(x)=x^4+3x^2-x+4$$ and $$q(x)=x+2$$.
We apply the remainder theorem. Notice that, in this case $$a=-2$$. $$$p(-2)=(-2)^4+3\cdot(-2)^2-(-2)+4=16+3\cdot4+2+4=34$$$
To verify it we use Ruffini:
| $$1$$ | $$0$$ | $$3$$ | $$-1$$ | $$4$$ | |
| $$-2$$ | $$-2$$ | $$4$$ | $$-14$$ | $$30$$ | |
| $$1$$ | $$-2$$ | $$7$$ | $$-15$$ | $$34$$ |
And, it is the same as the previous solution.
Calculate the remainder of the division $$\dfrac{p(x)}{q(x)}$$, where $$p(x)=x^5-2x^2+x+3$$ and $$q(x)=x+1$$.
We apply the remainder theorem. Notice that, in this case $$a=-1$$. $$$p(-1)=(-1)^5-2\cdot(-1)^2+(-1)+3=-1-2-1+3=-1$$$
To verify it we use Ruffini:
| $$1$$ | $$0$$ | $$0$$ | $$-2$$ | $$1$$ | $$3$$ | |
| $$-1$$ | $$-1$$ | $$1$$ | $$-1$$ | $$3$$ | $$-4$$ | |
| $$1$$ | $$-1$$ | $$1$$ | $$-3$$ | $$4$$ | $$-1$$ |
And it is the same than the previous solution.
Factor theorem
Its statement is the following one:
A polynomial $$p(x)$$ is divisible by another of the form $$x-a$$ if, and only if, $$p(a)=0$$. In this case, we will say that $$a$$ is a root or zero of the polynomial $$p(x)$$.
Calculate the remainder of the division $$\dfrac{p(x)}{q(x)}$$, where $$p(x)=x^5+2x^4-3x^3+x^2-1$$ and $$q(x)=x-1$$.
We apply the remainder theorem $$$p(1)=1^5+2\cdot1^4-3\cdot1^3+1^2-1=0$$$
We verify the result using Ruffini:
| $$1$$ | $$2$$ | $$-3$$ | $$1$$ | $$0$$ | $$-1$$ | |
| $$1$$ | $$1$$ | $$3$$ | $$0$$ | $$1$$ | $$2$$ | |
| $$1$$ | $$3$$ | $$0$$ | $$1$$ | $$1$$ | $$0$$ |
Indeed, the remainder is $$0$$. And so, according to the factor theorem, the division of $$p(x)$$ by $$q(x)$$ is exact.