1. Concept of the remainder theorem
The rest of the channelisation polynomials to be an interesting when the rest of the result of the substitution value of 0 divisor into the Shared Function. By that’s why then the mathematician examines the rest of the divisions in the remainder theorem.
a. Concept division remainder with divisor $(x-k)$.
On the division of polynomial by ordinary relationships have been known:
The divisor $P(x)$ will be replaced with the $(x-k)$, so:
If $ x$ replaced with $k$ then:
$f(k)=(k-k).H(k)+S$ or

Appear to be in the calculation of that if polynomials $f(x)$ divided by $(x-k)$ then the rest is $S=f(k)$

Example 1:
Determine the remainder of the division on the following functions:
a. $\quad f(x)=x^3-2x^2-5x+1$ divided $(x-2)$.
b. $\quad f(x)=x^4+3x^3-3$ divided $(x+1)$.
a. $\quad S=f(2)$
$\quad =2^3-2.(2)^2-5.(2)+1$
$\quad =-9$
b. $\quad S=f(-1)$
$\quad =(-1)^4+3.(-1)^3-3$
$\quad =-5$

Example 2:
Determine the value $t$ if:
a. $\quad x^3+tx^2+3x-(3+t)$ divided $(x-2)$ remainder 17
b. $\quad x^4+tx^3-3x^2+(t^2+1)x-16$ divided up $(x-2)$
a. $\quad S=f(2)=17$
$\quad 17=2^3+t.(2)^2+3.(2)-(3+t)$
$\quad 17=8+4t+6-3-t$
$\quad -3t=-6$
$\quad t=2$.
b. $\quad S=f(2)=0$
$\quad 0=2^4+t.(2)^3-3.(2)^2$
$\quad +(t^2+1)(2)-16$
$\quad 0=t^2+4t-5$
$\quad t=-5$ dan $t=1$.

b. Concept division remainder with divisor $ax+b$.
In the division of polynomial it has been known relationship:
Because by divisor if $ax+b$ then $P(x)$ become $ax+b$, so
If $x=-b/a$ then:

Appear to be in the calculation that if $f(x)$ divided $ax+b$ then remainder is:

Example 3:
Determine devision remainder from:
a. $\quad 2x^3-3x^2+4x-7$ by divisor $(2x+3)$.
b. $\quad x^3+x^2-2x+8$ by divisor $(3x-2)$.
a. $\quad S=f(-3/2)$
$\quad =2.(-3/2)^3-3.(-3/2)^2$
$\quad +4.(-3/2)-7=-53/2$
b. $\quad S=f(2/3)$
$\quad =(2/3)^3+(2/3)^2-2.(2/3)+8$
$\quad =200/27$

Example 4:
Determine the value $t$ if $f(x)=2x^3-(t+1)x^2-19x+20$ and $f(x)=2x^3-13x^2+17x+t$ divided by $(2x-3)$ produce remainder of the same.
Answer: For function $f(x)=2x^3-(t+1)x^2-19x+20$ then:
For function $f(x)=2x^3-13x^2+17x+t$ then:
$ (-9t-16)/4=(4t+12)/4$ or

c. Concept division remainder with the divisor degrees of two or more and have the divisor factors
Because of the divisor degrees of two or more diverse types, then for that direct author provides examples of about the division remainder of degrees of two or more that has the factors. Next, the following theory the division remainder this will be able to be understand of every examples.

Example 5:
Determine the division remainder if $f(x)=x^4-2x^3+3x^2+3$ divided by $x^2-x-2$.
Because divisor is degrees-2, then his remainder degrees-1. If that remainder is $S=mx+n$. So,
See that divisor $x^2-x-2=(x-2)(x+1)$ then if divided by $(x-2)$, his remainder is:
And if divided by $(x+1)$ then:
Then: $m=2$ and $n=11$.
So, $S=2x+11$.

Example 6:
Given polynomial $f(x)$. If $f(x)$ divided by $(x-1)$ then remainder is 5 and if divided by $(x+1)$ then remainder is $-3$. Determine the remainder if divided by $x^2-1$.
If divisor is degrees-2 then remainder degrees is one. If remainder is $S=mx+n$ then:
$f(1)=m+n=5$ and $f(-1)=-m+n=-3$. Then, $m=4$ and $n=1$. So, $S=4x+1$.

Example 7:
Polynomial $f(x)$ if divided by $(x-1)$, $(x+1)$ and $(x-3)$ in a row the remainder is 12, 4, and 16. Determine the remainder if divided by $(x^2-1)(x-3)$.
Because divisor is polynom of degrees-3 then the remainder is polynom of degrees-2. If the remainder is $S=kx^2+lx+m$, then:
$f(-1)=k-l+m=4$ and
If this is resolved then retrieved:
$k=-1/2,\quad l=4$ $\quad$ and $m=17/2$. So, $S=-x^2/2+4x+17/2$.

Example 8:
Given $f(x)=x^4+ax^3+6x+b$ divided up by $x^2+2x-3$. Determine velue $a$ and $b$.
Because $x^2+2x-3=(x-1)(x+3)$ then,
$f(1)=a+b+7=0$ or $a+b=-7$.
$f(-3)=-27a+b+63=0$ or $-27a+b=-63$.
So, $a=2$ and $b=-9$.

Example 9:
If $f(x)$ divided by $x^2-1$ then remainder is $2x+3$, if divided by $x^2-4$ then remainder is $x+5$. Determine the remainder if divided by $x^2+x-2$.
Because $x^2+x-2=(x+2)(x-1)$ and take $S=mx+n$ then:
$f(-2)$ there is on divisor $x^2-4$ and $f(1)$ there is on divisor $ x^2-1$, then:
$f(-2)=-2m+n=3$ and $f(1)=m+n=5$.
So, $m=2/3$ and $n=13/3$, produce $S=2/3x+13/3$.

2. Concept the factor theorem
Note the following theorem:

$(x-k)$ is a factor of polynomial $f(x)$ if and only if $f(k)=0$.

Example 1:
Write down $f(x)=x^3-3x^2+4$ in form linear factors!
See that $f(-1)=0$, then one of his factor is $(x+1)$. Next, by doing division of Horner, then:

Note: The number most left is the multiplier into the number of rows down.
Other factors is retrieved $x^2-4x+4=(x-2)^2$.
So, $f(x)=(x+1)(x-2)^2$.

Example 2:
One of the root equation $x^3-(3+2m)x^2+$
$(m^2+5m+3)x-2m(m+2)=0$ is 2. Calculate $m$ and the real constant roots of more!
$f(2)=-m+1=0$ or $m=1$
Substitution $m=1$, then $f(x)=x^3-5x^2+9x-6=0$, and for looking for the other root then needed way the division of Horner as follows:
Then other factor: $x^2-3x+3$ (this factor form is quadratic). Because the value of the discriminant is smaller zero, then its constant factor is imaginary. So, the factors there are only 1 that is 2.

Looking for the properties of the root polynomial.
Pay attention to an explanation in the following image:

Example 3:
Known $x_1$, $x_2$ and $x_3$ is roots of the equation $2x^3-bx^2-18x+36=0$. Determine:
a. $\quad x_1+x_2+x_3$
b. $\quad x_1.x_2+x_1.x_3+x_2.x_3$
c. $\quad x_1.x_2.x_3$.
Of the formula that is described in the picture above, then:
a. $\quad x_1+x_2+x_3=-(-b)/2=b/2$
b. $\quad x_1.x_2+x_1.x_3+x_2.x_3$
$\quad =-18/2=-9$
c. $\quad x_1.x_2.x_3=-36/2=-18$.