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As has been seen, polynomials can be added, subtracted and multiplied. As it turns out, they can also be divided. For this, the methods of long division or *synthetic division* can be used.

In long division notation, the numerator is placed underneath the division symbol, and the denominator is placed to the left. The expression below shows the division $213 .$
$2 13 $
When the division is complete, there is a number above the division sign. $2 16113 −1221.1 $
Sometimes, two values will not divide *evenly.* As a result, the quotient contains a remainder. Here, there is a remainder of $1,$ which is expressed as **$1$ divided by $2,$** or $21 .$ The quotient can be written as
$213 =6+21 .$

The concepts of numerator, denominator, quotient, and remainder are also used when working with polynomial long division. If, for example, $x_{2}+4$ is divided by $x−1,$ the division is written as follows. $x−1 x_{2}+4 $ When the division is finished, the quotient is $x+1$ and the remainder is $3$. Thus, the result is written as

$(x+1)+x−13 .$To divide expressions where both the numerator and denominator are polynomials, polynomial long division can be used. The process is similar to when dividing real numbers. Consider the following division. $x_{2}+x+11x_{3}+7x−2−3x_{2} $

Sort the terms and write as long division

Divide the first term in the numerator by the first term in the denominator

$x_{2}+x+11 x_{3}−3x_{2}+7x−2 $

$x_{2}x_{3} =x$

$x−3x_{2}+7x−2+.x_{2}+x+11 x_{3}−3x_{2}+7x−2 $

Multiply the quotient in Step $2$ by the denominator

Subtract the product in Step $3$ from the numerator

$x−3x_{2}+7x−2+x_{2}+x+11 x_{3}−3x_{2}+7x−2 $

Subtract $x(x_{2}+x+11)=x_{3}+x_{2}+11x$

$x−3x_{2}+7x−2+33x_{2}+x+11 −(x_{3}−3x_{2}+7x1−2 −(x_{3}+x_{2}+11x) $

RemoveParSignsRemove parentheses and change signs

$x−3x_{2}+7x−2+3x_{2}+x+11 −x_{3}−3x_{2}+7x−2 −x_{3}3−x_{2}−11x $

SimpTermsSimplify terms

$x+7x−13888x_{2}+x+11 -4x_{2}−4x−2 $

Simplify

The expression calculated from the long division can now be interpreted as $x+x_{2}+x+11-4x_{2}−4x−2 .$ Since the numerator in the remaining fraction has the same degree as the denominator, perform Steps $2$-$4$ again.

Repeat Steps to 2$−4$ until the numerator has a lower degree than the denominator

$x+7x−13888x_{2}+x+11 -4x_{2}−4x−2 $

$x_{2}-4x_{2} =-4$

$x−4−1138888x_{2}+x+11 -4x_{2}−4x−2 $

Subtract $-4(x_{2}+x+11)=-4x_{2}−4x−44$

$x−4−(-4x_{2}+12x_{2}+x+11 −(-4x_{2}−4x−2 −(-4x_{2}−4x−44) $

RemoveParSignsRemove parentheses and change signs

$x−4-4x_{2}+12x_{2}+x+11 -4x_{2}−4x−2 +4x_{2}+4x+44 $

SimpTermsSimplify terms

$x−411x_{2}+x+11 424575 $

The quotient of $x_{2}+x+11x_{3}+7x−2−3x_{2} $ is the sum the expression above the division symbol and the remainder divided by the denominator. $x−4+x_{2}+x+1142 $

When dividing polynomials by binomials of the form $x−k,$ where $x$ is the independent variable and $k$ is a real number, there is an alternative to polynomial long division. **Synthetic division** uses $k$ and the coefficients of the numerator to find the quotient.
$x−3-x_{3}+4x_{2}+9 $
The result of the example division above can be determined using this process.
### 1

The division symbol for synthetic division is L-shaped. The number $k$ is written to the left. Here it's $3.$ $3 $ Next, write the coefficients of the numerator to the right. The numerator is $-x_{3}+4x_{2}+9$ so the coefficients are $-1,$ $4,$ $0,$ and $9.$ Note there is no $x$-term so that coefficient is $0.$ $3 -1 409 $

### 2

First, bring down the left-most coefficient. Here, that is $-1,$ so $-1$ is written below the horizontal line.
### 3

The newest number under the line is now multiplied by $k.$ Here that means $3(-1)=-3.$ The result is placed under the second term in the numerator.
### 4

Now, the numbers in the column created by multiplying are added.
### 5

These steps are repeated through the last column.
Now that the division is complete, the numbers below the horizontal line are the coefficients of the quotient.
### 6

Keeping in mind that the quotient is one degree lower than the numerator, the quotient can be written using the results of the synthetic division above. $-x_{2}+x+3+x−318 $ Since the last number below the horizontal line was not $0,$ there is a remainder. The remainder is written as the quotient of the remainder and the denominator.

Set up the division using $k$ and the coefficients of the numerator

Bring down the first coefficient

$3 -1 409 $

Bring down $-1$

$3 -1 4 0 9 -1 4 0 9 $

Multiply $k$ and the new number below the line

$3 -1 4 0 9 -1 4 0 9 $

$3(-1)=-3$

$3 -1 4-3 0 9 -1 4 0 9 $

Add the numbers in the column above the line

$3 -1 4-3 0 9 -1 4 0 9 $

$4+(-3)=1$

$3 -1 4-3 0 9 -1 1 0 9 $

Repeat Steps $3$-$4$ through the last column

$3 -1 4-3 0 9 -1 1 0 9 $

$3⋅1=3$

$3 -1 4-3 03 9 -1 1 0 9 $

$0+3=3$

$3 -1 4-3 03 9 -1 1 3 9 $

$3⋅3=9$

$3 -1 4-3 03 99 -1 1 3 9 $

$9+9=18$

$3 -1 4-3 03 9191 -1 1 3 18 $

Write the quotient

If a polynomial $p(x)$ is divided by a binomial in the form $x−a$, where $a$ is a real number, the result will be the sum of another polynomial and a fraction where the numerator is the remainder, $r.$ $x−ap(x) =q(x)+x−ar $ Then, according to the Remainder Theorem, the remainder can be determined by $r=p(a). $ In other words, the remainder is given by the function value of $p(x)$ when $x=a.$

To show why the remainder can be given by $r=p(a),$ the division $x−ap(x) =q(x)+x−ar $
can be rewritten by multiplying both sides by the binomial $x−a.$
By evaluating $p(x)$ when $x=a$ the expression can be simplified to $r=p(a).$
Therefore, the remainder is $r=p(a).$

$x−ap(x) =q(x)+x−ar $

MultEqn$LHS⋅(x−a)=RHS⋅(x−a)$

$p(x)=(q(x)+x−ar )⋅(x−a)$

DistrDistribute $(x−a)$

$p(x)=q(x)(x−a)+r$

$p(x)=q(x)(x−a)+r$

Substitute$x=a$

$p(a)=q(a)(a−a)+r$

SubTermSubtract term

$p(a)=q(a)⋅0+r$

MultiplyMultiply

$p(a)=r$

RearrangeEqnRearrange equation

$r=p(a)$

Use synthetic division to evaluate $f(4)$ if $f(x)=3x_{4}−5x_{3}+x_{2}−x−3.$

Show Solution

To evaluate the polynomial, we could substitute $x=4$ into the rule and simplify. However, the Remainder Theorem tells us that $f(4)$ is the remainder when dividing $f(x)$ by $x−4.$ Thus, by determining the remainder of
$x−43x_{4}−5x_{3}+x_{2}−x−3 ,$
will give $f(4).$ As instructed, we'll use synthetic division. The coefficients are $3,$ $-5,$ $1,$ $-1,$ and $-3.$
The remainder of the division is $457,$ which means that $f(4)=457.$

$4 3 -51-1-3 $

Bring down $3$

$4 3 -5 1 -1 -3 3 -5 1 -1 -3 $

$4⋅3=12$

$4 3 -512 1 -1 -3 3 -5 1 -1 -3 $

$-5+12=7$

$4 3 -512 1 -1 -3 3 73 1 -1 -3 $

$4⋅7=28$

$4 3 -512 128 -1 -3 3 75 11 -1 -3 $

$1+28=29$

$4 3 -512 128 -1 -3 3 72 29 -1 -3 $

$4⋅29=116$

$4 3 -512 128 -1116 -3 3 71 29 111 -3 $

$-1+116=115$

$4 3 -512 128 -1116 -3 3 7 29 115 -3 $

$4⋅115=460$

$4 3 -512 128 -1116 -3460 3 73 29 115 133 $

$-3+460=457$

$4 3 -512 128 -1116 -3460 3 7 29 115 457 $

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