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Two polynomials can be added or subtracted by combining their like terms. Additionally, the polynomials can be multiplied by using either the FOIL method, the Box Method, or the Distributive Property. Notice that the result of any of these operations is always a polynomial. Therefore, the set of polynomials is closed under addition, subtraction, and multiplication.

### Catch-Up and Review

Operation | Methods | Result |
---|---|---|

Addition | Combining Like Terms | Polynomial |

Subtraction | Combining Like Terms | Polynomial |

Multiplication | The FOIL Method The Box Method The Distributive Property |
Polynomial |

This lesson completes the set of the four basic operations for polynomials by investigating the methods to *divide* two polynomials.

**Here are a few recommended readings before getting started with this lesson.**

Consider the polynomials $P(x)$ and $Q(x)$ and the constant $k.$

Let $H(x)$ be the product of the two given polynomials — that is, $H(x)=P(x)⋅Q(x).$ Find $H(x)$ and evaluate it at $x=k.$ Next, consider a different pair of polynomials and repeat the process. What conclusion can be made about $H(x)?$ Repeat the process as much as needed.

Adding, subtracting, and multiplying polynomials are operations that can be performed by applying the properties of real numbers and the product of powers property and then combining like terms. However, dividing polynomials may not be so intuitive.
*evenly*, the remainder is $0.$ Otherwise, the remainder is a natural number less than the denominator. The division of two integers can always be written in terms of the quotient and remainder as follows.
### Method

## Polynomial Long Division

$x_{2}+2x+53x_{3}+2x_{2}−7 =? $

The good news is that polynomial division can be performed similarly to dividing integer numbers. In long division notation, the numerator is placed underneath the division symbol and the denominator is placed to the left. As a refresher, the process for calculating $4117 $ is shown.
When the denominator divides the numerator

$DenominatorNumerator =Quotient+DenominatorRemainder ⇓4117 =29+41 $

This process can be imitated for dividing polynomials, but instead of having just numbers, the process involves algebraic expressions. Here, both the quotient and the remainder are polynomials and the division is complete once the remainder's degree is lower than the denominator's degree.
To perform the division of two polynomials, the degree of the numerator must be greater than or equal to the degree of the denominator. For example, consider the following division.
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To verify whether the division is correct, substitute the dividend, divisor, quotient, and remainder into the following equation and check if a true statement is obtained.
**not** always a polynomial. Therefore, the set of polynomials is **not** closed under division. ### Extra

What Happens When the Numerator's Degree Is Less Than the Denominator's Degree?

$x_{2}+2x+53x_{3}+2x_{2}−7 $

The process for finding the division of two polynomials is similar to the process of dividing integers and is summarized in the following five steps.
1

Sort the Terms and Write as Long Division

To begin, write both polynomials in standard form. If the numerator has missing terms, include them with coefficient $0.$ For example, the numerator will be written as $3x_{3}+2x_{2}+0x−7.$ Then, write the polynomials as in long division notation.

$x_{2}+2x+5 3x_{3}+2x_{2}+0x−7 $

2

Divide the Leading Terms

Divide the leading term of the numerator by the leading term of the denominator. In this case, divide $3x_{3}$ by $x_{2}.$

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

Write the resulting monomial above the horizontal bar.
$x_{2}+2x+5 3x−73x_{3}+2x_{2}+0x−7 $

3

Multiply the Result From Step $2$ by the Denominator

The result from step $2$ is $3x.$ Therefore, multiply $3x$ by the denominator $x_{2}+2x+5.$

$3x(x_{2}+2x+5)=3x_{3}+6x_{2}+15x $

Next, write the resulting expression below the numerator and try to align the common powers.
$x_{2}+2x+5 3x−73x_{3}+2x_{2}+0x−7 3x_{3}+6x_{2}+15x−7 $

4

Subtract the Product From Step $3$ From the Numerator

Subtract the polynomial obtained in step $3$ from the numerator.

$x_{2}+2x+5 3x−73x_{3}+2x_{2}+0x−7 −(3x_{3}+6x_{2}+15x) -4x_{2}−15x−7 $

The result of the subtraction — the remainder — is the polynomial $-4x_{2}−15x−7.$ This means that the division of the given polynomials can be written as follows.
$x_{2}+2x+53x_{3}+2x_{2}−7 =3x+x_{2}+2x+5-4x_{2}−15x−7 $

However, since the degree of $-4x_{2}−15x−7$ is the same as the degree of the denominator, the division is not done yet. Thus, repeat steps $2-4$ until the remainder's degree is less than the denominator's degree. 5

Repeat Steps $2-4$ Until the Remainder's Degree is Less Than the Denominator's Degree

Steps $2-4$ are repeated until the polynomial obtained in step $4$ has a **lower** degree than the denominator. Dividing the leading terms $-4x_{2}$ and $x_{2}$ gives $-4$ as result. Next, multiply the denominator by $-4$ and write the result at the bottom.

$x_{2}+2x+5 3x−43x_{3}+2x_{2}+0x−7 −(3x_{3}+6x_{2}+15x) -4x_{2}−15x−7 -4x_{2}−8x−20 $

Now, subtract the two polynomials at the bottom.
$x_{2}+2x+5 3x−43x_{3}+2x_{2}+0x−7 −(3x_{3}+6x_{2}+15x) -4x_{2}−15x−7 −(-4x_{2}−8x−20) -7x+13 $

This time, the resulting polynomial has a degree of $1,$ which is lower than the denominator's degree. Therefore, the division is complete. This implies that the quotient of the division is $3x−4$ and the remainder is $-7x+13.$
$Q(x)R(x) =3x−4=-7x+13 $

Finally, the initial polynomial division can be written as the $quotient$ plus the division of the $remainder$ and the $denominator.$
$x_{2}+2x+53x_{3}+2x_{2}−7 =3x−4+x_{2}+2x+5-7x+13 $

$Dividend=Divisor⋅Quotient+Remainder⇕P(x)=D(x)Q(x)+R(x) $

Notice that unless the remainder is equal to $0,$ the division of two polynomials is When the numerator has a lower degree than the denominator, the result from dividing the leading terms will have a negative exponent. Therefore, it will not be a term of a polynomial. For example, consider the following division.
**not** a monomial.

$x_{2}−34x+1 $

The result of dividing the leading terms is $4x_{-1},$ which is
Electrical power is defined as the product of the voltage of a circuit and the current flowing through it. In other words, electrical power equals voltage times current.
### Hint

### Solution

From this relation, a formula for finding the voltage of a circuit can be derived.

$Voltage=CurrentPower $

Consider a circuit for which the power is modeled by $P(x)=2x_{4}−x_{3}+3x+4$ and the current is modeled by $I(x)=x_{2}−4x+5.$ Perform the required division to find an algebraic expression that represents the voltage of the circuit. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.22222em;\">V<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["2x^2+7x+18 + \\dfrac{40x - 86}{x^2 - 4 x + 5}"]}}

Divide the polynomials using long division.

According to the formula, the voltage of a circuit is found by dividing the power by the current.

$Voltage=CurrentPower $

In this case, both the power and the current of the circuit are represented by polynomials.
$P(x)I(x) =2x_{4}−x_{3}+3x+4=x_{2}−4x+5 $

To find the voltage, these two polynomials have to be divided. Such division can be done following the steps of polynomial long division. First, write both polynomials in standard form, filling in any missing terms in the numerator with zeros. In $P(x),$ only the quadratic term is missing.
$P(x)I(x) =2x_{4}−x_{3}+0x_{2}+3x+4=x_{2}−4x+5 $

Now, write the dividend $P(x)$ under the division symbol and the divisor $I(x)$ to the left.
$x_{2}−4x+5 2x_{4}−x_{3}+0x_{2}+3x+4 $

The first term of the quotient is found by dividing the leading term of $P(x)$ by the leading term of $I(x).$
$x_{2}2x_{4} =2x_{2} $

Next, write the monomial above the horizontal bar.
$x_{2}−4x+5 2x_{2}+3x+42x_{4}−x_{3}+0x_{2}+3x+4 $

Multiply the divisor by $2x_{2}$ and write the resulting expression below the dividend. In this case, the multiplication gives $2x_{4}−8x_{3}+10x_{2}.$
$x_{2}−4x+5 2x_{2}+3x+42x_{4}−x_{3}+0x_{2}+3x+4 2x_{4}−8x_{3}+10x_{2}+3x+4 $

Subtract the resulting polynomial from the dividend.
$x_{2}−4x+5 2x_{2}+3x+42x_{4}−x_{3}+0x_{2}+3x+4 −(2x_{4}−8x_{3}+10x_{2}) +3x+47x_{3}−10x_{2}+3x+4 $

Since the remainder has degree $3$ and the divisor has degree $2,$ the division is not done yet. Therefore, the process will be repeated until the remainder's degree is less than $2.$
The division is complete. The polynomial above the line is the quotient and the polynomial at the very bottom is the remainder.

$Q(x)R(x) =2x_{2}+7x+18=40x−86 $

Finally, the voltage of the given circuit can be represented by the following expression.
The polynomial long division is a powerful method that helps dividing any two given polynomials. However, when the divisor is a binomial of the form $x−k,$ there is a shortcut that can be applied.

When dividing a polynomial by a linear binomial, a binomial of the form $x−k,$ there is an alternative method to the polynomial long division called the synthetic division. This method uses the constant term of the binomial and the coefficients of the numerator to compute the quotient. Consider the following division.
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The steps of synthetic division are shown below.

$x−2x_{3}−3x_{2}+7 $

This division can be found following the next six steps.
1

Set Up the Division Using $k$ and the Coefficients of the Numerator

Recall that the denominator is a linear binomial in the form $x−k.$ The denominator is $x−2,$ so the value of $k$ is $2.$ The division symbol for synthetic division is L-shaped. The number $k$ is written to the left.

$21 1 -307 $

After the numerator is written in standard form, its coefficients and constant term are written to the right of the division symbol. Fill in any missing terms with a zero.
$21 1 -307 $

The numerator does not have a linear term, so a $0$ was added between $-3$ and $7.$ Note that the number at the left of the division symbol is the opposite to the constant term of the divisor. 2

Bring Down the Leading Coefficient

Bring the leftmost coefficient down across the line. In this case, $1$ is written below the horizontal line.

$21 1↓ -3 0 7 1 $

3

Multiply $k$ by the New Number Below the Line

Multiply the number written under the line in the previous step by $k$ and write the product below the next coefficient above the line. In this case, $1$ is multiplied by $2.$ The product $2$ is written in the next column below $-3.$

$21 1 -32 0 7 1 $

4

Add the Numbers in the Column Above the Line

Now, add the numbers in the column above the horizontal line, then write the result below the horizontal line in the same column. In this case, $-3$ and $2$ are added and their sum, $-1,$ is written below them and under the horizontal line.

$21 1 -32 0 7 1 -1 $

5

Repeat Steps $3$ and $4$ Through the Last Column

Steps $3$ and $4$ are repeated through the last column. In this exercise, $-1$ is multiplied by $2$ and the product is written in the next column and above the line.

$21 1 -32 0-2 7 1 -1 $

The numbers in this column are added and the sum is written below the line.
$21 1 -32 0-2 7 1 -1 -2 $

Then $-2$ is multiplied by $2$ and the product is written in the next column and above the line.
$21 1 -32 0-2 7-4 1 -1 -2 $

Finally, the numbers in the last column are added together and the sum is written below the line.
$21 1 -32 0-2 7-4 1 -1 -2 -3 $

The division is now complete. 6

Write the Quotient and Remainder

The rightmost number below the line is the remainder of the division. The remaining numbers below the line represent the coefficients of the quotient.
*always* one degree lower than the numerator because the denominator is a linear binomial. For the same reason, the remainder is *always* a number.

$21 1 -32 0-2 7-4 1 -1 -2 3 $

In this case, the remainder is $3.$ The quotient is a quadratic polynomial with coefficients $1,$ $-1,$ and $-2.$
$Q(x)r =x_{2}−x−2=3 $

Notice that the quotient is
Once the quotient and the remainder are found, the original division can be written as the $quotient$ plus the division of the $remainder$ and the divisor.

$x−2x_{3}−3x_{2}+7 =x_{2}−x−2+x−23 $

Consider the following polynomial.
### Hint

### Solution

$P(x)=3x_{4}−4x_{2}−36 $

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b Find the quotient and remainder of the division $P(x)÷(x−4).$

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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">r<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["668"]}}

a Use synthetic division. Remember to fill in the missing terms with zeros.

b Use synthetic division. Remember to fill in the missing terms with zeros.

a Since the divisor is a binomial of the form $x−k,$ synthetic division can be used. First, note that $P(x)$ has neither a cubic term nor a linear term. Therefore, fill in these two terms with zeros.

$x+23x_{4}+0x_{3}−4x_{2}+0x−36 $

Next, set up the coefficients of $P(x)$ and the synthetic division symbol. Here, $k=-2.$
$-21 3 0-40-36 $

Bring down the first coefficient, multiply it by $-2,$ and write the result in the second column.
$-21 3 0-6 -40-36 3 $

Add the numbers in the second column and write the result below the line. Then multiply the new number below the line by $-2$ and write the result in the third column.
$-21 3 0-6 -412 0-36 3 -6 $

Next, add the numbers in the third column and write the result below the line. Multiply this number by $-2$ and write the result in the fourth column.
$-21 3 0-6 -412 0-16 -36 3 -6 -8 $

It is almost done! Add the numbers in the fourth column and write the result below the line. Then, multiply this number by $-2$ and write the result in the fifth column.
$-21 3 0-6 -412 0-16 -36-32 3 -6 -8 -16 $

Finally, add the numbers in the last column and write the result below the line.
$-21 3 0-6 -412 0-16 -36-32 3 -6 -8 -16 -4 $

The quotient of the division is formed by using the first four numbers below the line. The remainder is the very last number below the line. The quotient is one degree lower than the numerator.
$Q(x)r =3x_{3}−6x_{2}+8x−16=-4 $

b As in Part A, the denominator is a linear binomial, $x−4.$ For that reason, synthetic division can be used again. The coefficients at the right of the division symbol are the same, but this time the value of $k$ is $4.$

$41 3 0-40-36 $

The procedure to perform the division is similar to the one applied in Part A. First, bring the number $3$ down. Next, multiply $4$ by $3$ and write the result in the second column and below $0.$
$41 3 012 -40-36 3 $

Next, find $0+12,$ which gives $12,$ and write the result in the same column but below the line. After that, multiply $4$ by $12$ and write the result in the third column and below $-4.$
$41 3 012 -448 0-36 3 12 $

As in the previous step, add the numbers in the third column. The sum of $-4$ and $48$ is $44,$ so write