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5. Dividing Polynomials
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5. 

Dividing Polynomials

Polynomials, expressions involving terms raised to a power, are fundamental in mathematics. Dividing one polynomial by another is a common task that requires specific techniques, one of which is polynomial long division. This method mirrors the traditional long division we use with numbers but applies it to polynomial expressions. Understanding this technique is crucial for solving complex mathematical problems, especially in algebra. It aids in simplifying functions and evaluating certain limits, making tasks more manageable and solutions clearer.
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11 Theory slides
11 Exercises - Grade E - A
Each lesson is meant to take 1-2 classroom sessions
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Dividing Polynomials
Slide of 11
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.
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.

Catch-Up and Review

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

Explore

Analyzing the Product of a Polynomial and

Consider the polynomials and and the constant
Random Polynomial P(x) and random Binomial Q(x) of the form x-k. For example, P(x)=x^2+2x+8, Q(x)=x+8, k=-8
Let be the product of the two given polynomials — that is, Find and evaluate it at Next, consider a different pair of polynomials and repeat the process. What conclusion can be made about Repeat the process as much as needed.
Discussion

Dividing Polynomials

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.
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 is shown.
Dividing 117 by 4 step by step
When the denominator divides the numerator evenly, the remainder is 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.
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.
Method

Polynomial Long Division

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.
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
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To begin, write both polynomials in standard form. If the numerator has missing terms, include them with coefficient For example, the numerator will be written as Then, write the polynomials as in long division notation.
2
Divide the Leading Terms
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Divide the leading term of the numerator by the leading term of the denominator. In this case, divide by
Write the resulting monomial above the horizontal bar.
3
Multiply the Result From Step by the Denominator
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The result from step is Therefore, multiply by the denominator
Next, write the resulting expression below the numerator and try to align the common powers.
4
Subtract the Product From Step From the Numerator
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Subtract the polynomial obtained in step from the numerator.
The result of the subtraction — the remainder — is the polynomial This means that the division of the given polynomials can be written as follows.
However, since the degree of is the same as the degree of the denominator, the division is not done yet. Thus, repeat steps until the remainder's degree is less than the denominator's degree.
5
Repeat Steps Until the Remainder's Degree is Less Than the Denominator's Degree
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Steps are repeated until the polynomial obtained in step has a lower degree than the denominator. Dividing the leading terms and gives as result. Next, multiply the denominator by and write the result at the bottom.
Now, subtract the two polynomials at the bottom.
This time, the resulting polynomial has a degree of which is lower than the denominator's degree. Therefore, the division is complete. This implies that the quotient of the division is and the remainder is
Finally, the initial polynomial division can be written as the plus the division of the and the
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.
Notice that unless the remainder is equal to the division of two polynomials is 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?
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.
The result of dividing the leading terms is which is not a monomial.
Example

Voltage of a Circuit

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.
Power = Voltage · Current
From this relation, a formula for finding the voltage of a circuit can be derived.
Consider a circuit for which the power is modeled by and the current is modeled by Perform the required division to find an algebraic expression that represents the voltage of the circuit.

Hint

Divide the polynomials using long division.

Solution
According to the formula, the voltage of a circuit is found by dividing the power by the current.
In this case, both the power and the current of the circuit are represented by polynomials.
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 only the quadratic term is missing.
Now, write the dividend under the division symbol and the divisor to the left.
The first term of the quotient is found by dividing the leading term of by the leading term of
Next, write the monomial above the horizontal bar.
Multiply the divisor by and write the resulting expression below the dividend. In this case, the multiplication gives
Subtract the resulting polynomial from the dividend.
Since the remainder has degree and the divisor has degree the division is not done yet. Therefore, the process will be repeated until the remainder's degree is less than
Division of polynomials
The division is complete. The polynomial above the line is the quotient and the polynomial at the very bottom is the remainder.
Finally, the voltage of the given circuit can be represented by the following expression.
V(x) = 2x^2 + 7x + 18 + (40x - 86)/(x^2 - 4 x + 5)
Discussion

Dividing a Polynomial by

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 there is a shortcut that can be applied.

Method

Synthetic Division

When dividing a polynomial by a linear binomial, a binomial of the form 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.
This division can be found following the next six steps.
1
Set Up the Division Using and the Coefficients of the Numerator
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Recall that the denominator is a linear binomial in the form The denominator is so the value of is The division symbol for synthetic division is L-shaped. The number is written to the left.
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.
The numerator does not have a linear term, so a was added between and 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
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Bring the leftmost coefficient down across the line. In this case, is written below the horizontal line.
3
Multiply by the New Number Below the Line
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Multiply the number written under the line in the previous step by and write the product below the next coefficient above the line. In this case, is multiplied by The product is written in the next column below
4
Add the Numbers in the Column Above the Line
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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, and are added and their sum, is written below them and under the horizontal line.
5
Repeat Steps and Through the Last Column
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Steps and are repeated through the last column. In this exercise, is multiplied by and the product is written in the next column and above the line.
The numbers in this column are added and the sum is written below the line.
Then is multiplied by and the product is written in the next column and above the line.
Finally, the numbers in the last column are added together and the sum is written below the line.
The division is now complete.
6
Write the Quotient and Remainder
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The rightmost number below the line is the remainder of the division. The remaining numbers below the line represent the coefficients of the quotient.
In this case, the remainder is The quotient is a quadratic polynomial with coefficients and
Notice that the quotient is always one degree lower than the numerator because the denominator is a linear binomial. For the same reason, the remainder is always a number.
The steps of synthetic division are shown below.
Division of x^3-3x^2+7 by x-2 using synthetic division
Once the quotient and the remainder are found, the original division can be written as the plus the division of the and the divisor.
Example

Using Synthetic Division

Consider the following polynomial.
a Find the quotient and remainder of the division
b Find the quotient and remainder of the division

Hint
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.

Solution
a Since the divisor is a binomial of the form synthetic division can be used. First, note that has neither a cubic term nor a linear term. Therefore, fill in these two terms with zeros.
Next, set up the coefficients of and the synthetic division symbol. Here,
Bring down the first coefficient, multiply it by and write the result in the second column.
Add the numbers in the second column and write the result below the line. Then multiply the new number below the line by and write the result in the third column.
Next, add the numbers in the third column and write the result below the line. Multiply this number by and write the result in the fourth column.
It is almost done! Add the numbers in the fourth column and write the result below the line. Then, multiply this number by and write the result in the fifth column.
Finally, add the numbers in the last column and write the result below the line.
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.
b As in Part A, the denominator is a linear binomial, 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 is
The procedure to perform the division is similar to the one applied in Part A. First, bring the number down. Next, multiply by and write the result in the second column and below
Next, find which gives and write the result in the same column but below the line. After that, multiply by and write the result in the third column and below
As in the previous step, add the numbers in the third column. The sum of and is so write in the same column but below the line. Then multiply by and write the resulting number in the fourth column and below
One more time, add the numbers in the fourth column and write the result in the same column but below the line. In this case, the sum is Then, multiply by and write the result in the fifth column and below
Finally, find the sum of the numbers in the last column and write the result below the line.
The first four numbers below the line are the coefficients of the quotient and the fifth number is the remainder.
Discussion

The Remainder of a Polynomial Division

From synthetic division, the division can be written as the quotient plus and the division of the remainder and Here, is a real number.
However, this relation can be further manipulated. For example, the fractions can be eliminated entirely by multiplying both sides of the equation by
Two important results follow from the last equation.
Rule

The Remainder Theorem

If a polynomial is divided by a binomial of the form then the remainder of the division is equal to

Proof
Consider the division of and a binomial of the form
The division can be written in terms of the quotient and the remainder by synthetic division or polynomial long division.
Since the divisor has degree and the degree of the remainder has to be lower, the remainder has degree This implies that is a constant. Thus, write instead of
Now, multiply both sides of the last equation by
MultEqn

Distr

Distribute

Finally, evaluate the last equation at
Substitute

SubTerm

Subtract term

ZeroPropMult

Zero Property of Multiplication

RearrangeEqn

Rearrange equation

Note that the Remainder Theorem gives the remainder of the division without the need to performing the division. For example, let To find the remainder of simply evaluate the numerator at
Conversely, this theorem also says that the process of synthetic division can be used to evaluate a polynomial function at a given value. For example, to find synthetic division can be used.
From the above,
Discussion

Remainder

As when dividing integers, there is special significance when the remainder of a polynomial division is zero. First, it implies that the denominator divides the numerator perfectly, and, consequently, that the result of the division is a polynomial.
Additionally, if the divisor is a linear binomial the fact that the remainder is gives a connection between the denominator and the roots of the numerator. The following theorem expands on this concept.
Rule

The Factor Theorem

Let be a polynomial and a real number. The binomial is a factor of if and only if

The binomial is a factor of if and only if

Notice that means that is a zero of Therefore, the theorem can also be stated as follows.

The binomial is a factor of if and only if is a zero of

The Factor Theorem is a special case of the Remainder Theorem and establishes a connection between the zeros of a polynomial and its factors.

Proof

Since the theorem is a biconditional statement, the proof will consists of two parts.

  • Part I: If is a factor of then
  • Part II: If then is a factor of

Part I

If is a factor of then can be written as the product of and a certain polynomial
Now, evaluate the equation at
Substitute

SubTerm

Subtract term

ZeroPropMult

Zero Property of Multiplication

Consequently, is a zero of This completes the proof of the first part.

Part II

Consider the division of and
The division can be rewritten in terms of the quotient and the remainder by using polynomial long division.
By the Remainder Theorem, the remainder of the previous division is equal to Since the remainder of the division is Therefore, the rightmost term of the previous equation is
Finally, multiply both sides of the equation by
Consequently, is a factor of This completes the proof of the second part.
Example

Applying The Remainder Theorem

a What is the remainder when is divided by
b Consider the polynomial Without evaluating the polynomial, find

Hint
a Use the Remainder Theorem.
b Use the process of synthetic division to find

Solution
a According to the Remainder Theorem, the remainder when a polynomial is divided by a binomial of the form is equal to the polynomial evaluated at
In the given case, the binomial is which means that Therefore, instead of performing the division, the remainder can be found simply by evaluating at
▼
Substitute for and evaluate
Substitute

CalcPowProd

Calculate power and product

Multiply

Multiply

AddSubTerms

Add and subtract terms

Consequently, the remainder is — that is, To verify that the result is correct, perform the polynomial long division.
Division of polynomials
As shown, the remainder of the division is as previously concluded. The remainder could also be found using synthetic division.
b Synthetic division can be used to find without evaluating the polynomial. Rather, the process of synthetic substitution will be used. First, set up the division symbol and the coefficients of Remember to fill in the missing terms with zeros. Then, write to the left.
Next, follow the steps of synthetic division. Bring down the number Multiply by and write the result below
Add the numbers in the second column and write the result below the line.
The same steps are repeated over and over until the last column is reached.
At the end of the process, it was concluded that Therefore, is a zero of Consequently, by the Factor Theorem, is a factor of In fact, can be written as follows.
Pop Quiz

Finding Remainders

Find the remainder of the given polynomial division.

Division of a random polynomial by a random binomial of the form (x-k)
Closure

Comparing Polynomial Long Division and Synthetic Division

When it comes to dividing polynomials, there are two main methods that can be applied — namely, polynomial long division and synthetic division. The table below lists some pros and cons of each method.

Pros Cons
Polynomial Long Division Works for every polynomial division Involves variables, powers, and many computations
Synthetic Division Involves only numbers Works only when the divisor has the form
Keep in mind that synthetic division is a shortcut of polynomial long division. In fact, synthetic division can be derived from the process of polynomial long division. Consider, for example, the following division.
This calculation can be solved by using long division.
Polynomial long division

To clarify or simplify the calculations, the minus signs can be removed by distributing them to the parentheses. Also, the remaining terms in the dividend can be hidden in the third, fifth, and seventh rows until the terms are needed. This cleaner look can be seen below.

Polynomial long division
Now, take a look at the process for computing the same division using synthetic division.
Notice that the numbers below the horizontal line are the leading coefficients of the polynomials in the first, third, fifth, seventh, and ninth rows of the polynomial long division. Furthermore, the additions performed in synthetic division can also be located in the long division process.
Polynomial long division. Highlighting leading coefficients and additions
In fact, the long division process can be transformed into synthetic division by removing the polynomial above the horizontal line and all the variables.
Transforming a polynomial long division into synthetic division
In summary, when the divisor has the form both polynomial long division and synthetic division lead to the same result. However, synthetic division requires fewer steps and is faster.



Level 1
Level 2
Level 3
Dividing Polynomials
Consider the following synthetic division.
Write the polynomial divisor.
Write the polynomial dividend.
Write the quotient.
Write the remainder.

Let's start by remembering that a synthetic division can be performed only when the denominator, or divisor, is a linear binomial of the form x-k. Therefore, the divisor we are looking for has this form. D(x) = x- k Now, at the moment of setting up the division, we draw the L-shaped division symbol and place the value of k at the left. rl IR-0.15cm r k & |rr & & & & Comparing the previous setup to the given one, we can identify the value of k. rl IR-0.15cm r 4 & |rr 2 & -4 & -13 & 0 & 9 & 8 & 16 & 12 & 48 & c 2 & 4 & 3 & 12 & 57 From the above, the value of k is 4. This means that D(x)=x-4.


When we perform a synthetic division, the number of columns inside the division symbol equals the degree of the dividend plus 1. In the given synthetic division there are 5 columns inside the division symbol. Therefore, the dividend is a polynomial of degree 4. P(x) = ax^4 + bx^3 + cx^2 + dx + e Additionally, we place the coefficients of the dividend inside the division symbol on the top row. rl IR-0.15cm r & |rr a& b& c& d& e Comparing the last setup to the given synthetic division, we can determine the value of each coefficient. rl IR-0.15cm r 4 & |rr 2 & -4 & -13 & 0 & 9 & 8 & 16 & 12 & 48 & c 2& 4& 3& 12 & 57 ⇓ a = 2 b = -4 c = -13 d = 0 e = 9 Now, let's substitute these values into P(x) and simplify the right-hand side.


When performing a synthetic division, the degree of the quotient equals the number of columns below the division symbol minus 2. In the given synthetic division there are 5 columns below the division symbol. Therefore, the dividend is a polynomial of degree 3. Q(x) = ax^3 + bx^2 + cx + d In addition, the coefficients of the quotient are the numbers below the division symbol except for the rightmost number. rl IR-0.15cm r & |rr & & & & & & c a& b& c& d& Comparing this setup to the given synthetic division, we can determine the value of each coefficient. rl IR-0.15cm r 4 & |rr 2 & -4 & -13 & 0 & 9 & 8 & 16 & 12 & 48 & c 2 & 4 & 3 & 12 & 57 ⇓ a = 2 b = 4 c = 3 d = 12 Now, let's substitute these values into Q(x) and simplify the right-hand side.


When performing a synthetic division, the divisor is a linear polynomial. This implies that the degree of the remainder is 0. In other words, the remainder is a number. r = a Additionally, the remainder is the rightmost number below the division symbol. rl IR-0.15cm r & |rr & & & & & & c & & & a Comparing this setup to the given synthetic division, we can determine the remainder of the division. rl IR-0.15cm r 4 & |rr 2 & -4 & -13 & 0 & 9 & 8 & 16 & 12 & 48 & c 2 & 4 & 3 & 12 & 57 From the above, the value of a is 57. This means that r=57. Using the information found in this entire exercise, we can write an equation involving the dividend, divisor, quotient, and remainder. 2x^4 - 4x^3 - 13x^2 + 9/x-4 = 2x^3 + 4x^2 + 3x + 12 + 57/x-4

Electrical power 🔌 is defined as the product of the voltage ⚡ of a circuit and the current flowing through it. From this relation, the following formula can be derived.
Consider a circuit for which the power is modeled by and the voltage is modeled by Perform the corresponding division to find an algebraic expression that represents the current of the circuit.

According to the formula, we can find the current of a circuit by dividing the power by the voltage. Current = Power/Voltage In our case, both the power and the voltage of the circuit are represented by polynomials. P(x) &= 2x^4 - 20x^3 + 14x^2 - 72 V(x) &= 2x^2-4x+6 Let's divide these two polynomials to find the current of the circuit. We can do this by following the steps of polynomial long division. First, we write both polynomials in standard form, filling in any missing terms in the numerator with zeros. In P(x), only the linear term is missing. P(x) &= 2x^4 - 20x^3 + 14x^2 + 0x - 72 V(x) &= 2x^2-4x+6 Now, we write the dividend P(x) under the division symbol and the divisor V(x) to the left. 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 To find the first term of the quotient, we divide the leading term of P(x) by the leading term of V(x). 2x^4/2x^2 = x^2 Next, we write the resulting monomial above the horizontal bar. rr & x^2 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 After this, we multiply the divisor by x^2 and write the resulting expression below the dividend. In this case, the multiplication gives 2x^4-4x^3+6x^2. rr & x^2 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 & 2x^4 - 4x^3 + 6x^2 Let's subtract the resulting polynomial from the dividend. rr & x^2 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 & -( 2x^4-4x^3 + 6x^2) & -16x^3 + 8x^2 +0x-72 Since the result has degree 3 and the divisor has degree 2, the division is not done yet. Therefore, we will repeat the process until the result's degree is less than 2. rr & x^2 - 8x-12 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 & -( 2x^4-4x^3 + 6x^2) & -16x^3 + 8x^2 +0x-72 & -( -16x^3 + 32x^2 - 48x) & -24x^2 +48x-72 & -( -24x^2-48x-72) & We completed the division. The polynomial above the line is the quotient and the polynomial at the very bottom is the remainder. Q(x) &= x^2 - 8x-12 R(x) &= 0 Since R(x)=0, the polynomial V(x) divides P(x) exactly. Therefore, we can write the following equation. 2x^4 - 20x^3 + 14x^2 - 72/2x^2-4x+6 = x^2 - 8x-12 From this, we conclude that the current of the given circuit is represented by the polynomial I(x)= x^2 - 8x-12.

Consider the following polynomial.
Find the quotient of the division
Find the remainder of the division

Since the divisor is a binomial of the form x-k, we can perform the required division using synthetic division. The first thing we note is that P(x) has no quadratic term. Therefore, we fill in this term with a zero. 2x^4 - 4x^3 + 0x^2+ 12x - 24/x+3 Now, let's set up the coefficients of P(x) and the synthetic division symbol. Here, k= -3. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 Next, we bring down the first coefficient, multiply it by -3, and write the result inside the division symbol in the second column. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 &-6 & c 2 & The next step is to add the numbers in the second column and write the result below the line. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 & -6 & c 2 & -10 We repeat the steps through the last column. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 & -6 & 30 & -90 & 234 & c 2 & -10 & 30 & -78 & 210 We are done! Remember, in synthetic division, the quotient's degree equals the degree of P(x) minus one. Therefore, the degree of Q(x) is 3. Additionally, the quotient is formed by using the first four numbers below the line. Q(x) = 2x^3 - 10x^2 + 30x - 78


As in Part A, we will perform the required division using synthetic division. The initial setup for the synthetic division is the same as in Part A except for the number to the left of the division symbol. Here, the divisor is x-2, therefore, k= 2. rl IR-0.15cm r 2 & |rr 2 & -4 & 0 & 12 & -24 Now, we bring down the first coefficient, multiply it by 2, and write the result in the second column above the horizontal line. After that, we add the numbers in the second column and write the result below the line. rl IR-0.15cm r 2 & |rr 2 & -4 & 0 & 12 & -24 & 4 & c 2 & 0 Let's repeat the previous steps until we reach the last column. rl IR-0.15cm r 2 & |rr 2 & -4 & 0 & 12 & -24 & 4 & 0 & 0 & 24 & c 2 & 0 & 0 & 12 & 0 We are done! Since the divisor has degree 1, the degree of the remainder is 0. Therefore, the remainder is a number. In fact, the remainder of the division is the rightmost number below the line. r = 0

Consider the following polynomial.
Find the sum of the remainders obtained when is divided by and

We start by noticing that the divisors are linear binomials of the form x-k. Therefore, each division can be performed using synthetic division. However, since we are not asked for the quotients but only for the remainders, instead of applying the synthetic division three times, we can use the Remainder Theorem.

The Remainder Theorem |- If a polynomial P(x) is divided by a binomial of the form x-k, then the remainder of the division is equal to P(k).

This theorem tells us that we can find the remainder of each division simply by evaluating P(x) at the corresponding value. For instance, the remainder of P(x)÷ (x-1) is equal to P(1). Thus, let's find it.

Similarly, we can find the remainders of the other two divisions. The results are summarized in the table.

Divisor k P(k) Remainder
x+1 -1 P( -1) = 2( -1)^8 + 3( -1)^6 - ( -1)^5 - 2( -1)^4 + 3( -1)^2 - 2( -1) + 5 14
x-2 2 P( 2) = 2( 2)^8 + 3( 2)^6 - 2^5 - 2( 2)^4 + 3( 2)^2 - 2( 2) + 5 653

Now that we know the three remainders, let's add them up. Sum &= 8 + 14 + 653 &⇓ Sum &= 675 ✓

Heichi plans to solve some exercises on dividing polynomials later today to prepare for a test. To help himself, he wrote the following note on a piece of paper.

Note: P(x)/D(x) = Q(x)D(x) + R(x), where Q(x) is the quotient and R(x) is the remainder
However, Heichi's note is not correct. Write the correct right-hand side of the equation.

Let's begin by verifying that Heichi's note is incorrect. We can do it by using integer numbers — after all, integers are 0-degree polynomials. For example, let's consider the division 11 ÷ 4. & 2 4) & l 11 & -8 & 3 The quotient of the division is 2 and the remainder is 3. Now, let's substitute these values into the equation that Heichi wrote.

As seen, the left-hand side is not equal to the right-hand side. The same will happen for any polynomial division. To write the correct statement, notice that in the previous computations, the right-hand side resulted to be equal to the numerator. That is, the right-hand side in Heichi's note is equal to P(x). P(x) = Q(x)D(x) + R(x) The previous equation is true for integers and also for polynomials. Now, let's divide both sides of this equation by D(x) and simplify.

From the above, the correct right-hand side is equal to the quotient of the division plus the division of the remainder and the divisor. P(x)/D(x) = Q(x) + R(x)/D(x) We can test this equation with the numbers we used before.

As seen, we got a true statement and this is true for any polynomial division.

When practicing polynomial division, Magdalena divided a polynomial by and got as the quotient and as the remainder. Find the polynomial

When we divide a polynomial P(x) by another polynomial D(x), we get a quotient Q(x) and a remainder R(x). P(x)/D(x) → & Q(x) D(x) & | l P(x) & ... & R(x) Moreover, once we know the quotient and remainder, we can write the original division as the quotient plus the division of the remainder and the divisor. P(x)/D(x) = Q(x) + R(x)/D(x) In the given situation, the divisor, quotient, and remainder are known. Thus, to find the dividend, we can solve the previous equation for P(x) and then substitute the corresponding polynomials.

We are told that the divisor is x-2, therefore D(x)=x-2. Similarly, Q(x)=3x^3-2x+7 and R(x)=-3. Let's substitute these expressions into the last equation to finally get P(x).

Dividing Polynomials

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