Topics
2. Polynomial Expressions and Equations
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United States of America
Grades 9-12

2.04 Polynomial division

Lesson
Practice

Introduction

In the previous lesson, we reviewed how to add, subtract, and multiply polynomials. Now, we will explore various methods of dividing polynomials. In some cases, knowing the remainder of a division problem is useful, and we will learn how to determine the remainder without actually performing the division.

Polynomial division

Recall from our work with rational numbers, that when we divide a sum by a real number, we can use the fact that:\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}for any real numbers, a, b or c.

We can extend this concept to the division of polynomial expressions. For any polynomial expressions A, B or C:\dfrac{A+B}{C} = \dfrac{A}{C} + \dfrac{B}{C}

The simplest form of division of polynomials is when the divisor is a monomial. The process involves dividing each term of the polynomial by the monomial then simplifying each individual fraction using the properties of exponents.

Exploration

Use the process of long division to divide the following:

578 \div 5

2\,199 \div 13

\left(6x^3-10x^2+8x-7\right)\div 2x

\left(3x^5-x^3-4x^2+9\right)\div x^2

  1. What patterns do you notice?
  2. How are the properties of exponents used?
  3. When will a problem have a remainder?

When dividing by a monomial that contains a variable, we use the quotient of powers property of exponents, \dfrac{x^m}{x^n}=x^{m-n}, to simplify each term.

In general, when dividing polynomials where the divisor is not a monomial, we can use the process of polynomial long division.

When we divide a dividend, p\left(x\right), by a divisor, b\left(x\right), we can write the expression as the sum of the quotient, q\left(x\right), and the remainder, r\left(x\right), divided by the divisor. The notation for this is {\dfrac{p\left(x\right)}{b\left(x\right)}=q\left(x\right)+\dfrac{r\left(x\right)}{b\left(x\right)}}

If our division was done correctly, then p\left(x\right) = q\left(x\right)b\left(x\right) + r\left(x\right). Polynomial long division works in a very similar way to long division with whole numbers where we:

  1. Divide the first terms of the dividend and divisor
  2. Multiply the answer by the divisor
  3. Subtract the product from the first part of the dividend
  4. Bring down the next part of the dividend
  5. Repeat steps 1-4 until there are no terms left to bring down

Note: Before performing long division, the terms of the divisor and dividend should first be arranged in descending order of exponents. In cases where there is no term corresponding to an exponent in the dividend or divisor, we use a placeholder term with a coefficient of 0.

For example, the long division of (2x^3 - 5x + 7)\div(x - 1) is shown below.

A figure showing the polynomial long division for 2 x cubed plus 0 x squared minus 5 x plus 7 divided by x minus 1. The expression 2 x cubed plus 0 x squared minus 5 x plus 7 is labeled Dividend, x minus 1 labeled Divisor, 2 x squared plus 2 x minus 3 labeled Quotient, and 4 labeled Remainder. Speak to your teacher for more information.

Step 1: 2x^3 \div x=2x^2

Step 2: 2x^2\left(x-1\right)=2x^3-2x^2

Step 3: \left(2x^3+0x^2\right)-\left(2x^3-2x^2\right)=2x^2

Step 4: Bring down -5x

Steps 1-4 are repeated two more times to find the quotient and the remainder.

For the example above, the dividend is p\left(x\right)=2x^3-5x+7, the divisor is b\left(x\right) =x-1, and we found the quotient to be q(x) =2x^2+2x-3 and remainder r(x) = 4. So, our final answer is \dfrac{2x^3-5x+7}{x-1}=2x^2+2x-3+\dfrac{4}{x-1}

Examples

Example 1

The rectangle has an area of 4 x^{4} - 12 x square units, and its width is 4x units. Find the length of the rectangle.

Worked Solution
Create a strategy

We can begin by identifying the known values and rearranging the formula for the area of the rectangle to highlight the unit we need to solve for.

\displaystyle \text{Area}\displaystyle =\displaystyle \text{Length} \cdot \text{Width}Formula for the Area of a Rectangle
\displaystyle \text{Length}\displaystyle =\displaystyle \dfrac{\text{Area}}{\text{Width}}Solve the formula for length

\text{Area}=4x^4-12x and \text{Width}=4x

Apply the idea
\displaystyle \text{Length}\displaystyle =\displaystyle \dfrac{4x^4 - 12x}{4x}Substitute expressions for area and width
\displaystyle =\displaystyle \dfrac{4x^4}{4x}-\dfrac{12x}{4x}Divide each term by 4x
\displaystyle =\displaystyle x^{3} - 3Simplify the expression

Since there are no negative exponents and the expression is already in standard form, the final answer is x^{3} - 3. The length of the rectangle is x^{3} - 3 units.

Reflect and check

Notice the quotient of powers property was used in step 2 when dividing each term by 4x.\dfrac{4x^4}{4x}=x^{4-1}=x^3

\dfrac{-12x}{4x}=-3x^{1-1}=-3

To check the answer we found for the length, we can write the dividend as the product of the quotient and divisor.

\displaystyle \text{Length} \times \text{Width}\displaystyle =\displaystyle \text{Area}Formula for the Area of a Rectangle
\displaystyle \left(x^{3} - 3\right)\left(4x\right)\displaystyle =\displaystyle 4x^4 - 12xDistributive property

The product of the quotient and the divisor does result in the dividend, so our answer is correct.

Example 2

Use algebraic manipulation to divide \dfrac{x^2+8x-1}{x^2-1}

Worked Solution
Create a strategy

There are different ways to manipulate an expression algebraically. One way is to factor the expression in the numerator and denominator, but the expression in the numerator of this expression is prime, or not factorable.

Another way to manipulate an expression algebraically is by using the commutative property to rearrange terms.

Apply the idea
\displaystyle \dfrac{x^2+8x-1}{x^2-1}\displaystyle =\displaystyle \dfrac{x^2-1+8x}{x^2-1}Rearrange the terms in the numerator
\displaystyle =\displaystyle \dfrac{x^2-1}{x^2-1}+\dfrac{8x}{x^2-1}Separate the fractions
\displaystyle =\displaystyle 1+\dfrac{8x}{x^2-1}Simplify the first fraction
Reflect and check

If you had not noticed that x^2 and -1 were in both the numerator and denominator, you still could have gotten the answer by using the fact that \dfrac{a+b+c}{d}=\dfrac{a}{d}+\dfrac{b}{d}+\dfrac{c}{d}, then rearranging the terms that way.

\displaystyle \dfrac{x^2+8x-1}{x^2-1}\displaystyle =\displaystyle \dfrac{x^2}{x^2-1}+\dfrac{8x}{x^2-1}+\dfrac{-1}{x^2-1}
\displaystyle =\displaystyle \dfrac{x^2}{x^2-1}+\dfrac{-1}{x^2-1}+\dfrac{8x}{x^2-1}
\displaystyle =\displaystyle \dfrac{x^2-1}{x^2-1}+\dfrac{8x}{x^2-1}
\displaystyle =\displaystyle 1+\dfrac{8x}{x^2-1}

Example 3

Write \dfrac{x^3 + 7 x^2 + 14 x + 3}{x + 2} in terms of the quotient and remainder by using long division.

Worked Solution
Create a strategy

First, we need to check that the terms are arranged in descending order of exponents. In this problem, the terms in the dividend and divisor are in the correct order.

Next, we need to check that there are no missing terms. To do that, we look at the exponents. No terms are missing because the exponents in the dividend are 3,2,1,0 and the exponents in the divisor are 1, 0.

Apply the idea

We set up the long division with the dividend under the table and the divisor outside the table.

A figure showing a polynomial long division set up: the dividend x cubed plus 7 x squared plus 14 x plus 3 under the table and the divisor x plus 2 outside the table.

To begin the division, we only need to consider the first terms of the dividend and divisor.

A figure showing partial steps of the polynomial long division for x cubed plus 7 x squared plus 14 x plus 3 divided by x plus 2. Speak to your teacher for more information.

Dividing these, we get x^3\div x=x^2, and we write this above the table.

Then, we multiply the result by the divisor, x^2\left(x+2\right)=x^3+2x^2, and write this below the dividend.

Next, we subtract and bring down the next term. When subtracting, remember to distribute the negative sign across all terms.

Now, we repeat the steps from above.

A figure showing partial steps of the polynomial long division for x cubed plus 7 x squared plus 14 x plus 3 divided by x plus 2. Speak to your teacher for more information.

Dividing the first terms, we get 5x^2 \div x =5x.

Multiplying this result by the divisor, we have 5x\left(x+2\right)=5x^2+2x.

Next, we subtract and bring down the next term.

Since we brought down the last term in the dividend, this will be the last time we repeat the steps.

A figure showing the polynomial long division for x cubed plus 7 x squared plus 14 x plus 3 divided by x plus 2. Speak to your teacher for more information.

Dividing the first terms gives us 4x \div x=4.

Multiplying this by the divisor gives us 4\left(x+2\right)=4x+8

Subtracting that from 4x+3 leaves a remainder of -5.

The long division gives us a result of x^2 + 5x + 4 as the quotient and -5 as the remainder. So we can rewrite the expression as \frac{x^3 + 7 x^2 + 14 x + 3}{x + 2} = x^2 + 5x + 4 + \frac{-5}{x + 2}

Reflect and check

The parentheses included in each of the subtraction steps are very important because it reminds us to distribute the negative sign to all terms. Not distributing the negative and adding the terms instead is a common mistake, so remember to always include the parentheses.

Idea summary

To divide 2 polynomials, we can use algebraic manipulation or long division. For algebraic manipulation, we can use the fact that \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c} or rearrange the terms. We can also try factoring the dividend and divisor and canceling like terms.

For polynomial long division, we first check for any missing terms in the dividend and divisor. If there is a missing term, we add a 0 coefficient in its place. Then, we follow these steps for the division:

  1. Divide the first terms of the dividend and divisor
  2. Multiply the answer by the divisor
  3. Subtract the product from the first part of the dividend
  4. Bring down the next part of the dividend
  5. Repeat steps 1-4 until there are no terms left to bring down

Synthetic division

There is an efficient method of polynomial division known as synthetic division that can be used only when the divisor is a linear expression in the form x\pm a, where a is a constant.

Consider the long division and synthetic division of \dfrac{3x^3+17x^2+6x-26}{x+5}:

A figure showing the polynomial long division for 3 x cubed plus 17 x squared plus 6 x minus 26 divided by x plus 5. Speak to your teacher for more information.
Long division
A figure showing a synthetic division with negative 5 in the middle top corner. Top row has digits 3, 17, 6, negative 26. Middle row has negative 15 under the 17, negative 10 under the 6, and 20 under negative 26. Bottom row has 3, 2, negative 4, and negative 6.
Synthetic division

Synthetic division is a short-hand notation version of long division where only the coefficients are used. To begin the synthetic division, we write the coefficients of the terms in the divinded inside an upside down division table. To find the number that goes outside the table, we set the linear divisor equal to zero and solve for x.\begin{aligned}x+5&=0\\x&=-5\end{aligned}

Once the synthetic division is set up, we always bring the first number down. Then, we follow these steps:

  1. Multiply the first number below the table by the number outside the table
  2. Write the result under the second coefficient from the dividend
  3. Add the numbers in the second column
  4. Repeat until the last column has been added
A figure showing the partial steps of synthetic division for 3 x cubed plus 17 x squared plus 6 x minus 26 divided by x plus 5. Speak to your teacher for more information.

Step 1: Multiply 3\times -5

Step 2: Write -15 under 17

Step 3: Add 17+-15

A figure showing the partial steps of synthetic division for 3 x cubed plus 17 x squared plus 6 x minus 26 divided by x plus 5. Speak to your teacher for more information.

Step 1: Multiply 2\times -5

Step 2: Write -10 under 6

Step 3: Add 6+-10

A figure showing the steps of synthetic division for 3 x cubed plus 17 x squared plus 6 x minus 26 divided by x plus 5. Speak to your teacher for more information.

Step 1: Multiply -4\times -5

Step 2: Write 20 under -26

Step 3: Add -26+20

The answers beneath the table will be the coefficients of the quotient, and the last number is always the remainder. Because we are dividing by a linear term, the degree of the quotient will be one less than the degree of the divisor.

Therefore, the quotient for the example is q\left(x\right)=3x^2+2x+4 and the remainder is r\left(x\right)=-6, just like we found when we solved using long division.

Examples

Example 4

Rewrite \dfrac{2x^3 - 3x^2 + 4x - 1}{x + 1}as the sum of the quotient and a remainder fraction by using synthetic division.

Worked Solution
Create a strategy

For synthetic division, we still need to check that the terms are arranged in descending order of exponents and there are no missing terms. If there is a missing term, we write a 0 coefficient in its place.

This problem does not have missing terms because the exponents in the dividend are 3,2,1,0 and the exponents in the divisor are 1,0.

Apply the idea

We set up the synthetic division by writing the coefficients of the dividend inside an upside down division table. Then, we find the number that goes outside the table by setting the divisor equal to zero and solving for x:

\begin{aligned}x+1&=0\\x&=-1\end{aligned}

Now, we bring down the first number and follow these steps until the last column has been added:

  1. Multiply the number below the table by the number outside the table
  2. Write the result under the next coefficient from the dividend
  3. Add the numbers in the next column
A figure showing a synthetic division with negative 1 in the middle left corner. Top row has digits 2, negative 3, 4, negative 1. Middle row has negative 2 under the negative 3, 5 under the 4, and negative 9 under negative 1. Bottom row has 2, negative 5, 9, and negative 10.

These numbers below the table are the coefficients of the quotient. Remember the quotient will be one degree less than the degree of the dividend. The degree of the dividend is 3, so the exponents of the variable will begin from 2 and decrease by one until we get to the remainder.

Writing in the variables, we get a result of 2x^2 - 5x + 9 as the quotient and -10 as the remainder. So we can rewrite the expression as \frac{2x^3 - 3x^2 + 4x - 1}{x + 1} = 2x^2 - 5x + 9 - \frac{10}{x + 1}

Reflect and check

We can check the answer by determining if the dividend is equal to the product of the quotient and the divisor plus the remainder.

\displaystyle \left( \text{Quotient} \times \text{Divisor} \right) + \text{Remainder}\displaystyle =\displaystyle \text{Dividend}Formula
\displaystyle \left[\left(2x^{2} - 5x + 9\right) \left(x + 1\right) \right]- 10\displaystyle =\displaystyle \left(2x^3+2x^2-5x^2-5x+9x+9\right)-10Expand
\displaystyle =\displaystyle 2x^3-3x^2+4x-1Combine like terms

This polynomial is the same as the dividend, so we know we did the synthetic division correctly.

Example 5

Determine an appropriate method for dividing the following expressions. Explain your choice.

a

\dfrac{3x^4-6x^3+19x^2-8x+20}{3x^2+4}

Worked Solution
Create a strategy

The denominator of this expression is non-linear, and it cannot be factored. Therefore, long division is the most efficient method that can be used to evaluate this division.

Apply the idea

The first thing we need to do is make sure the terms are in descending order and no terms are missing. The dividend is not missing any terms, but the divisor is missing an x-term. We need to add a 0 term, so the divisor becomes 3x^2+0x+4.

Now, we can divide the polynomials as follows:

A figure showing the polynomial long division for 3 x raised to the fourth power minus 6 x cubed plus 19 x squared minus 8 x plus 20 divided by 3 x squared plus 0 x plus 4. Speak to your teacher for more information.

Therefore, \dfrac{3x^4-6x^3+19x^2-8x+20}{3x^2+4}=x^2-2x+5.

Reflect and check

If we multiply both sides of the answer by 3x^2+4, we can show that the dividend is equal to the product of the divisor and the quotient. 3x^4-6x^3+19x^2-8x+20=\left(3x^2+4\right)\left(x^2-2x+5\right)

This division could have been done with algebraic manipulation, but it would have been difficult to see.

\displaystyle \dfrac{3x^4-6x^3+19x^2-8x+20}{3x^2+4}\displaystyle =\displaystyle \dfrac{3x^4+19x^2+20-6x^3-8x}{3x^2+4}Rearrange the terms
\displaystyle =\displaystyle \dfrac{3x^4+19x^2+20-2x\left(3x^2+4\right)}{3x^2+4}Factor -2x from last two terms
\displaystyle =\displaystyle \dfrac{3x^4+4x^2+15x^2+20-2x\left(3x^2+4\right)}{3x^2+4}Rewrite 19x^2 as {4x^2+15x^2}
\displaystyle =\displaystyle \dfrac{x^2\left(3x^2+4\right)+5\left(3x^2+4\right)-2x\left(3x^2+4\right)}{3x^2+4}Factor by grouping
\displaystyle =\displaystyle \dfrac{x^2\left(3x^2+4\right)}{3x^2+4}+\dfrac{5\left(3x^2+4\right)}{3x^2+4}+\dfrac{-2x\left(3x^2+4\right)}{3x^2+4}Separate the fractions
\displaystyle =\displaystyle x^2+5-2xReduce by 3x^2+4
\displaystyle =\displaystyle x^2-2x+5Rearrange the terms
b

\left(-x^4+16x^2+2x\right)\div \left(x-4\right)

Worked Solution
Create a strategy

All terms have a GCF of -x, so we can try factoring this: \dfrac{-x\left(x^3-16x-2\right)}{x-4}

However, the resulting trinomial cannot be factored further, making the algebraic manipulation more difficult and therefore, less efficient. Because the divisor is linear, synthetic division is the most efficient method for dividing these polynomials.

Apply the idea

To set up the synthetic division, we need to determine which terms are missing in the dividend and add a 0 coefficient in their place. Notice we are missing the x^3-term and the constant, so the coefficients in the table will be -1,0,16,2,0.

Next, we set the divisor equal to zero and solve for x:\begin{aligned}x-4&=0\\x&=4\end{aligned}

Now, we can perform the synthetic division:

A figure showing a synthetic division with 4 in the top left corner. Top row has digits negative 1, 0, 16, 2, 0. Middle row has negative 4 under the leftmost 0, negative 16 under the 16, 0 under the 2, and 8 under the rightmost 0. Bottom row has negative 1, negative 4, 0, 2, and 8.

The degree of the quotient will be one less than the degree of the dividend. Since the degree of the dividend is 4, the exponents of the terms in the quotient will begin from 3 and decrease by one.

Therefore, the quotient is -x^3-4x^2+2 and the remainder is -8. We can also write this as \dfrac{-x^4+16x^2+2x}{x-4}=-x^3-4x^2+2+\dfrac{8}{x-4}

Reflect and check

As we discussed when creating a strategy for this problem, algebraic manipulation for this one is possible. The easiest way to begin is to factor -x^2 from the first two terms:

\displaystyle \dfrac{-x^4+16x^2+2x}{x-4}\displaystyle =\displaystyle \dfrac{-x^2\left(x^2-16\right)+2x}{x-4}Factor -x^2 from first two terms
\displaystyle =\displaystyle \dfrac{-x^2\left(x-4\right)\left(x+4\right)+2x}{x-4}Factor difference of squares
\displaystyle =\displaystyle \dfrac{-x^2\left(x-4\right)\left(x+4\right)}{x-4}+\dfrac{2x}{x-4}Separate the fractions
\displaystyle =\displaystyle -x^2\left(x+4\right)+\dfrac{2x}{x-4}Reduce by x-4

Looking at the remainder, we see the degrees of the numerator and denominator are the same. This tells us that the remainder can actually be divided further. Let's use synthetic division to further simplify the remainder.

A figure showing a synthetic division with 4 in the top left corner. Top row has digits 2 and 0. Middle row has 8 under the 0. Bottom row has 2 and 8.

In other words, \dfrac{2x}{x-4}=2+\dfrac{8}{x-4}. Continuing where we left off:

\displaystyle -x^2\left(x+4\right)+\dfrac{2x}{x-4}\displaystyle =\displaystyle -x^2\left(x+4\right)+2+\dfrac{8}{x-4}Rewrite \dfrac{2x}{x-4}
\displaystyle =\displaystyle -x^3-4x^2+2+\dfrac{8}{x-4}Distributive property
c

\dfrac{4x^2-24x+36}{2x-6}

Worked Solution
Create a strategy

The first method we should check for is factoring, since that is usually the quickest method. Always look for a GCF before attempting to factor using any other method.

The numerator has a GCF of 4 and a GCF of 2 is in the denominator. We will factor these out, then try to factor the numerator further.

Apply the idea
\displaystyle \dfrac{4x^2-24x+36}{2x-6}\displaystyle =\displaystyle \dfrac{4\left(x^2-6x+9\right)}{2\left(x-3\right)}Factor out the GCFs

Once the GCF has been factored out of the numerator, it is easier to see that this quadratic can be factored further. This is a perfect square trinomial, so we can use that identity to factor:

\displaystyle \dfrac{4\left(x^2-6x+9\right)}{2\left(x-3\right)}\displaystyle =\displaystyle \dfrac{4\left(x-3\right)\left(x-3\right)}{2\left(x-3\right)}Factor the perfect square trinomial
\displaystyle =\displaystyle 2\left(x-3\right)Reduce by common factor of 2\left(x-3\right)
\displaystyle =\displaystyle 2x-6Distributive property

When 4x^2-24x+36 is divided by 2x-6, the result is 2x-6.

Reflect and check

From this result, we know 4x^2-24x+36=\left(2x-6\right)^2 which we can use to check the answer. Using the perfect square trinomial identity \left(A+B\right)^2=A^2+2AB+B^2 with A=2x and B=-6:

\displaystyle \left(2x-6\right)^2\displaystyle =\displaystyle 4x^2-24x+36Perfect square trinomial expansion

We find the product of the quotient and the divisor is, in fact, equal to the dividend.

d

\dfrac{4x^2+6x-8}{2x+3}

Worked Solution
Create a strategy

This one can be solved using any method, so let's use long division.

Apply the idea

We set up and solve this as follows:

A figure showing the partial steps of the polynomial long division for 4 x squared plus 6 x minus 8 divided by 2 x plus 3. Speak to your teacher for more information.

At this point, we are supposed to repeat the steps of long division and divide -8 by the divisor. However, -8 is a constant which has a degree of 0 and cannot be divided by 2x+3 which has a degree of 1. So, we write a zero in the constant place value in the quotient, and continue with the remaining steps.

A figure showing the polynomial long division for 4 x squared plus 6 x minus 8 divided by 2 x plus 3. Speak to your teacher for more information.

The result of the given division is 2x+\dfrac{-8}{2x+3}.

Reflect and check

Show that synthetic can be used, but is more difficult because of leading coefficient

\displaystyle \dfrac{4x^2+6x-8}{2x+3}\div \dfrac{2}{2}\displaystyle =\displaystyle \dfrac{\frac{4x^2+6x-8}{2}}{\frac{2x+3}{2}}Divide by \frac{2}{2}
\displaystyle =\displaystyle \dfrac{2x^2+3x-4}{x+\frac{3}{2}}Evaluate the division

Now, we can use synthetic division with x=-\frac{3}{2} outside the table.

A figure showing a synthetic division with negative 3 halves in the top left corner. Top row has digits 2, 3, negative 4. Middle row has negative 3 under the 3 and 0 under negative 4. Bottom row has 2, 0, and negative 4.

This gives us an answer of 2x+\dfrac{-4}{x+\frac{3}{2}}. Normally, we do not leave fractions in the denominator, so we will rewrite the remainder by multiplying the numerator and denominator by 2:

\displaystyle \dfrac{-4}{x+\frac{3}{2}}\times \dfrac{2}{2}\displaystyle =\displaystyle \dfrac{-4\cdot 2}{\left(x+\frac{3}{2}\right)\cdot 2}Multiply the remainder by \frac{2}{2}
\displaystyle =\displaystyle \dfrac{-8}{2x+3}Evaluate the multiplication

So our final answer is 2x+\dfrac{-8}{2x+3}, just like we found before.

Idea summary

Sythetic division is a shot-hand notation of division that can only be used when the divisor is a linear expression in the form x\pm a where a is a constant.

We set up the sythetic divison by writing the coefficients of the divident in an upside down division table, and adding 0 coefficients for any missing terms. Then, we set the divisor equal to zero, solve for x, and write the result outside the table. Next, we follow these steps until the last column has been added:

  1. Multiply the number below the table by the number outside the table
  2. Write the result under the next coefficient from the dividend
  3. Add the numbers in the next column

The remainder theorem

Exploration

Divide the following polynomials:

  • \left(x^3+2x^2-5x-6\right)\div \left(x-1\right)
  • \left(x^3-7x^2-7x+20\right)\div \left(x+4\right)

Evaluate the following:

  • f\left(1\right) for f\left(x\right)=\left(x^3+2x^2-5x-6\right)
  • f\left(-4\right) for f\left(x\right)=\left(x^3-7x^2-7x+20\right)

Explain the relationship between the remainder of the division and the value of the function at the point.

In previous sections, when dividing p\left(x\right) by a linear divisor, x\pm a where a is a constant, we wrote our answers in the form \dfrac{p\left(x\right)}{x-a}=q\left(x\right)+\dfrac{r\left(x\right)}{x-a}

To check the answers we found for the quotient and remainder, we multiplied the quotient by the divisor and added the remainder. That is, p\left(x\right)=q\left(x\right)\left(x-a\right)+r\left(x\right)

Suppose x=a and the remainder is a constant R. If we substitute this into the above equation, we find \begin{aligned}p\left(a\right)&=q\left(a\right)\left(a-a\right)+R\\&=q\left(a\right)\cdot 0+R\\&=R\end{aligned}

This result shows that we can find the remainder of p\left(x\right)\div \left(x-a\right) without performing the division. It is known as the remainder theorem.

Remainder theorem

For a polynomial p\left(x\right) and a number a, the remainder on division by {x - a} is p\left(a\right).

Examples

Example 6

Using the remainder theorem, determine the remainder when p\left(x\right)=2x^3-4x^2+3x-1 is divided by 2x+1.

Worked Solution
Create a strategy

To use the remainder theorem, we need to substitute the value of x that makes 2x+1 equal to 0 into the polynomial.

\begin{aligned}2x+1&=0\\2x&=-1\\x&=-\dfrac{1}{2}\end{aligned}

By the remainder theorem, the answer of p\left(-\dfrac{1}{2}\right) is the remainder when p\left(x\right) is divided by 2x+1.

Apply the idea
\displaystyle p\left(x\right)\displaystyle =\displaystyle 2\left(-\dfrac{1}{2}\right)^3-4\left(-\dfrac{1}{2}\right)^2+3\left(-\dfrac{1}{2}\right)-1Substitute x=-\dfrac{1}{2}
\displaystyle =\displaystyle 2\left(-\dfrac{1}{8}\right)-4\left(\dfrac{1}{4}\right)+3\left(-\dfrac{1}{2}\right)-1Evaluate the exponents
\displaystyle =\displaystyle -\dfrac{1}{4}-1-\dfrac{3}{2}-1Evaluate the multiplication
\displaystyle =\displaystyle -\dfrac{1}{4}-\dfrac{4}{4}-\dfrac{6}{4}-\dfrac{4}{4}Create common denominators
\displaystyle =\displaystyle -\dfrac{15}{4}Evaluate the subtraction

When p\left(x\right) is divided by \left(2x+1\right), the remainder is -\dfrac{15}{4}.

Reflect and check

The second part of the remainder theorem tells us if p\left(a\right)=0, then x-a is a factor. From that, we can also determine that if p\left(a\right)\neq 0, then x-a is not a factor of p\left(x\right).

Applying that to this problem, we can determine that \left(2x+1\right) is not a factor of p\left(x\right). This concept will be explored more in the next lesson.

Example 7

When the polynomials P\left(x\right)=x^4+5x^3-mx+n and Q\left(x\right)=mx^2+nx-1 are each divided by D\left(x\right)=x-1, the remainders are 7 and -6 respectively. Find the values of m and n.

Worked Solution
Create a strategy

By the remainder theorem, P\left(1\right)=7 and Q\left(1\right)=-6. We can use this to create two equations, then solve them for m and n.

Apply the idea

\begin{aligned}P\left(1\right)&=7\\\left(1\right)^4+5\left(1\right)^3-m\left(1\right)+n&=7\\1+5-m+n&=7\\-m+n&=1\end{aligned}

\begin{aligned} Q\left(1\right)&=-6\\m\left(1\right)^2+n\left(1\right)-1&=-6\\m+n&=-5\end{aligned}

The two equations we found are -m+n=1 and m+n=-5. This created a system of equations that we can solve by adding the equations together:

\begin{aligned}-m+n&=1\\ + m+n&=-5 \\ \hline 2n&=-4\\n&=-2 \end{aligned}

We can now substitute n=-2 into one of the equations and solve for m:\begin{aligned}m+\left(-2\right)&=-5\\m&=-3\end{aligned}

Reflect and check

If we describe this problem and the answer in context, we can say:

When x^4+5x^3+3x-2 is divided by x-1, the remainder is 7. And when -3x^2-2x-1 is divided by x-1, the remainder is -6.

Idea summary

To find the remainder of \dfrac{p\left(x\right)}{x-a} without performing the division, we can evaluate p\left(a\right). The result will be the value of the remainder.

Outcomes

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.APR.B.2

Know and apply the remainder theorem: for a polynomial p(x) and a number a, the remainder on division by x - a is p(a), so p(a) = 0 if and only if (x - a) is a factor of p(x).

A.APR.D.6

Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.

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