Dividing by Binomials and Polynomials

Learning Objectives

Introduction

Dividing a polynomialA monomial or the sum or difference of two or more monomials. by a monomial can be handled by dividing each term in the polynomial separately. This can’t be done when the divisor has more than one term. However, the process of long division can be very helpful with polynomials.

Dividing by a Binomial

First, recall how you can use long division to divide two whole numbers, say `900` divided by `37`.

An image shows a long division problem. The dividend is 900. The divisor is 37.

First, you would think about how many `37`s are in `90`, as `9` is too small. (Note: you could also think, how many `40`s are there in `90`.)

 

An image shows a long division problem. The dividend is 900. The divisor is 37. The quotient is 2. Multiply the quotient, 2, by the divisor, 37, which equals 74 and is placed below 90, the first two digits in the dividend.

There are two `37`s in `90`, so write `2` above the last digit of `90`. Two `37`s is `74`; write that product below the `90`.

An image shows a long division problem. The dividend is 900. The divisor is 37. The quotient is 2. Multiply the quotient, 2, by the divisor, which equals 74 and is placed below the 90 in the dividend. 74 is subtracted from 90, the first two digits of the dividend, which equals 16.

Subtract: `90 - 74` is `16`. (If the result is larger than the divisor, which is `37`, then you need to use a larger number for the quotient.)

An image shows a long division problem. The dividend is 900. The divisor is 37. The quotient is 2. Multiply the quotient, 2, by the divisor, which equals 74 and is placed below the 90 in the dividend. 74 is subtracted from 90, the first two digits of the dividend, which equals 16. Bring down the zero from the dividend to make 160.

Bring down the next digit (`0`) and consider how many `37`s are in `160`.

A long division problem is shown. The dividend is 900. The divisor is 37. The final quotient is 24. Multiply the first part of the quotient, 2, by the divisor, which equals 74 and is placed below the 90 in the dividend. 74 is subtracted from 90, the first two digits of the dividend, which equals 16. Bring down the zero from the dividend to make 160. Next, 4 is added to the quotient, making it 24. The 4 in the quotient is multiplied by the divisor, resulting in 148, which is subtracted from 160 in the line above it.

There are four `37`s in `160`, so write the `4` next to the two in the quotient. Four `37`s is `148`; write that product below the `160`.

An image shows a long division problem. The dividend is 900. The divisor is 37. The final quotient is 24. Multiply the first part of the quotient, 2, by the divisor, which equals 74 and is placed below the 90 in the dividend. 74 is subtracted from 90, the first two digits of the dividend, which equals 16. Bring down the zero from the dividend to make 160. Next, 4 is added to the quotient, making it 24. The 4 in the quotient is multiplied by the divisor resulting in 148, which is subtracted from 160 in the line above it. This equals 12, which is the remainder.

Subtract: `160 - 148` is `12`. This is less than `37` so the `4` is correct. Since there are no more digits in the dividend to bring down, you’re done.

The final answer is `24"R"12`, or `24 12/37`. You can check this by multiplying the quotient (without the remainder) by the divisor, and then adding in the remainder. The result should be the dividend:

`24 * 37 + 12 = 888 + 12 = 900`

To divide polynomials, use the same process. This example shows how to do this when dividing by a binomial A polynomial with exactly two terms, such as `5y^2 - 4x` and `x^5 + 6`. .

Example

Problem

Divide: `(x^2-4x-12) -:(x+2)`

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2.

 

How many `x`’s are there in `x^2`? That is, what is `x^2/x`?

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor, which equals X squared plus 2 X.

 

 

 

 

 

`x^2/x=x`. Put `x` in the quotient above the `-4x` term. (These are like terms, which helps to organize the problem.)

Write the product of the divisor and the part of the quotient you just found under the dividend. Since `x(x + 2) = x^2+2x`, write this underneath, and get ready to subtract.

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor. Rewrite the result, X squared plus 2 X, as its opposite, negative X squared minus 2 X. Add to the dividend.

 

Rewrite `-(x^2 + 2x)` as its opposite `-x^2 - 2x` so that you can add the opposite. (Adding the opposite is the same as subtracting, and it is easier to do.)

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor. Rewrite the result, X squared plus 2 X, as its opposite, negative X squared minus 2 X. Add to the dividend, which equals negative 6 X.

 

 

Add `-x^2` to `x^2`, and `-2x` to `-4x`.

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor. Rewrite the result, X squared plus 2 X, as its opposite, negative X squared minus 2 X. Add to the dividend, which equals negative 6 X. Bring down minus 12 from the dividend.

 

Bring down `-12`.

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor. Rewrite the result, X squared plus 2 X, as its opposite, negative X squared minus 2 X. Add to the dividend, which equals negative 6 X. Bring down minus 12 from the dividend.

 

Repeat the process. How many times does `x` go into `-6x`? In other words, what is `(-6x)/x`?

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor. Rewrite the result, X squared plus 2 X, as its opposite, negative X squared minus 2 X. Add to the dividend, which equals negative 6 X. Bring down minus 12 from the dividend. Next, 6 is subtracted from X in the quotient making it X minus 6. Negative 6 in the quotient is multiplied by the divisor resulting in, open parenthesis, negative 6 X minus 12, close parenthesis, which is subtracted from the line above it.

 

Since `(-6x)/x=-6`, write `-6` in the quotient. Multiply `-6` and `x + 2` and prepare to subtract the product.

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor. Rewrite the result, X squared plus 2 X, as its opposite, negative X squared minus 2 X. Add to the dividend, which equals negative 6 X. Bring down minus 12 from the dividend. Next, 6 is subtracted from X in the quotient making it X minus 6. Negative 6 in the quotient is multiplied by the divisor resulting in, open parenthesis, negative 6 X minus 12, close parenthesis, which is rewritten as its opposite, 6 X plus 12. Add 6 X plus 12 to the line above it.

 

Rewrite `-(-6x - 12)` as `6x + 12`, so that you can add the opposite.

 

An image shows a long division problem. The dividend is X squared minus 4 X minus 12. The divisor is X plus 2. The first part of the quotient is X. Multiply the quotient, X, by the divisor. Rewrite the result, X squared plus 2 X, as its opposite, negative X squared minus 2 X. Add to the dividend, which equals negative 6 X. Bring down minus 12 from the dividend. Next, 6 is subtracted from X in the quotient making it X minus 6. Negative 6 in the quotient is multiplied by the divisor resulting in, open parenthesis, negative 6 X minus 12, close parenthesis, which is rewritten as its opposite, 6 X plus 12. Add 6 X plus 12 to the line above it. There is no remainder.

 

Add. In this case, there is no remainder, so you’re done.

Answer

`(x^2-4x-12)-:(x+2)=x-6`

 

 

Check this by multiplying:

`(x - 6)(x + 2) = x^2 + 2x - 6x - 12 = x^2 - 4x - 12`

Let’s try another example. In this example, a term is “missing” from the dividend.

Example

Problem

Divide: `(x^3-6x-10)-:(x-3)`

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3.

 

In setting up this problem, notice that there is an `x^3` term but no `x^2` term. Add `0x^2` as a “place holder” for this term. (Since `0` times anything is `0`, you’re not changing the value of the dividend.)

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3. The first part of the quotient is X squared. Multiply the first part of the quotient, X squared, by the divisor, which equals open parenthesis, X cubed minus 3 X squared, close parenthesis. This line is subtracted from the dividend.

 

 

 

Focus on the first terms again: how many `x`’s are there in `x^3`? Since `x^3/x=x^2`, put `x^2` in the quotient.

 

Multiply `x^2(x - 3) = x^3-3x^2` and write this underneath the dividend and prepare to subtract.

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3. The first part of the quotient is X squared. Multiply the first part of the quotient, X squared, by the divisor, which equals minus X cubed plus 3 X squared (the result of distributing the negative across x cubed minus 3 x squared). This line is subtracted from the dividend. Then add, which results in 3 X squared.

 

Rewrite the subtraction using the opposite of the expression `x^3 - 3x^2`. Then add.

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3. The first part of the quotient is X squared. Multiply the first part of the quotient, X squared, by the divisor, which equals negative X cubed plus 3 X squared. This line is subtracted from the dividend. Then add, which results in 3 X squared. Bring down the rest of the dividend, resulting in 3 X squared minus 6 X minus 10.

 

Bring down the rest of the expression in the dividend. It’s helpful to bring down all of the remaining terms.

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3. The first part of the quotient is X squared. Multiply the first part of the quotient, X squared, by the divisor, which equals open parenthesis, X cubed minus 3 X squared, close parenthesis. This line is subtracted from the dividend. Rewrite the subtraction using the opposite of the expression. Then add, which results in 3 X squared. Bring down the rest of the dividend, resulting in 3 X squared minus 6 X minus 10. Next, 3 X is added to the quotient making it X squared plus 3 X. The 3 X in the quotient is multiplied by the divisor resulting in, open parenthesis, 3 X squared minus 9 X, close parenthesis, which is subtracted from the line above it.

 

Now, repeat the process with the remaining expression as the dividend: `3x^2 - 6x - 10`.

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3. The first part of the quotient is X squared. Multiply the first part of the quotient, X squared, by the divisor, which equals open parenthesis, X cubed minus 3 X squared, close parenthesis. This line is subtracted from the dividend. Rewrite the subtraction using the opposite of the expression. Then add, which results in 3 X squared. Bring down the rest of the dividend, resulting in 3 X squared minus 6 X minus 10. Next, 3 X is added to the quotient making it X squared plus 3 X. The 3 X in the quotient is multiplied by the divisor resulting in, open parenthesis, 3 X squared minus 9 X, close parenthesis, which is subtracted from the line above it. This equals 3 X minus 10.

 

Remember to watch the signs!

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3. The first part of the quotient is X squared. Multiply the first part of the quotient, X squared, by the divisor, which equals open parenthesis, X cubed minus 3 X squared, close parenthesis. This line is subtracted from the dividend. Rewrite the subtraction using the opposite of the expression. Then add, which results in 3 X squared. Bring down the rest of the dividend, resulting in 3 X squared minus 6 X minus 10. Next, 3 X is added to the quotient making it X squared plus 3 X. The 3 X in the quotient is multiplied by the divisor resulting in, open parenthesis, 3 X squared minus 9 X, close parenthesis, which is subtracted from the line above it. This equals 3 X minus 10. Next, 3 is added to the quotient making it X squared plus 3 X plus 3. The 3 in the quotient is multiplied by the divisor resulting in, open parenthesis, 3 X minus 9 X, close parenthesis, which is subtracted from the line above it.

 

How many `x`’s are there in `3x`? Since there are `3`, multiply `3(x - 3) = 3x-9`, write this underneath the dividend and prepare to subtract.

 

An image shows a long division problem. The dividend is X cubed plus 0 X squared minus 6 X minus 10. The divisor is X minus 3. The first part of the quotient is X squared. Multiply the first part of the quotient, X squared, by the divisor, which equals open parenthesis, X cubed minus 3 X squared, close parenthesis. This line is subtracted from the dividend. Rewrite the subtraction using the opposite of the expression. Then add, which results in 3 X squared. Bring down the rest of the dividend, resulting in 3 X squared minus 6 X minus 10. Next, 3 X is added to the quotient making it X squared plus 3 X. The 3 X in the quotient is multiplied by the divisor resulting in, open parenthesis, 3 X squared minus 9 X, close parenthesis, which is subtracted from the line above it. This equals 3 X minus 10. Next, 3 is added to the quotient making it X squared plus 3 X plus 3. The 3 in the quotient is multiplied by the divisor resulting in, open parenthesis, 3 X minus 9 X, close parenthesis, which is subtracted from the line above it. This equals a remainder of negative 1.

 

 

 

 

Continue until the degreeThe value of an exponent. of the remainder is less than the degree of the divisor. In this case the degree of the remainder, `-1`, is `0`, which is less than the degree of `x - 3`, which is `1`.

Also notice that you have brought down all the terms in the dividend, and that the quotient extends to the right edge of the dividend. These are other ways to check whether you have completed the problem.

 

 

 

 

Answer

`(x^3-6x-10)-:(x-3)=x^2+3x+3, "R"-1`

 

`x^2+3x+3+(-1)/(x-3)`, or `x^2+3x+3-1/(x-3)`

 

You can write the remainder using the symbol `R`, or as a fraction added to the rest of the quotient with the remainder in the numerator and the divisor in the denominator. In this case, since the remainder is negative, you can also subtract the opposite.

Check the result:

`(x - 3)(x^2 + 3x + 3)`

`=`

`x(x^2 + 3x + 3) - 3(x^2+3x + 3)`

 

`=`

`x^3 + 3x^2 + 3x - 3x^2 - 9x - 9`

 

`=`

`x^3 - 6x - 9`

 

 

 

`x^3 - 6x - 9 + (-1)`

`=`

`x^3 - 6x - 10`

 

Divide:

`(x^2-8x+3)-:(x-5)`

 

A) `x-3-12/(x-5)`

 

B) `x-13+68/(x-5)`

 

C) `x-13-62/(x-5)`

 

D) `x-15`

 

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Dividing Polynomials by Polynomials

The process above works for dividing any polynomials, no matter how many terms are in the divisor or the dividend. The main things to remember are:

Example

Problem

Divide: `(3x^3+2x^2-3x+4)-:(x^2-3x+5)`

 

An image shows a long division problem. The dividend is 3 X cubed plus 2 X squared minus 3 X plus 4. The divisor is X squared minus 3 X plus 5. The quotient is 3 X plus 11. Multiply the first part of the quotient, 3 X, by the divisor, which equals open parenthesis, 3 X cubed minus 9 X squared plus 15 X, close parenthesis. This line is subtracted from the dividend, which equals 11 X squared minus 18 X. Bring down the 4 from the dividend, resulting in 11 X squared minus 18 X plus 4. Next, multiply the second part of the quotient by the divisor, which equals open parenthesis, 11 X squared minus 33 X plus 55, close parenthesis. This is subtracted from the line above it which equals the remainder 15 X minus 51.

 

 

 

 

Focus on the first terms: What is `(3x^3)/x^2`? Since `(3x^3)/x^2=3x`, start by putting `3x` in the quotient. Follow the process as above. Watch the signs!

The degree of the remainder is `1`, which is less than the degree of the divisor, `2`. You can stop.

Answer

`(3x^3+2x^2-3x+4)-:(x^2-3x+5)=3x+11"R"51x-51`

or `3x+11+(15x-51)/(x^2-3x+5)`

 

 

Check:

`(x^2 - 3x + 5)(3x + 11)`

`=`

`(x^2 - 3x + 5)(3x) + (x^2 - 3x + 5)(11)`

 

`=`

`3x^3 - 9x^2 + 15x + 11x^2 - 33x + 55`

 

`=`

`3x^3 + 2x^2 - 18x + 55`

 

 

 

`3x^3 + 2x^2 - 18x + 55 + (15x - 51)`

`=`

`3x^3 + 2x^2 - 3x + 4`

 

Divide:

`(x^3-2x^2+3x+7)-:(x^2+2x-1)`

 

A) `x+(2x+7)/(x^2+2x-1)`

 

B) `x+8+23/(x^2+2x-1)`

 

C) `x-4+(12x+3)/(x^2+2x-1)`

 

D) `13x-1`

 

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Summary

Dividing polynomials by polynomials of more than one term can be done using a process very much like long division of whole numbers. You must be careful to subtract entire expressions, not just the first term. Stop when the degree of the remainder is less than the degree of the divisor. The remainder can be written using R notation, or as a fraction added to the quotient with the remainder in the numerator and the divisor in the denominator.