Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 2

Algebra 2 Dividing Polynomials

Name: Algebra 2: Dividing Polynomials
Brand: Educator.com
Price: 35 USD
Availability: InStock

Section 6: Polynomial Functions: Lecture 3 | 31:11 min

Lecture Description

Next Lecture

Previous Lecture

Discussion
Answer Engine
Study Guides
Practice Questions
Download Lecture Slides
Table of Contents
Transcription
Related Books

Lecture Comments (9)

	1 answer Last reply by: Dr Carleen Eaton Fri Mar 8, 2019 1:54 PM Post by Kenneth Geller on February 14, 2019 In example 3 is the remainder 1232 or 1232/y-4 ?
	1 answer Last reply by: Angel La Fayette Wed Jan 2, 2013 4:11 PM Post by Angel La Fayette on January 2, 2013 How come the -15 in the dividend was not divided by the 3?
	1 answer Last reply by: Angel La Fayette Wed Jan 2, 2013 4:11 PM Post by Daniel Cuellar on October 26, 2012 to fix you mr. Jeff, 6 divided by 3 is 2...
	2 answers Last reply by: Angel La Fayette Wed Jan 2, 2013 4:06 PM Post by Jeff Mitchell on March 21, 2011 approx 16:42 into lecture you show 3x^3+12x^2-15x+6 and divide by 3 with result x^3+4x^2-15X+3 but I believe it should be x^3+4x^2-5x+3 ~Jeff

Answer EngineGet answers to any question!Ask any question related to Algebra 2

Working on the solution...

Dividing Polynomials

When dividing a polynomial by a binomial, any missing terms of the dividend must be written explicitly, using a coefficient of 0.
In synthetic division, the coefficient of the binomial divisor must be 1. If it is not, the division problem must be rewritten so that it is 1.

Dividing Polynomials

Divide: [(25x⁵y⁴ − 15xy³ + 4x³y²)/(5x²y³)]

Break down the divident into three terms. Each term must get a copy of the divisor.
[(25x⁵y⁴)/(5x²y³)] − [(15xy³)/(5x²y³)] + [(4x³y²)/(5x²y³)]
Start by simplifying the coefficients of each term if possible.
[(5x⁵y⁴)/(x²y³)] − [(3xy³)/(x²y³)] + [(4x³y²)/(5x²y³)]
Simplify by using properties of exponents.
5x³y − 3x^{− 1} + [(4xy^{− 1})/5]
Eliminate negative exponents by bringing the variable to denominator, thus making exponents positive

5x³y − [3/x] + [4x/5y]

Divide: [(21x⁴y⁴ − 14x²y⁵ + 7x³y)/(7x⁶y⁷)]

Break down the divident into three terms. Each term must get a copy of the divisor.
[(21x⁴y⁴)/(7x⁶y⁷)] − [(14x²y⁵)/(7x⁶y⁷)] + [(7x³y)/(7x⁶y⁷)]
Start by simplifying the coefficients of each term if possible.
[(3x⁴y⁴)/(x⁶y⁷)] − [(2x²y⁵)/(x⁶y⁷)] + [(x³y)/(x⁶y⁷)]
Simplify by using properties of exponents.
3x^{− 2}y^{− 3} − 2x^{− 4}y^{− 2} + x^{− 3}y^{− 6}
Eliminate negative exponents by bringing the variable to denominator, thus making exponents positive

[3/(x²y³)] − [2/(x⁴y²)] + [1/(x³y⁶)]

Divide [(n³ + 9n² + 18n + 10)/(n + 1)]

Always check that there are no missing terms. If there are any missing terms, always use a place holder such as 0x³. In this case, there are no missing terms, so everything can proceed.
Divide n³ by n = n², then multiply n² by n and 1 and change the sign of the result (This is the same as adding the opposite as explained in the lesson). Add, then bring down the next term (18n) .
n²
n+1 n³ 9n² 18n 10
+ −n³ −n² ↓
0 8n² 18n
Divide 8n² by n = 8n . Multiply 8n by n and 1, change the sign of the result. Add, then bring the next term 10.
n² 8n
n+1 n³ 9n² 18n 10
+ −n³ −n² ↓ ↓
0 8n² 18n ↓
+ −8n² −8n ↓
0 10n 10
Divide 10n by n = 10. Multiply 10 by n and 1, change the sign of the result. Add and notice how there is no reminder.
n² 8n
n+1 n³ 9n² 18n 10
+ −n³ −n² ↓ ↓
0 8n² 18n ↓
+ −8n² −8n ↓
0 10n 10
+ −10n −10

n² + 8n + 10

Divide [(n³ − 9n + 21)/(n + 4)]

Always check that there are no missing terms. If there are any missing terms, always use a place holder such as 0x³. In this case, use 0n² since the square terms is missing.
Divide n³ by n = n², then multiply n² by n and 4 and change the sign of the result(This is the same as adding the opposite as explained in the lesson). Add, then bring down the next term ( − 9n) .
n²
n+4 n³ 0n² −9n 21
+ −n³ −4n² ↓
0 −4n² −9n
Divide − 4n² by n = − 4n . Multiply − 4n by n and 4, change the sign of the result. Add, then bring the next term 21.
n² -4n
n+4 n³ 0n² −9n 21
+ −n³ −4n² ↓ ↓
0 −4n² −9n ↓
+ +4n² +16n ↓
0 7n 21
Divide 7n by n = 7. Multiply 7 by n and 4, change the sign of the result. Add and notice how the remainder is 7.
n² -4n
n+4 n³ 0n² −9n 21
+ −n³ −4n² ↓ ↓
0 −4n² −9n ↓
+ +4n² +16n ↓
0 7n 21
−7n −28
0 −7

n² − 4n + 7 − [7/(n + 4)]

Divide using Synthetic Division [(x⁴ + 7x³ + 14x² + 9x + 9)/(x + 3)]

As is the case with long division, always check that there are no missing terms.
In this case, there are no missing terms, so you may continue.
Also, check that the divisor is in the format (x − r), if it happens to be in the
format (ax − r), then everything needs to be divided by a in order to proceed with synthetic division.
In this case, x + 3 meets the pattern, therefore, you can divide.
Get coeffieients and remember, the number outside is always the opposite sign, in this case, − 3
-3 1 7 14 9 9
Bring the first number down.
-3 1 7 14 9 9

1
Multiply (1) by ( - 3), result goes below 7. Add, result goes below
-3 1 7 14 9 9
-3
1 4
Multiply (4) by ( - 3), result goes below 14. Add, result goes below
-3 1 7 14 9 9
-3 -12
1 4 2
Multiply (2) by ( - 3), result goes below 9. Add, result goes below
-3 1 7 14 9 9
-3 -12 -6
1 4 2 3
Multiply (3) by ( - 3), result goes below 9. Add, result goes below
-3 1 7 14 9 9
-3 -12 -6 -9
1 4 2 3 0
Remember, the result are the coefficients. The last number is always the remainder, in this case, it's 0.

x³ + 4x² + 2x + 3

Divide using Synthetic Division [(p⁴ − 3p³ − 9p² − 12p − 7)/(p + 1)]

As is the case with long division, always check that there are no missing terms.
In this case, there are no missing terms, so you may continue.
Also, check that the divisor is in the format (x − r), if it happens to be in the
format (ax − r), then everything needs to be divided by a in order to proceed with synthetic division.
In this case, p + 1 meets the pattern, therefore, you can divide.
Get coeffieients and remember, the number outside is always the opposite sign, in this case, − 1
-1 1 -3 -9 -12 -7
Bring the first number down.
-1 1 -3 -9 -12 -7

1
Multiply (1) by ( − 1), result goes below − 3. Add, result goes below the line
-1 1 -3 -9 -12 -7
1 -1
1 -4
Multiply ( − 4) by ( − 1), result goes below − 9. Add, result goes below the line.
-1 1 -3 -9 -12 -7
1 -1 4
1 -4 -5
Multiply ( − 5) by ( − 1), result goes below − 12. Add, result goes below the line.
-1 1 -3 -9 -12 -7
1 -1 4 5
1 -4 -5 -7
Multiply (7) by ( − 1), result goes below − 7. Add, result goes below the line.
-1 1 -3 -9 -12 -7
1 -1 4 5 7
1 -4 -5 -7 0
Remember, the result are the coefficients. Start with one power less than the divident. The last number is always the remainder, in this case, it's 0.

x³ − 4x² − 5x − 7

Divide using Synthetic Division [(p⁴ − 3p³ + p − 3)/(p − 3)]

As is the case with long division, always check that there are no missing terms.
In this case, the p² is missing , therefore, add a place holder in it's place.
Also, check that the divisor is in the format (x − r), if it happens to be in the
format (ax − r), then everything needs to be divided by a in order to proceed with synthetic division.
In this case, p − 3 meets the pattern, therefore, you can divide.
Get coeffieients and remember, the number outside is always the opposite sign, in this case, + 3
+3 1 -3 0 1 -3
Bring the first number down.
+3 1 -3 0 1 -3

1
Multiply (1) by (3), result goes below − 3. Add, result goes below the line
+3 1 -3 0 1 -3
3
1 0
Multiply (0) by (3), result goes below 0. Add, result goes below the line.
+3 1 -3 0 1 -3
3 0
1 0 0
Multiply (0) by (3), result goes below 1. Add, result goes below the line.
+3 1 -3 0 1 -3
3 0 0
1 0 0 1
Multiply (1) by ( + 3), result goes below − 3. Add, result goes below the line.
+3 1 -3 0 1 -3
3 0 0 3
1 0 0 1 0
Remember, the result are the coefficients. Start with one power less than the divident. The last number is always the remainder, in this case, it's 0.

p³ + 0p² + 0p + 1 = p³ + 1

Divide using Synthetic Division [(2b⁴ − 6b³ + 5b − 17)/(b − 3)]

As is the case with long division, always check that there are no missing terms.
In this case, the b² is missing , therefore, add a place holder in its place.
Also, check that the divisor is in the format (x − r), if it happens to be in the
format (ax − r), then everything needs to be divided by a in order to proceed with synthetic division.
In this case, b − 3 meets the pattern, therefore, you can divide.
Get coeffieients and remember, the number outside is always the opposite sign, in this case, + 3
+3 2 -6 0 5 -17
Bring the first number down.
+3 2 -6 0 5 -17

2
Multiply (2) by (3), result goes below − 6. Add, result goes below the line
+3 2 -6 0 5 -17
6
2 0
Multiply (0) by (3), result goes below 0. Add, result goes below the line.
+3 2 -6 0 5 -17
6 0
2 0 0
Multiply (0) by (3), result goes below 5. Add, result goes below the line.
+3 2 -6 0 5 -17
6 0 0
2 0 0 5
Multiply (5) by ( + 3), result goes below − 17. Add, result goes below the line.
+3 2 -6 0 5 -17
6 0 0 15
2 0 0 5 −2
Remember, the result are the coefficients. Start with one power less than the divident. The last number is always the remainder, in this case, it's − 2.
2b³ + 0b² + 0b + 5 − [2/(b − 3)] = 2b³ + 5 − [2/(b − 3)]

2b³ + 5 − [2/(b − 3)]

Divide using Synthetic Division [(4n³ − 2n² − 4n)/(4n − 2)]

As is the case with long division, always check that there are no missing terms.
In this case, the constant term is missing , therefore, add a place holder in its place.
Also, check that the divisor is in the format (x − r), if it happens to be in the
format (ax − r), then everything needs to be divided by a in order to proceed with synthetic division.
In this case, 4n − 2 does not meets the pattern, therefore, you have to make it by dividing everything by 4.
[(4n³)/4] − [(2n²)/4] − [4n/4] + [0/4] ÷[4n/4] − [2/4]
Simplify as much as possible
n³ − [(1n²)/2] − n + 0 ÷n − [1/2]
Get coeffieients and remember, the number outside is always the opposite sign, in this case, + [1/2]
[1/2] 1 −[1/2] -1 0
Bring the first number down.
[1/2] 1 −[1/2] -1 0

1
Multiply (1) by (1/2), result goes below − 1/2. Add, result goes below the line
[1/2] 1 −[1/2] -1 0
[1/2]
1 0
Multiply (0) by (1/2), result goes below − 1. Add, result goes below the line.
[1/2] 1 −[1/2] -1 0
[1/2] 0
1 0 -1
Multiply ( − 1) by (1/2), result goes below 0. Add, result goes below the line.
[1/2] 1 −[1/2] -1 0
[1/2] 0 −[1/2]
1 0 -1
−[1/2]
Remember, the result are the coefficients. Start with one power less than the divident. The last number is always the remainder, in this case, it's − 1/2.
Using some algebra one to reduce the complex fraction [([1/2])/(n − [1/2])] you get:
n² − n − [([1/2])/(n − [1/2])] = n² − n − [([1/2])/([2n/2] − [1/2])]
n² − n − [([1/2])/([2n/2] − [1/2])] = n² − n − [([1/2])/([(2n − 1)/2])]
n² − n − [([1/2])/([(2n − 1)/2])] = n² − n − ( [1/2] ÷[(2n − 1)/2] )
n² − n − ( [1/2] ÷[(2n − 1)/2] ) = n² − n − ( [1/2]*[2/(2n − 1)] )
n² − n − ( [1/2]*[2/(2n − 1)] ) = n² − n − [1/(2n − 1)]

n² − n − [1/(2n − 1)]

Divide using Synthetic Division [(3r³ − 7r² − 8r + 14)/(3r − 4)]

As is the case with long division, always check that there are no missing terms.
In this case, the constant term is missing , therefore, add a place holder in its place.
Also, check that the divisor is in the format (x − r), if it happens to be in the
format (ax − r), then everything needs to be divided by a in order to proceed with synthetic division.
In this case, 3r − 4 does not meets the pattern, therefore, you have to make it by dividing everything by 3.
[(3r³)/3] − [(7r²)/3] − [8r/3] + [14/3] ÷[3r/3] − [4/3]
Simplify as much as possible
r³ − [(7r²)/3] − [8r/3] + [14/3] ÷r − [4/3]
Get coeffieients and remember, the number outside is always the opposite sign, in this case, + [4/3]
[4/3] 1 −[7/3] −[8/3] [14/3]
Bring the first number down.
[4/3] 1 −[7/3] −[8/3] [14/3]

1
Multiply (1) by (4/3), result goes below − 7/3. Add, result goes below the line
[4/3] 1 −[7/3] −[8/3] [14/3]
[4/3]
1 −[3/3]
Multiply ( − 1) by (4/3), result goes below − 8/3. Add, result goes below the line.
[4/3] 1 −[7/3] −[8/3] [14/3]
[4/3] −[4/3]
1 −[3/3] = −1 −[12/3]=−4
Multiply ( − 4) by (4/3), result goes below 14/3. Add, result goes below the line.
[4/3] 1 −[7/3] −[8/3] [14/3]
[4/3] −[4/3] −[16/3]
1 −[3/3] = −1 −[12/3]=−4
−[2/3]
Remember, the result are the coefficients. Start with one power less than the divident. The last number is always the remainder, in this case, it's − 2/3.
Using some algebra one to reduce the complex fraction [([1/2])/(n − [1/2])] you get:
r² − r − 4 − [([2/3])/(r − [4/3])] = r² − r − 4 − [([2/3])/([3r/3] − [4/3])]
r² − r − 4 − [([2/3])/([3r/3] − [4/3])] = r² − r − 4 − [([2/3])/([(3r − 4)/3])]
r² − r − 4 − [([2/3])/([(3r − 4)/3])] = r² − r − 4 − ( [2/3] ÷[(3r − 4)/3] )
r² − r − 4 − ( [2/3] ÷[(3r − 4)/3] ) = r² − r − 4 − ( [2/3]*[3/(3r − 4)] )
r² − r − 4 − ( [2/3]*[3/(3r − 4)] ) = r² − r − 4 − [2/(3r − 4)]

r² − r − 4 − [2/(3r − 4)]

« Previous Question | Next Question »

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Dividing Polynomials

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Dividing by a Monomial

Long Division

Synthetic Division

Divisor in Synthetic Division

Example: Coefficient to 1

Example 1: Divide Polynomials

Example 2: Divide Polynomials

Example 3: Synthetic Division

Example 4: Synthetic Division

Intro 0:00
Dividing by a Monomial 0:13

Example: Numbers
Example: Polynomial by a Monomial

Long Division 2:28

Remainder Term
Example: Dividing with Numbers
Example: With Polynomials
Example: Missing Terms

Synthetic Division 11:44

Restriction
Example: Divisor in Form

Divisor in Synthetic Division 15:54

Example: Coefficient to 1

Example 1: Divide Polynomials 17:10
Example 2: Divide Polynomials 19:08
Example 3: Synthetic Division 21:42
Example 4: Synthetic Division 25:09

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Dividing Polynomials

Welcome to Educator.com.0000

Today, we are going to be talking about dividing polynomials, starting out with a review of techniques learned in Algebra I,0002

and then going on to learn a new technique called synthetic division.0008

First, we are discussing dividing a polynomial by a monomial.0013

The technique is to divide each term of the polynomial by the monomial.0017

And if you think about just dividing with regular numbers, if you have something like this,0023

you could handle it by saying, "15 divided by 3, plus 6 divided by 3, plus 24 divided by 3," and splitting that up.0030

This would give you 5; 6 divided by 3 is 2; and then, 24 divided by 3 is 8; adding this up gives you 15.0041

You may also have done this and added 15, 6, and 24; and those add up to 45/3, to get 15.0054

So, these two are equivalent; and that is what allows you to handle dividing a polynomial by a monomial, by dividing each term by the monomial.0068

For example, 10x⁴ + 8x³ - 12x, all divided by 2x:0078

looking at this up here as my example of how to handle it, I am going to say,0088

"OK, this is equivalent to 10x⁴ divided by 2x, plus 8x³ divided by 2x, minus 12x divided by 2x.0092

And this is a step you might do in your head, or you might write it out.0106

10 divided by 2 is 5; x⁴ divided by x would be the same as saying x^{4 - 1}, or x³.0111

8 divided by 2 gives me 4; x³ divided by x is x².0122

Here, I have -12 divided by 2 (is 6); and the x's cancel.0129

My result is 5x³ + 4x² - 6.0135

And I handled that by dividing each term in the polynomial by the monomial, separately.0140

OK, now when you are working with dividing a polynomial by another polynomial, one technique is long division.0148

And we talked about this in Algebra I, and you can also review those lectures; and I will review it here, as well.0154

Just as with regular division with numbers (long division) you might end up with a remainder.0161

So, first, just reviewing long division (which I am sure you know well, but just to think about the steps you are taking):0166

you probably do this so much that you don't even think about the steps,0173

but you want to realize what you are doing each step of the way,0175

so that you can apply it when you are dividing polynomials.0180

If you were asked to divide something like 513 by 2 by long division, think about what you would be doing.0184

First, you divide with the first term, using the first term in the divisor: 5 divided by 2.0192

OK, so that is going to give me 2.0202

Next, I am going to multiply--I am going to multiply 2 by the divisor 2 to get 4.0204

And then, I am going to subtract that product: 2 times 2 is 4, so I am going to subtract that--that is going to give me 1.0213

Next, I will bring down the next number; in the case of polynomials, I will bring down the next term.0224

OK, 2 goes into 11 five times; this gives me 5 times 2 (is 10): 11 minus 10 is 1; now I am bringing down this 3.0235

2 goes into 13 six times, and this is going to give me 12, because now I am multiplying and then subtracting; and I have a remainder of 1.0255

Remember that you can always check your answer by realizing that the dividend equals the divisor, times the quotient, plus the remainder.0265

Looking at this with these numbers, my dividend here is 513; so it equals the divisor (which is 2), times 256, plus the remainder of 1, which equals 513.0283

So, that checks out.0297

Now, with polynomials, it's the same concept.0298

Look at this example, which is going to be 5x² + 4x - 7, all divided by x + 3.0301

Using long division: it is the same steps that I took over here.0316

I am going to start out with x and divide that into 5x²; I have 5x² divided by x;0324

and you can see that this is going to give me 5x, because one of the x's will cancel.0331

OK, so that gives me 5x; now, my next step is to multiply 5x times x, which gives me 5x².0337

5x times 3 is 15x; now, I divided; I multiplied; the next step is to subtract.0350

The 5x's cancel out; 4x - 15x is -11x.0362

The next step: bring down the next number--and here, that would be -7, the next term.0370

OK, so I go back up; I was down here--I go back to #1 again and divide.0378

-11x divided by x; the x's cancel, so that gives me -11.0383

So, I am going to put that up here; I divided; now I need to multiply.0391

-11 times x is -11x; -11 times 3 gives me -33, minus (I subtract--subtracting is the same as adding the opposite,0395

so I am going to add, and the opposite of these two would be positive terms)...-11x and 11x cancels out.0409

And then, I have -7 + 33, which is going to give me 26; and that is my remainder.0417

So, my answer is the quotient, 5x - 11, plus the remainder of 26.0424

Now, to check this out, I am going to say, "OK, the dividend equals the divisor, times the quotient, plus the remainder."0429

Let's make sure this checks out: that is 5x² - 11x + 15x - 33 (just using the FOIL method) + 26.0446

Simplify to get 5x²...-11x + 15x gives me 4x; -33 + 26 is -7.0462

And this does check out; so again, this is really just using the same techniques as you used for long division with numbers.0470

Now, one thing to be aware of is that sometimes there are missing terms.0477

And if that is the case, you need to use a coefficient of 0 and represent those terms when you are dividing.0483

Now, if you think about it, when we have numbers like, say, 107, we have a 0 here as a placeholder.0490

When we are doing long division with polynomials, we need to do the same thing.0500

For example, let's say you were asked to divide 3x³ + 6x - 4: you were asked to divide that by x + 2.0505

If you look here, there is a missing term: I have an x³ term, but I have no x² term.0517

And then, I have an x and a constant.0524

So, I have a missing x² term; and really, I can represent that x² term by giving it a coefficient of 0.0525

And I need to do that before I divide.0537

So, if I have x + 2, that is my divisor; my dividend--what I am going to write here--is0539

3x³ + (my missing term) 0x² + 6x - 4.0546

And then, I am going to go about long division like I usually do.0555

So, I take 3x³, and I divide that by x; and that is going to give me 3x²,0559

because 3x³ divided by x is going to give me 3x².0566

OK, so I divided; the next step is to multiply: 3x² times x is 3x³; 3x² times 2 is 6x².0574

The next step--subtract: these cancel out; this is 0x² - 6x².0584

And you can see how, if you didn't have that missing term in there, you would get a completely different answer.0592

OK, so now I have -6x²; I am going to bring down this next term, 6x.0599

OK, now x goes into -6x² -6x times; so I am going to put -6x up here.0605

I divided; now multiply: -6x times x gives me -6x²; -6x times 2 is -12x.0620

Now, I am subtracting; but remember, that is the same as adding the opposite--I am going to change these signs.0629

These cancel out; and then, I end up with 6x + 12x, which is 18x.0636

Bring down the next term, -4.0644

x goes into 18x 18 times, so up here, I am going to have an 18.0648

18 times x is 18x; 18 times 2 is 36; now I subtract.0659

18x - 18x: that cancels; this negative applies to here as well, so I have -4 minus 36 to get -40.0670

So, the remainder equals -40; and so, my answer is 3x² - 6x + 18, with a remainder of -40.0679

Again, the key point here was to make sure that, when you notice that there is a missing term0693

(the x² term is missing) to represent that using the coefficient of 0.0699

OK, there is another way to divide that is much faster and easier than long division.0704

But it is only applicable to dividing a polynomial by a binomial.0709

So, when you have those cases, synthetic division is an excellent method to use.0715

Now, there is another restriction, and that is that the divisor (which, in this case, is the binomial) must be in the form x - r, where r is a constant.0721

So, the divisor has to be in this form; if it is not in that form, you have to get it into that form.0731

OK, let's take a look at an example where the divisor is already in this form, to keep it simple for now.0738

3x³ - 2x² + 7x + 4; if I am asked to divide that by x - 2,0747

and it is in this form of x minus a constant, I could use long division.0757

But you will see that this is a much faster method.0763

So, first, draw a symbol like this; now, you put the r term right here (in this case, it is 2).0766

And it says - 2, but you actually use the opposite sign; if this were to say x + 3, then I would put a negative here.0777

So, this is x - 2; I use the opposite sign, so I am going to put a 2 right here.0785

Now, what I write in here are the coefficients of the dividend: 3, -2, 7, and 4.0792

And just as with long division, if there is a missing term (let's say my x² term was missing),0805

I would have represented that here with a coefficient of 0--the same idea as with long division.0810

OK, so here I have the constant; here I just have the coefficients from up here.0815

The first step is to bring down that first coefficient and just put it here.0821

OK, the second step is to multiply 2 times 3; I am going to multiply 2 times 3, and my result will be 6.0826

All I am going to do is add this to -2; -2 + 6--that is going to give me 4.0839

Then, I repeat that process: I am going to multiply -2 by 4 to get 8, and I am going to add: 8 + 7 is 15.0849

I am going to repeat that again: -2 times 15 is going to give me 30; and I am going to add again: 4 + 30 is 34.0865

Now, what do these numbers represent? Well, they represent the quotient and the remainder.0876

And to write them out, what you are going to do is look at the dividend.0888

And the quotient is going to be a degree one less than the degree of the dividend.0892

So, the degree of the dividend was 3; so the degree of the quotient is going to be 2.0897

I am going to write this out as 3x² + 4x + 15, with a remainder of 34.0901

And you can see how much quicker and easier this is than long division.0915

Again, write the constant from the divisor here; write the coefficients here, making sure to use 00919

if you have a missing term, using 0 as the coefficient; bring down the first term.0924

Then, multiply the divisor constant by this number and get a product.0929

Add that to the next column; then multiply 2 times this number, 4, to get 8; add to this column, 15.0937

2 times 15 is 30; add to this column to get 34, which is the remainder.0946

OK, as I mentioned, the divisor must be in the form x - r; if the coefficient of x is not 1,0954

you need to rewrite it so that the coefficient is 1 in order to use this method.0962

For example, if I was given 3x³ + 12x² - 15x + 6, divided by 3x - 9,0968

I can see that this coefficient is not 1; in order to make it 1, I need to divide each term0981

in both the dividend and the divisor by this coefficient.0987

So, I am going to divide each term by 3; and that is going to give me x³ + 4x² - 15x + 2, divided by x - 3.0992

Now, I can go about synthetic division in the usual way.1007

And as you will see here, I got lucky, and dividing things by 3 still kept everything as integers.1010

However, it is very possible that, when you divide by this number, you are going to end up with some fractions.1016

So, that is a drawback, and it makes it more complicated; but it is necessary to do this in order to use synthetic division.1022

OK, first we are going to practice dividing a polynomial by a monomial.1030

And you will recall that the technique is to divide each term in the polynomial by the monomial.1035

So, I am rewriting this, dividing each term by the monomial.1042

So, 20 divided by 4 gives you 5; the x's cancel out; and then, y⁴ divided by y²...this is 4 - 2...is y², 5y².1059

12 divided by 4 (and there is a negative sign in front of that) gives me 3.1073

x to the first minus 2...this is going to give me x to the -1, and I am going to have to take care of that in a minute to put it in my final form.1080

But for right now, we will leave it like that.1091

y³ divided by y²...3 - 2 just gives me y, plus...I'll just leave this as 5/4...1093

x³ divided by x² is x; and then here, I have y, which is really y¹,1103

divided by y², is 1 - 2; and this is y^-1.1113

Now, remember: you want to simplify things, and they are not in simplest form if you have negative powers.1116

So, recalling that rule that a^-n equals 1/aⁿ, I can simplify by moving this x^-1 to the denominator.1122

And I can do the same thing here: x stays up here; y moves to the denominator.1136

And this is my answer: dividing a polynomial by a monomial, and then simplifying.1143

OK, this next problem, dividing a polynomial by a monomial, could be done by synthetic division, because it is in this form x - r.1150

But just to get a little more practice on long division, let's do long division on this one.1159

x goes into 3x³...this becomes 2, so this is 3x²; so it goes into it 3x² times.1172

So, I divided; and now I am going to multiply: 3x² times x is 3x³;1187

3x² times -3 is -9x².1193

Now, I am subtracting; and that is the same as adding the opposite, so this becomes negative; this becomes positive.1197

These cancel out, so I have -2x² + 9x² is 7x².1205

Bring down the next term, and then divide again.1213

I have 7x² divided by x equals 7x; 7x times x; 7x, -3 gives me -21x.1217

Subtract (which means I am adding the opposite): this becomes a negative; this one becomes a positive.1233

This cancels; 4x + 21x is 25x; now, divide again: 25x divided by x...the x's cancel; that is 25.1240

So, I am going to have 25 up here.1255

Multiply 25 times x; that is 25x; and bring down this next term; I have -8 there.1257

So, 25 times -3 is going to give me -75; I am going to subtract, which is adding the opposite.1269

So, that is -25 and positive 75; these cancel; -8 + 75 is 67, and that is my remainder.1280

So, the answer is 3x² + 7x + 25, with a remainder of 67.1289

That was long division; now, this next example specifies to divide using synthetic division.1298

And I am checking, and the divisor is in the form x - r; so I can go ahead and do it without any further manipulation of the expression.1307

Recall: in synthetic division, set it up as follows: you are going to put the constant here, with the opposite sign.1321

This is -4; I am going to make it a 4.1329

Now, before I proceed with putting the coefficients in, it is very important to check for missing coefficients.1332

And I am looking, and I have y⁴; I do not have y³; I have y², y, and a constant.1339

So, I am going to rewrite this using 0 for my coefficient with the missing term.1347

I have a missing y³ term, and I am going to rewrite it; and the coefficient is 0 (to use that as a placeholder).1358

So, this is really what I am going to do; OK.1369

Now, I am going to use these coefficients: I have 5, 0, -3, 2, and then my constant, -8.1372

The first step is just to bring down that first term, 5; the second step is to multiply.1386

4 times 5 is 20; after multiplying, add: 0 and 20 is 20; multiply again: 4 times 20 is 80; -3 and 80 is 77.1392

4 times 77--if you calculated that out, you would find that it is 308; 308 + 2 is 310.1417

4 times 310--if you work that out, you will find that it is 1240; 1240 and -8 is going to give you 1232.1431

So again, write the coefficients here, being careful to realize you have a missing term,1444

so you need to represent the coefficient for that missing term as 0.1448

Bring down the first term, and then multiply and add that product to the next column.1452

Find the sum; multiply; add the product to the next column; find the sum; and continue on.1460

Now, for my quotient: this is the quotient; this is going to be the remainder.1467

The quotient is going to have a degree one less than the degree of the dividend;1479

so the degree of the dividend is 4, so the degree of the quotient is going to be 3;1483

so I am going to write this out as 5y³ + 20y² + 77y + 310;1488

the remainder is 1232, which is a big remainder, but this is correct.1498

In this example, we also are asked to do synthetic division; and I check, and I have a couple of things going on here.1510

I have a missing coefficient, and this is not in the correct form of x - r.1518

So, dealing with the missing term--I have a missing term, which I am going to represent with a coefficient of 0.1526

So, first addressing the missing term: I am missing a z² term; I have 6z⁴ - 8z³.1533

Since I have no z² term, I am going to use the coefficient of 0 for that: 0z² - 4z + 8.1543

OK, that is taken care of; now, the other problem I have is that this is not in the correct form.1552

In order to have this z have a coefficient of 1, I need to divide all the terms in the divisor and the dividend by 2, so I am going to do that.1558

So, I divide by 2--divide each term by 2 to get the form...I am going to say z instead of x...the form z - r.1569

OK, so 6z⁴: dividing that by 2 is going to give me 3z⁴.1590

8z³ divided by 2 is -4z³; 0z² divided by 2 is 0z².1596

-4z divided by 2 is -2z; 8 divided by 2 is 4; so far, pretty good.1611

2z divided by 2 gives me z, which is what I wanted; now, here, when I divide -1 by 2, I am going to get a fraction.1620

And it makes it more difficult to work with, but you can still do the synthetic division.1628

Now, I am ready to set this up: here, I am going to take -1/2; I am going to take the opposite sign and write it here; that is 1/2.1634

Then, I am going to put the coefficients here--my new coefficients, after dividing: 3, -4, 0, -2, and 4.1644

OK, bring down the 3 and multiply 3 by 1/2; 3 times 1/2 is just 3/2.1659

1/2 times 3 is 3/2; now, you might need to work out the arithmetic on the side, and that is fine.1670

I have -4 and 3/2; -4 is equal to -8/2; I want to get a common denominator.1677

Adding that to 3/2 is going to give me -5/2, so this is -5/2.1685

Now, multiplying 1/2 by -5/2 is going to give me -5/4; this times this is going to give me -5/4.1695

0 and -5/4 is just -5/4; now, I have to multiply 1/2 by -5/4, and this is going to give me -5/8; I have -5/8 here.1709

I want to get a common denominator; and so, I am going to do -2 times 8, which is going to give me -16/8.1724

So, this is -16/8 + -5/8; and it is just -16 and -5, is -21/8; I am combining these to get -21/8.1737

OK, now 1/2 times -21/8 equals -21/16; this times this is -21/16.1753

I have to add that to 4; so I want to get a common denominator: 4 times 16 is going to give me 64, so 4 = 64/16.1771

So, I want to take 64/16 - 21/16; and that is going to be...64 - 21 is 43/16.1788

OK, this was kind of messy to do; but synthetic division does work, and will end up giving you the correct answer.1801

So, what you need to look at here is that this has a degree of 4, so this is actually going to be a degree of 3.1809

So, rewriting this up here, my quotient is going to be 3z³ - 5/2z² - 5/4z - 21/8.1816

And I have a remainder of 43/16.1841

So, two things to notice: we had a missing term here--the z² term was missing--1846

so I had to use a coefficient of 0; and the second is that this was not in this form.1852

So, I had to divide every term in the numerator and the denominator, the divisor and the dividend, by 2, and then proceed as usual with synthetic division.1856

That concludes this lesson on dividing polynomials for Educator.com.1865

And I will see you again soon!1871

Related Books

Algebra and Trigonometry: Structure and Method, Book 2

Author: McDougal Littell

ISBN: 395977258

Publisher: McDougal Littell

Year: 1999

This classic textbook does a great job of explaining concepts and includes lots of in-depth, worked out examples in every section as well as practice problems.

Related Books

Name	Description	Link
BookRenter.com	BookRenter.com is simply the most reliable online textbook rental service.	Visit BookRenter.com
PhysicsForums.com Homework Help	Physics Forums is a scientific community for students looking for math & science help.	Visit PhysicsForums.com Homework Help

Dr. Carleen Eaton

Dividing Polynomials

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro