Difference between revisions of "Dividing Polynomials"
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− | ==Resources | + | ==Polynomial Long Division== |
+ | Suppose we would like to divide one polynomial by another. The procedure is similar to long division of numbers and is illustrated in the following example: | ||
+ | |||
+ | ===Example 1=== | ||
+ | Divide <math>x^2-2x-15</math> (the dividend or numerator) by <math>x+3</math> (the divisor or denominator)}} | ||
+ | Similar to long division of numbers, we set up our problem as follows: | ||
+ | :<math>\begin{array}{rl}\\ | ||
+ | x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} | ||
+ | \hline | ||
+ | \,x^2-2x-15 | ||
+ | \end{array}\end{array}</math> | ||
+ | |||
+ | First we have to answer the question, how many times does <math>x+3</math> go into <math>x^2</math>? To find out, divide the leading term of the dividend by leading term of the divisor. So it goes in <math>x</math> times. We record this above the leading term of the dividend: | ||
+ | :<math>\begin{array}{rl}&~~\,x\\ | ||
+ | x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} | ||
+ | \hline | ||
+ | \,x^2-2x-15 | ||
+ | \end{array}\\ | ||
+ | \end{array}</math> | ||
+ | |||
+ | , and we multiply <math>x+3</math> by <math>x</math> and write this below the dividend as follows: | ||
+ | :<math>\begin{array}{rl}&~~\,x\\ | ||
+ | x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} | ||
+ | \hline | ||
+ | \,x^2-2x-15 | ||
+ | \end{array}\\ | ||
+ | &\!\!\!\!-\underline{(x^2+3x)~~~}\\ | ||
+ | \end{array}</math> | ||
+ | |||
+ | Now we perform the subtraction, bringing down any terms in the dividend that aren't matched in our subtrahend: | ||
+ | :<math>\begin{array}{rl}&~~\,x\\ | ||
+ | x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} | ||
+ | \hline | ||
+ | \,x^2-2x-15 | ||
+ | \end{array}\\ | ||
+ | &\!\!\!\!-\underline{(x^2+3x)~~~}\\ | ||
+ | &\!\!\!\!~~~~~~-5x-15~~~\\ | ||
+ | \end{array}</math> | ||
+ | |||
+ | Now we repeat, treating the bottom line as our new dividend: | ||
+ | :<math>\begin{array}{rl}&~~\,x-5\\ | ||
+ | x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} | ||
+ | \hline | ||
+ | \,x^2-2x-15 | ||
+ | \end{array}\\ | ||
+ | &\!\!\!\!-\underline{(x^2+3x)~~~}\\ | ||
+ | &\!\!\!\!~~~~~~-5x-15~~~\\ | ||
+ | &\!\!\!\!~~~-\underline{(-5x-15)~~~}\\ | ||
+ | &\!\!\!\!~~~~~~~~~~~~~~~~~~~0~~~\\ | ||
+ | \end{array}</math> | ||
+ | |||
+ | In this case we have no remainder. | ||
+ | |||
+ | ===Example 2=== | ||
+ | What about a non-divisible polynomials like this? | ||
+ | |||
+ | :<math>(3x^2 + 3x - 4) / (x - 4)</math> | ||
+ | |||
+ | {| class="wikitable" cellpadding="3" style="text-align: left;" | ||
+ | !colspan="4"|Long division method | ||
+ | |- | ||
+ | |1 | ||
+ | |style="padding:10px;"|We first consider the highest-degree terms from both the dividend and divisor, the result is the first term of our quotient. | ||
+ | |style="text-align: center;"|<math>(3x^2) / (x) = 3x</math> | ||
+ | |rowspan="7"|<math>\begin{array}{r|ccc} | ||
+ | x - 4 & 3x^2 & 3x & -4 \\ | ||
+ | \hline | ||
+ | \hline | ||
+ | & 3x^2 & 3x & \\ | ||
+ | 3x(x - 4) & 3x^2 & -12x & \\ | ||
+ | \hline | ||
+ | & & 15x & -4 \\ | ||
+ | 15(x - 4) & & 15x & -60 \\ | ||
+ | \hline | ||
+ | & & & 56 \\ | ||
+ | \end{array}</math> | ||
+ | |- | ||
+ | |2 | ||
+ | |style="padding:10px;"|Then we multiply this by our divisor. | ||
+ | |style="text-align: center;"|<math>(3x) \times (x - 4) = 3x^2 - 12x</math> | ||
+ | |- | ||
+ | |3 | ||
+ | |style="padding:10px;"|And subtract the result from our dividend. | ||
+ | |style="text-align: center;"|<math>(3x^2 + 3x - 4) - (3x^2 - 12x) = 15x - 4</math> | ||
+ | |- | ||
+ | |4 | ||
+ | |style="padding:10px;"|Now once again with the highest-degree terms of the remaining polynomial, and we got the second term of our quotient. | ||
+ | |style="text-align: center;"|<math>(15x) / (x) = 15</math> | ||
+ | |- | ||
+ | |5 | ||
+ | |style="padding:10px;"|Multiplying... | ||
+ | |style="text-align: center;"|<math>(15) \times (x - 4) = 15x - 60</math> | ||
+ | |- | ||
+ | |6 | ||
+ | |style="padding:10px;"|Subtracting... | ||
+ | |style="text-align: center;"|<math>(15x - 4) - (15x - 60) = 56</math> | ||
+ | |- | ||
+ | |7 | ||
+ | |style="padding:10px;"|We are left with a constant term - our remainder: | ||
+ | |style="text-align: center;"|<math>\begin{array}{lcr} Q(x) = 3x + 15 & & R = 56 \end{array}</math> | ||
+ | |} | ||
+ | |||
+ | So finally: | ||
+ | :<math>(3x^2 + 3x - 4) = (3x + 15) \times (x - 4) + 56</math> | ||
+ | |||
+ | ==Resources== | ||
===Dividing Polynomials With Long Division=== | ===Dividing Polynomials With Long Division=== | ||
* [https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:quad-div-by-linear/v/polynomial-division Intro to Polynomial Long Division], Khan Academy | * [https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:quad-div-by-linear/v/polynomial-division Intro to Polynomial Long Division], Khan Academy | ||
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* [https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut37_syndiv.htm Synthetic Division], WTAMU VirtualMathLab | * [https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut37_syndiv.htm Synthetic Division], WTAMU VirtualMathLab | ||
* [https://www.youtube.com/watch?v=bZoMz1Cy1T4 Synthetic Division Example], patrickJMT | * [https://www.youtube.com/watch?v=bZoMz1Cy1T4 Synthetic Division Example], patrickJMT | ||
+ | |||
+ | == Licensing == | ||
+ | Content obtained and/or adapted from: | ||
+ | * [https://en.wikibooks.org/wiki/Calculus/Algebra Algebra, Wikibooks: Calculus] under a CC BY-SA license | ||
+ | * [https://en.wikibooks.org/wiki/High_School_Mathematics_Extensions/Supplementary/Polynomial_Division Polynomial Division, Wikibooks: High School Mathematics Extensions] under a CC BY-SA license |
Latest revision as of 13:17, 21 October 2021
Contents
Polynomial Long Division
Suppose we would like to divide one polynomial by another. The procedure is similar to long division of numbers and is illustrated in the following example:
Example 1
Divide (the dividend or numerator) by (the divisor or denominator)}} Similar to long division of numbers, we set up our problem as follows:
First we have to answer the question, how many times does go into ? To find out, divide the leading term of the dividend by leading term of the divisor. So it goes in times. We record this above the leading term of the dividend:
, and we multiply by and write this below the dividend as follows:
Now we perform the subtraction, bringing down any terms in the dividend that aren't matched in our subtrahend:
Now we repeat, treating the bottom line as our new dividend:
In this case we have no remainder.
Example 2
What about a non-divisible polynomials like this?
Long division method | |||
---|---|---|---|
1 | We first consider the highest-degree terms from both the dividend and divisor, the result is the first term of our quotient. | ||
2 | Then we multiply this by our divisor. | ||
3 | And subtract the result from our dividend. | ||
4 | Now once again with the highest-degree terms of the remaining polynomial, and we got the second term of our quotient. | ||
5 | Multiplying... | ||
6 | Subtracting... | ||
7 | We are left with a constant term - our remainder: |
So finally:
Resources
Dividing Polynomials With Long Division
- Intro to Polynomial Long Division, Khan Academy
- Polynomial Long Division, Lumen Learning
- Polynomial Long Division, Purple Math
- Long Division With Polynomials, The Organic Chemistry Tutor
- Long Division of Polynomials, patrickJMT
Dividing Polynomials with Synthetic Division
- Synthetic Division of Polynomials, Khan Academy
- Synthetic Division, Purple Math
- Synthetic Division, WTAMU VirtualMathLab
- Synthetic Division Example, patrickJMT
Licensing
Content obtained and/or adapted from:
- Algebra, Wikibooks: Calculus under a CC BY-SA license
- Polynomial Division, Wikibooks: High School Mathematics Extensions under a CC BY-SA license