1. Goal: to divide polynomials using LONG and SYNTHETIC methods.
11.3 Dividing Polynomials
Goal: to divide polynomials using LONG and SYNTHETIC methods.
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2. Remember:
Long Division: without a calculator we can divide the following 4 9 - 92 21 2 -
207
Keep going and Finish!! 5
Answer: 49251
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3. When Dividing Polynomials:
This long division technique can also be used to divide polynomials
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4. Divide: - x² +3x -18 / (x-3) - x + 6 R 0 x - 3 x2 + 3x - 18 ( x2 - 3x
- 18 - ( 6x
- 18 )
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5. Divide: - x² +3x -18 / (x-3) - -16 2x - 3 R : -16 x - 8 2x2 - 19x + 8
+ 8 - (
- 3x
+ 24 )
-16
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6. Divide: - - (5x³ -13x² +10x -8) / (x-2) - 5x² - 3x + 4 R: 0 x - 2
+ 4
R: 0
x - 2
5x³ x² x - 8 - (
5x³
- 10x² )
-3x²
+ 10x - (
-3x²
+ 6x ) 4x
- 8 - ( 4x
- 8 )
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7. So in other words… 5x³ -13x² +10x - 8 x-2 OR 5x³ -13x² +10x - 8=
x-2 OR
5x³ -13x² +10x - 8 =
(5x² -3x + 4)
(x-2)
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8. When Dividing Polynomials:
SYNTHETIC division uses the coefficients of each term. Make sure you include the zeros for the powers that are not present.
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9. Synthetic division (5x³ -13x² +10x -8) / (x-2)
Opposite
of number in
divisor
x³ x² x x0
Use the coefficients of each term. 2 10 -6 8 5 -3 4
Start with one power less than the original equations.
5x² -3x + 4
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10. Synthetic division (2w³ +3w -15) / (w-1)
Opposite
of number in
divisor
w³ w² w w0
Use the coefficients of each term. 1 2 2 5
Notice: -10 is the remainder. 2 2 5
-10
Start with one power less than the original equations.
2w² w /(w-1)
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11. Synthetic division: (x³ -13x +12 ) / (x+4) x2 -4x + 3 1 0 -13 12
Use the coefficients of each term. Include the ones not present  0x2
Opposite
of number in
divisor
x³ x² x x0 -4 -4
+16
-12 1 -4 3
Start with one power less than the original equations.
x x + 3
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12. A Couple of Notes
Use synthetic division when the coefficient in front of x is 1
(x- 2) (2x-3)
YES NO
To test so see if a binomial is a factor, you want to see if you get a remainder of zero. If yes, it is a factor. If you get a remainder, the answer is no.
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13. In this case, x-3 is not a factor because there was a remainder of 6
+ 6
R 6
x - 3
2x² -19x - (
x² - 3x ) 6x
-12 - ( 6x ) 6
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14. From this example, x-8 IS a factor because the remainder is zero
(2x² -19x + 8) / (x-8)
2x - 1
R: 0
x - 8
2x² x + 8
-( 2x² x )
- x
- ( x )
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15. In this case, x-3 is not a factor because there was a remainder of 6
+ 6
R 6
x - 3
x² + 3x - (
x² - 3x ) 6x
-12 - ( 6x ) 6
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