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5.5 Synthetic & Long Division I.. Synthetic Substitution and Remainders. A) When you use synthetic substitution, the last number you get after performing.

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Presentation on theme: "5.5 Synthetic & Long Division I.. Synthetic Substitution and Remainders. A) When you use synthetic substitution, the last number you get after performing."— Presentation transcript:

1 5.5 Synthetic & Long Division I.. Synthetic Substitution and Remainders. A) When you use synthetic substitution, the last number you get after performing the math is the “y” coordinate for the specific value for “x” you tested. 1) This “y” value is also called the remainder. B) If the remainder is zero, then your “x” value is a solution to the polynomial (it is a root, an x-intercept, an answer, etc.). 1) You can factor out this value using synthetic division.

2 5.5 Synthetic & Long Division II.. Synthetic Division and Remainders. A) Writing “x” values as factors. Change its sign “(x – h)”. B) To factor (x – h) from the original polynomial, change the sign of the “h” term and use synthetic substitution. 1) This is called Synthetic Division. C) The #s below the synthetic division bar are the coefficients of the new smaller polynomial you get after factoring by (x – h) 1) The smaller poly will have a degree (exp) one less than the original poly. 2) Try to factor the new smaller polynomial, using any factoring method until you cannot factor anymore. a) All parenthesis now say (x – h) or if h is a fraction b / a then move the bottom number in front of x (ax – b).

3 5.5 Synthetic & Long Division II.. Synthetic Division and Remainders. Examples: (3x 3 – 4x 2 – 28x – 16) / (x + 2) – 2 3 – 4 – 28 – 16 remainder = 0 – 6 20 16 so x + 2 is a perfect factor. 3 – 10 – 8 0 decrease the degree by 1 to get the new smaller poly. (3x 3 – 4x 2 – 28x – 16) / (x + 2) = 3x 2 – 10x – 8 – Or – (x + 2) (3x 2 – 10x – 8) = 3x 3 – 4x 2 – 28x – 16 Now factor using Reverse FOIL (3 terms)

4 5.5 Synthetic & Long Division III.. Finding possible Synthetic Division values for polys. A) Since polys. must have integer coefficients, we need to find all the possible values that might work. 1) ± (the factors of the last term) / (factors of 1st term) a) “The last shall come first, and the first shall be last.” 2) Any fractional “x” factor can be written as … a) If (x – b / a ) is a factor, then it’s equal to (ax – b). B) Shortcut to finding possible values to test. Use the graph on a graphing calculator to find which of the possible values look like x-intercepts. 1) Test these values using synthetic division.

5 5.5 Synthetic & Long Division IV.. Writing Factors of Synthetic Division with remainders ≠ 0. A) If we factor and the remainder is not zero, we call this end number the remainder. 1) We can factor this poly, and its remainder, like this… a) (x – h) (one degree smaller poly + remainder) (x – h) Example: Use synthetic division on (x 3 – x 2 + 4x – 10) ÷ (x + 2) – 2 1 –1 4 –10 –2 6 –20 1 –3 10 –30 has a remainder of – 30. So its factors are … (x + 2)(x 2 – 3x + 10 + –30 / (x + 2) )

6 5.5 Synthetic & Long Division V.. Factoring fully using synthetic division. A) When you factor using synthetic division and get a remainder of 0, the values below the division line are the coefficients of the one degree smaller poly. 1) Write the factor (x – h) and the new one degree smaller poly and try to factor the one degree smaller poly using any factoring method. (Grouping, Reverse FOIL, etc.) Homework: Synthetic Division


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