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6.8 Synthetic Division

6.8 Synthetic Division. Polynomial Division, Factors, and Remainders. In this section, we will look at two methods to divide polynomials: long division (similar to arithmetic long division) synthetic division (a quicker, short-hand method). Rewrite in long division form. 2x. + 7.

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6.8 Synthetic Division

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  1. 6.8 Synthetic Division

  2. Polynomial Division, Factors, and Remainders In this section, we will look at two methods to divide polynomials: long division (similar to arithmetic long division) synthetic division (a quicker, short-hand method)

  3. Rewrite in long division form... 2x + 7 divisor Think, how many times does x go into 2x2 ? dividend 2x2 – 4x Example: Divide (2x2 + 3x – 4) ÷ (x – 2) Multiply by the divisor. 7x – 4 Subtract. 7x – 14 Think, how many times does x go into 7x ? 10 remainder (x – 2) 2x2 + 3x – 4 Write the result like this...

  4. p2 + p + 1 p3 – p2 Example: Divide (p3 – 6) ÷ (p – 1) Be sure to add “place-holders” for missing terms... p2 + 0p p2 – p p – 6 (p – 1) p3 + 0p2 + 0p – 6 p – 1 –5

  5. Synthetic division can be used when the divisor is in the form (x – k). k Bring down the first coefficient. 8 4 -16 Example: Use synthetic division for the following: (2x3– 7x2– 8x + 16) ÷ (x – 4) First, write down the coefficients in descending order, and k of the divisor in the form x – k : 4 2 –7 –8 16 These are the coefficients of the quotient (and the remainder) 2 1 –4 0 Multiply this by k Add the column. Repeat the process.

  6. place-holder Notice that k is –1 since synthetic division works for divisors in the form (x – k). Example: Divide (5x3 + x2 – 7) ÷ (x + 1) –5 4 –4 5 –4 4 –11 –1 5 1 0 –7

  7. Now, let f(x) = 2x4 + x3 – 3x2 – 5 What is f(2)? f(2) = 2(2)4 + (2)3 – 3(2)2 – 5 f(2) = 2(16) + 8 – 3(4) – 5 f(2) = 32 + 8 – 12 – 5 f(2) = 23 This is the same as the remainder when f(x) is divided by (x – 2): 28 2 2 1 –3 0 –5 4 10 14 2 5 7 14 23 f(2) = 23

  8. Example: Use synthetic substitution to find f(4) if f(x) = x4 – 6x3 + 8x2 + 5x + 13 4 –8 0 20 1 –2 0 5 33 4 1 –6 8 5 13 f(4) = 33

  9. You can also use synthetic division to find factors of a polynomial... Example: Given that (x + 2) is a factor of P(x), factor the polynomial P(x) = x3 – 13x2 + 24x + 108 We can use synthetic division to find the other factors... Since P(–2) = 0, then (x+2) is a factor of P(x) –2 30 –108 1 –15 54 0 –2 1 –13 24 108 This means that you can write x3 – 13x2 + 24x + 108 = (x + 2)(x2 – 15x + 54) Factor this... = (x + 2)(x – 9)(x – 6) The complete factorization is:(x + 2)(x – 9)(x – 6)

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