 # 6.8 Synthetic Division. Polynomial Division, Factors, and Remainders In this section, we will look at two methods to divide polynomials: long division.

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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)

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

Example: Divide (p 3 – 6) ÷ (p – 1) (p – 1) p 3 + 0p 2 + 0p – 6 Be sure to add “place-holders” for missing terms... p2p2 p 3 – p 2 p 2 + 0p + p p 2 – p p – 6 + 1 p – 1 –5

Synthetic division can be used when the divisor is in the form (x – k). Example: Use synthetic division for the following: (2x 3 – 7x 2 – 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 k 2 Bring down the first coefficient. 8 Multiply this by k 1 Add the column. 4 –4 -16 0 These are the coefficients of the quotient (and the remainder) Repeat the process.

Example: Divide (5x 3 + x 2 – 7) ÷ (x + 1) –1 5 1 0 –7 Notice that k is –1 since synthetic division works for divisors in the form (x – k). place-holder 5 –5 –4 4 4 –11

2 2 1 –3 0 –5 f(2) = 23 2 4 5 10 7 14 Now, let f(x) = 2x 4 + x 3 – 3x 2 – 5 28 23 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):

4 1 –6 8 5 13 f(4) = 33 1 4 –2 –8 0 0 5 Example: Use synthetic substitution to find f(4) if f(x) = x 4 – 6x 3 + 8x 2 + 5x + 13 20 33

–2 1 –13 24 108 This means that you can write x 3 – 13x 2 + 24x + 108 = (x + 2)(x 2 – 15x + 54) 1 –2 –15 30 54 –108 0 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) = x 3 – 13x 2 + 24x + 108 We can use synthetic division to find the other factors... Factor this... = (x + 2)(x – 9)(x – 6) The complete factorization is: (x + 2)(x – 9)(x – 6) Since P(–2) = 0, then (x+2) is a factor of P(x)

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