1. Pre Calc Lesson 2.2 Synthetic Division ‘Remainder’ and ‘Factor’ Theorems Review Long Division: 5365 ÷ 27 Now review ‘long division’ of polynomials: (2x 4 - 15x 2 – 10x + 5) ÷ (x – 3)

2. Now take the same problem and do the problem again only with ‘synthetic division’ this time. (2x 4 – 15x 2 – 10x + 5) ÷ (x – 3) 2 0 - 15 - 10 5 6 18 9 -3 x = 3 ) 2 6 3 -1 2 Therefore our result is the same: 2x 3 + 6x 2 + 3x – 1; R = 2 Or P(3) = 2 ‘The Remainder Theorem’: If a polynomial P(x) is divided by (x – a), the remainder is always P(a).

3. Example 1: Divide P(x) = x 3 + 5x 2 + 5x – 2 by (x + 2) ***(1 st —take (x + 2) = 0  solve for ‘x’  x = - 2 Now pull out the coefficients of P(x)   1 5 5 -2 -2 -6 2 x = - 2) 1 3 -1 0 Therefore  Quotient is: 1x 2 + 3x – 1 Remainder is : 0 Which leads us into the ‘factor theorem’ For a polynomial P(x), x – a is a factor iff  (if and only if) P(a) = 0

4. Example 2: If P(x) = 2x 4 + 5x 3 - 8x 2 – 17x – 6, determine whether each is a factor of P(x). a)x – 1b) x – 2 a) (since we are checking x = 1, the simplest method here is direct substitution. When substituting ‘1’ in for ‘x’, all you need to do is add up the coefficients as they stand! 2 + 5 + -8 + -17 + -6 = -24 Here P(1) = - 24 b) 2 5 -8 -17 -6 4 18 20 6 2) 2 9 10 3 0 Here P(2) = 0 so ‘x – 2’ is a factor!!!!

5. Example 3: If x = 2 is a root of 2x 3 + 5x 2 – 23x + 10 = 0 Find the remaining roots! 1 st – Break it down synthetically 2 + 5 - 23 + 10 4 18 - 10 2) 2 9 - 5 0 Therefore we have 1 st of all re-inforced that x = 2 is a root. Thus (x – 2) is a ‘linear factor’ but that means (2x 2 + 9x – 5) would be a ‘quadratic factor’. The remaining ‘two’ roots would have to come from this ‘quadratic’ factor. So now all we need to do is to find the other ‘two’ roots from this quadratic factor: (2x 2 + 9x – 5) (2x – 1)(x + 5) 2x – 1 = 0 ; x + 5 = 0 2x = 1 ; x = - 5 Therefore all 3 roots are: x = 2; x = ½ ; x = - 5 Hw: pg 61 #1-25 odd