Pre Calc Lesson 2.2 Synthetic Division ‘Remainder’ and ‘Factor’ Theorems

Review Long Division:

5365 ÷ 27

Now review ‘long division’ of polynomials:

(2x4 - 15x2 – 10x + 5) ÷ (x – 3)

Now take the same problem and do the problem againonly with ‘synthetic division’ this time.

(2x4 – 15x2 – 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: 2x3 + 6x2 + 3x – 1; R = 2

Or P(3) = 2

‘The Remainder Theorem’: If a polynomial P(x) is divided by (x – a), the remainderis always P(a).

Example 1: Divide P(x) = x3 + 5x2 + 5x – 2 by (x + 2)***(1st—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: 1x2 + 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

Example 2: If P(x) = 2x4 + 5x3 - 8x2 – 17x – 6 , determinewhether each is a factor of P(x).

a) x – 1 b) x – 2

a) (since we are checking x = 1, the simplest method here is direct substitution. When substituting ‘1’ in for ‘x’, allyou 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!!!!

Example 3: If x = 2 is a root of 2x3 + 5x2 – 23x + 10 = 0Find the remaining roots!1st – Break it down synthetically 2 + 5 - 23 + 10 4 18 - 10 2) 2 9 - 5 0

Therefore we have 1st of all re-inforced that x = 2 is a root.Thus (x – 2) is a ‘linear factor’ but that means (2x2 + 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: (2x2 + 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