Factors of PoLynomials

1. Factors of PoLynomials

2. What is to be learned • How to find factors of poLynomials of degree 3 or more. • How we use this to solve poLynomial equations

3. Factors Factors of 16?  1, 2, 4, 8, 16 A divisor will be a factor if there is no remainder

4. ÷ (x + 2) – 2 Time for The Big L x2 + 5x + 6 1 5 6 D -6 -2 Remainder = 0 3 1 0 Q (x + 2) is a Factor x2 + 5x + 6 = (x + 3) (x + 2) + 0 D Q R

5. ÷ (x + 1) -1 x3 - x2 + 2x - 2 1 -1 2 -2 2 -1 -4 Remainder = -6 -2 1 4 -6  (x + 1) is not a Factor If at first you don’t succeed try, try again…..

6. ÷ (x - 1) 1 x3 - x2 + 2x - 2 1 -1 2 -2 D 0 1 2 Remainder = 0 0 1 2 0 Q (x - 1) is a Factor May be able to factorise this bit x3 - x2 + 2x - 2= (x2 + 2) (x – 1) D Q

7. Factorisation of PoLynomialsusing The Big L (also known as the remainder/factor theorem) A divisor is a factor if Using Big L  remainder = 0 For poLynomials of degree 3, or over there is no remainder

8. Solutions are factor of this Try ÷ (x - 1) 1 x3 + 2x2 – x – 2 1 2 -1 -2 D 3 1 2 Remainder = 0 3 1 2 0 Q (x - 1) is a Factor factorise this bit x3 + 2x2 – x – 2= (x2 +3x + 2) (x – 1) D Q

9. Solutions are factor of this Try ÷ (x - 1) 1 x3 + 2x2 – x – 2 1 2 -1 -2 D 3 1 2 Remainder = 0 3 1 2 0 Q (x - 1) is a Factor factorise this bit x3 + 2x2 – x – 2= (x2 +3x + 2) (x – 1) =(x – 1)(x + 2)(x + 1)

10. If (x+ 1) is a factor, find p ÷ (x + 1) p -1 7 2x3 + px2 + 2x – 3 2 2 -3 -5 -2 3 Remainder = 0 5 2 -3 0 p = 7

11. Sammy Scientist has invented a machine that produces power (P Kw) The amount of power depends on how many hundred mice (m), he can persuade to run on the big wheel attached to it. Formula is P = 2m3 – 17m2 + 40m. How many mice will it take for the power to be 16Kw? 2m3 – 17m2 + 40m = 16 2m3 – 17m2 + 40m – 16 = 0 PoLynomial Equation send for The Big L

12. Solutions known as roots 2m3 – 17m2 + 40m – 16 = 0 Try ÷ (m - 4) 4 2 -17 40 -16 8 -36 16 Remainder = 0 -9 2 4 0 (m - 4) is a Factor 2m3 – 17m2 + 40m – 16 = 0 m = 4 (2m2 – 9m + 4) = 0 (m – 4) or m = ½ (m – 4)(2m )(m ) = 0 – – 1 4 400 or 50 mice m – 4 = 0 m – 4 = 0 2m – 1 = 0

13. Solutions known as roots 2m3 – 17m2 + 40m – 16 = 0 Try ÷ (m - 4) 4 2 -17 40 -16 8 -36 16 Remainder = 0 -9 2 4 0 (m - 4) is a Factor 2m3 – 17m2 + 40m – 16 = 0 m = 4 (2m2 – 9m + 4) = 0 (m – 4) or m = ½ (m – 4)(2m )(m ) = 0 – – 1 4 400 or 50 mice m – 4 = 0 2m – 1 = 0

14. Solutions known as roots 2m3 – 17m2 + 40m – 16 = 0 Try ÷ (m - 4) 4 2 -17 40 -16 8 -36 16 Remainder = 0 -9 2 4 0 (m - 4) is a Factor 2m3 – 17m2 + 40m – 16 = 0 m = 4 (2m2 – 9m + 4) = 0 (m – 4) or m = ½ (m – 4)(2m )(m ) = 0 – – 1 4 400 or 50 mice m – 4 = 0 2m – 1 = 0

15. Solutions known as roots 2m3 – 17m2 + 40m – 16 = 0 Try ÷ (m - 4) 4 2 -17 40 -16 8 -36 16 Remainder = 0 -9 2 4 0 (m - 4) is a Factor 2m3 – 17m2 + 40m – 16 = 0 m = 4 (2m2 – 9m + 4) = 0 (m – 4) or m = ½ (m – 4)(2m )(m ) = 0 – – 1 4 400 or 50 mice m – 4 = 0 2m – 1 = 0

16. Solving PoLynomial Equations • Make one side = 0 • Find factor using The Big L • Fully factorise • Find solutions (roots)

17.  x3 - 2x2 – x + 2 = 0 1 Ex. Show that 1 is a solution and solve x3 - 2x2 – x = -2 1 -2 -1 2 ÷ (x - 1) -1 1 -2 Remainder = 0 -1 1 -2 0 (x - 1) is a Factor x3 - 2x2 – x + 2= 0 x = 1,x = 2, x = -1 (x2 - x - 2) = 0 (x – 1) (roots) (x – 1)(x – 2)(x + 1) = 0 x-1=0,x-2=0,x+1=0