1. 2.3 Synthetic Substitution OBJ:  To evaluate a polynomial for given values of its variables using synthetic substitution

2. Top P 38 EX :  P ( 2 ) P = 3 x 3 + 10 x 2 – 5x – 4 2| 3 10 -5 -4 ↓_____________ 3 2| 3 10 -5 -4 ↓ 6_________ 3 16 2| 3 10 -5 -4 ↓ 6 32____ 3 16 27 2| 3 10 -5 -4 ↓ 6 32 54 3 16 27 50 3 x 3 + 10 x 2 – 5x – 4 3(2) 3 + 10(2) 2 –5(2)–4 3(8) + 10(4) – 10 – 4 24 + 40 – 10 – 4 = 50 P ( 2 ) = 50

3. P 38 EX 2:  2 x 4 – x 3 + 5x + 3 x = -2 -2| 2 -1 0 5 3 ↓ | 2________________ -2| 2 -1 0 5 3 ↓ -4 | 2 -5___________ -2| 2 -1 0 5 3 ↓ -4 10 | 2 -5 10_______ -2| 2 -1 0 5 3 ↓ -4 10 -20 | 2 -5 10 15 33 x = 3 3| 2 -1 0 5 3 ↓________________ | 2 3| 2 -1 0 5 3 ↓ 6___________ | 2 5 3| 2 -1 0 5 3 ↓ 6 15_______ | 2 5 15 3| 2 -1 0 5 3 ↓ 6 15 45____ 2 5 15 50 153

4. EX 3:  3 x 4 – 2x 2 – 6x + 10 2| 3 0 -2 -6 10 ↓ | 3_______________ 2| 3 0 -2 -6 10 ↓ 6 | 3 6_____________ 2| 3 0 -2 -6 10 ↓ 6 12 | 3 6 10_________ 2| 3 0 -2 -6 10 ↓ 6 12 20 28 | 3 6 10 14 |38| -2| 3 0 -2 -6 10 | ↓________________ 3 -2| 3 0 -2 -6 10 | ↓ -6_____________ 3 -6 -2| 3 0 -2 -6 10 | ↓ -6 12________ 3 -6 10 -2| 3 0 -2 -6 10 | -6 12 -20 52 3 -6 10 -26 |62|

5. DEF:  Remainder Theorem When synthetic division is done to a polynomial for a specific value, the remainder is the same as if the polynomial was evaluated at that value. The remainder when a polynomial is synthetically divided by a # is equal to the value when the polynomial is evaluated with the #.

6. EX 4:  P(x)=x 5 – 3x 4 + 3x 3 – 5x 2 +12 2| 1 -3 +3 -5 0 +12 ↓ | 1________________ 2| 1 -3 +3 -5 0 +12 ↓ 2 | 1 -1_____________ 2| 1 -3 +3 -5 0 +12 ↓ 2 -2 | 1 -1 1__________ 2| 1 -3 +3 -5 0 +12 ↓ 2 -2 2 -6 -12 | 1 -1 1 -3 -6__| 0 |__ -1| 1 -1 1 -3 -6 |0| ↓ 1__________________ -1| 1 -1 1 -3 -6 |0| ↓ -1 1-2_______________ -1| 1 -1 1 -3 -6 |0| ↓ -1 2 1 -2 3___________ -1| 1 -1 1 -3 -6 |0| ↓ -1 2 -3 6 1 -2 3 -6__|0| ____ 1x 3 –2x 2 +3x – 6 = 0

7. DEF:  Factor Theorem When synthetic division is done to a polynomial for a specific value (c) and the remainder is 0, then (x – c) is a factor. If a polynomial is synthetically divided by a # and the remainder is 0, then x– # is a factor.

8. 15.8 Higher-Degree Polynomial Equations OBJ:  To find the zeros of an integral polynomial  To factor an integral polynomial in one variable into first-degree factors  To solve an integral polynomial equation of degree >2

9. DEF:  Integral Polynomial DEF:  Zero of a polynomial DEF:  Integral Zero Theorem A polynomial with all integral coefficients A # that if evaluated in a polynomial would result in a 0 as the remainder The integral zeros of a polynomial are the integral factors of the constant term, called “p”

10. EX 1:  If P(x) = 2x 4 + 5x 3 – 11x 2 – 20x +12 has two first degree factors (x –2) and (x + 3), find the other two. Top P 409 2| 2 5 -11 -20 +12 2________________ 2| 2 5 -11 -20 +12 ↓ 4 2 9_____________ 2| 2 5 -11 -20 +12 ↓ 4 18 2 9 7 _________ 2| 2 5 -11 -20 +12 ↓ 4 18 14 2 9 7 -6____ 2| 2 5 -11 -20 +12 ↓ 4 18 14 -12 2 9 7 -6 |0| -3| 2 9 7 -6 |0| ↓ 2_________________ -3| 2 9 7 -6 ↓ -6 2 3______________ -3| 2 9 7 -6 ↓ -6 -9 2 3 -2 |0|____ 2x 2 + 3x – 2 (2x – 1)(x + 2)

11. P 410 HW 2 P 411 (1-19 odd) EX 1: P(x) =x 4 – x 3 – 8x 2 + 2x +12 ± 1, 2, 3, 4, 6, 12 Use calculator table to find the zeros 3| 1 -1 -8 +2 +12 ↓________________ 1________________ 3| 1 -1 -8 +2 +12 3______________ 1 2______________ 3| 1 -1 -8 +2 +12 3 6__________ 1 2 -2__________ 3| 1 -1 -8 +2 +12 3 6 -6_____ 1 2 -2 -4_____ -2| 1 2 -2 -4 0 ↓_________________ 1_________________ -2| 1 2 -2 -4 0 ↓ -2_____________ 1 0_____________ -2| 1 2 -2 -4 0 ↓ -2 0_________ 1 0 -2_________ -2| 1 2 -2 -4 0 ↓ -2 0 4_____ 1 0 -2 0_____ x 2 – 2 = 0 x =± √2

12. P410 EX 2:  x 4 – 5x 2 –36 = 0 ± 1, 2, 3, 4, 6, 9, 12, 18, 36 Use calculator table to find the zeros. 3| 1 0 -5 0 -36 1 ↓ 3 1 3 3| 1 0 -5 0 -36 ↓ 3 9 1 3 4 3| 1 0 -5 0 -36 ↓ 3 9 12 1 3 4 12 -3 1 3 4 12 ↓ 1 ↓ -3 1 0 -3 1 3 4 12 ↓ -3 0 1 0 4 0 x 2 + 4 = 0 x = ± 2i

13. EX 4:  P(x) = 2x 3 –17x 2 + 40x –16 ±1, 2, 4, 8, 16 Use calculator table to find the zeros. 4| 2 -17 +40 -16 ↓ 2 ↓ 8 2 -9 4| 2 -17 +40 -16 ↓ 8 -36 2 -9 4 2 -9 4 |0| 2x 2 – 9x + 4 (2x – 1)(x – 4)

14. Note the following 3 facts: 1) Degree of polynomial is 3 and 3 factors 2) 2 distinct zeros, 4 a multiplicity of two 3) 3 unique factors 2(x – 1/2)(x – 4)(x – 4); constant of 2, coefficient of first term

15. 15.9 Rational Zero Theorem OBJ:  To find the zeros of an integral polynomial using the rational zero theorem

16. DEF:  Rational Zero Theorem If p is a factor of the constant term and q is a factor of the highest degree term, than p/q are the possible rational zeros of polynomial.

17. P 413 HW 3 P 415 (5-8) EX 1 :P(x) = 6x 4 – x 3 – 14x 2 + 2x +4 ± 1, 2, 4, ½, ⅓, ⅔, 1 ⅓, 1/6 -1/2| 6 -1 -14 +2+4 -3 2 6 -4 2/3| 6 -4 -12 8 |0| 4 0 -8 6 0 -12 |0| 6x 2 – 12 x = ± √2

18. EX 2:  6x 4 – 7x 3 + 3x 2 –7x –3 ± 1, 3, ½, ⅓, 1/6, 1½ 3/2| 6 -7 3 -7 -3 9 3 9 3 -⅓| 6 2 6 2 |0| -2 0 -2 6 0 6 0 6x 2 + 6 = 0 x = ± i

19. P414 HW 4 P415(10–16, 22–26e) EX 3:  3x 4 –5x 3 +10x 2 –20x –8 = 0 ± 1, 2, 4, 8, ⅓, ⅔, 1 ⅓, 2 ⅔ 2| 3 -5 +10 -20 -8 6 2 24 8 -⅓| 3 1 12 4 |0| -1 0 -4 3 0 12 |0| 3x 2 + 12 = 0 x = ± 2i