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

Top P 38 EX :  P ( 2 ) P = 3 x x 2 – 5x – 4 2| ↓_____________ 3 2| ↓ 6_________ | ↓ 6 32____ | ↓ x x 2 – 5x – 4 3(2) (2) 2 –5(2)–4 3(8) + 10(4) – 10 – – 10 – 4 = 50 P ( 2 ) = 50

P 38 EX 2:  2 x 4 – x 3 + 5x + 3 x = -2 -2| ↓ | 2________________ -2| ↓ -4 | 2 -5___________ -2| ↓ | _______ -2| ↓ | x = 3 3| ↓________________ | 2 3| ↓ 6___________ | 2 5 3| ↓ 6 15_______ | | ↓ ____

EX 3:  3 x 4 – 2x 2 – 6x | ↓ | 3_______________ 2| ↓ 6 | 3 6_____________ 2| ↓ 6 12 | _________ 2| ↓ | |38| -2| | ↓________________ 3 -2| | ↓ -6_____________ | | ↓ -6 12________ | | |62|

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 #.

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

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.

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

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”

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| ↓ 4 2 9_____________ 2| ↓ _________ 2| ↓ ____ 2| ↓ |0| -3| |0| ↓ 2_________________ -3| ↓ ______________ -3| ↓ |0|____ 2x 2 + 3x – 2 (2x – 1)(x + 2)

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________________ 3| ______________ 1 2______________ 3| __________ __________ 3| _____ _____ -2| ↓_________________ 1_________________ -2| ↓ -2_____________ 1 0_____________ -2| ↓ -2 0_________ _________ -2| ↓ _____ _____ x 2 – 2 = 0 x =± √2

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 ↓ ↓ x = 0 x = ± 2i

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

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

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

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.

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| /3| |0| |0| 6x 2 – 12 x = ± √2

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

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| ⅓| |0| |0| 3x = 0 x = ± 2i