2.3 Synthetic Substitution

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Presentation transcript:

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 3 + 10 x 2 – 5x – 4 P ( 2 ) = 50 2| 3 10 -5 -4 ↓_____________ 3 ↓ 6_________ 3 16 ↓ 6 32____ 3 16 27 ↓ 6 32 54 3 16 27 50 3 x3 + 10 x2 – 5x – 4 3( )3+ 10( )2–5( )–4 3(2)3+ 10(2)2–5(2)–4 3(8) + 10(4) – 10 – 4 24 + 40 – 10 – 4 = 50 P ( 2 ) = 50

EX 3:  3 x 4 – 2x 2 – 6x + 10 ↓ ______________ ↓ -6_____________ 2| 3 0 -2 -6 10_ ↓ _____________ 3 2| 3 0 -2 -6 10_ ↓ 6_____________ 3 6 ↓ 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 ↓ -6_____________ 3 -6 ↓ -6 12________ 3 -6 10 ↓ -6 12 -20 52 3 -6 10 -26 |62|

DEF:  Remainder Theorem The remainder, when a polynomial is synthetically divided by a number, is equal to the value when the polynomial is evaluated with the number.

EX 4:P(x)=x5 – 3x4 + 3x3– 5x2 +12 -1| 1 -1 1 -3 -6 |0| 2| 1 -3 +3 -5 0 +12 ↓________________ 1________________ ↓ 2_____________ 1 -1_____________ ↓ 2 -2__________ 1 -1 1__________ ↓ 2 -2 2 -6 -12 1 -1 1 -3 -6__| 0 |__ -1| 1 -1 1 -3 -6 |0| ↓ _________________ 1__________________ ↓ -1_______________ 1 -2_______________ ↓ -1 2___________ 1 -2 3___________ ↓ -1 2 -3 6____ 1 -2 3 -6__|0| ___ 1x3–2x2+3x – 6 = 0

DEF:  Factor Theorem 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”

P410 EX 2:  x 4 – 5x 2 –36 = 0 -3 1 3 4 12 ↓ _________ 1 ↓ -3_______ ± 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 9_________ 1 3 4 ↓ 3 9 12 36 1 3 4 12 0 -3 1 3 4 12 ↓ _________ 1 ↓ -3_______ 1 0 ↓ -3 0__12__ 1 0 4 0 x2 + 4 = 0 x = ± 2i

P 410 HW 2 P 411 (1-19 odd) EX 1: P(x) =x4 – x3 – 8x2 + 2x +12 ± 1, 2, 3, 4, 6, 12 Use calculator table to find the zeros 3| 1 -1 -8 +2 +12 ↓________________ 1________________ 3______________ 1 2______________ 3 6__________ 1 2 -2__________ 3 6 -6___-12 1 2 -2 -4___0_ -2| 1 2 -2 -4 0 ↓_________________ 1_________________ ↓ -2_____________ 1 0_____________ ↓ -2 0_________ 1 0 -2_________ ↓ -2 0 4_____ 1 0 -2 0_____ x2 – 2 = 0 x =± √2

EX 1:  If P(x) = 2x4 + 5x3 – 11x2 – 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________________ ↓ 4 2 9_____________ 2| 2 5 -11 -20 +12 ↓ 4 18 2 9 7 _________ ↓ 4 18 14 2 9 7 -6____ ↓ 4 18 14 -12 2 9 7 -6 |0| -3| 2 9 7 -6 |0| ↓ 2_________________ -3| 2 9 7 -6 ↓ -6 2 3______________ ↓ -6 -9 6 2 3 -2 |0|____ 2x2 + 3x – 2 (2x – 1)(x + 2)

EX 4:  P(x) = 2x3 –17x2 + 40x –16 ±1, 2, 4, 8, 16 Use calculator table to find the zeros. 4| 2 -17 +40 -16 ↓______________ 2 ↓ 8__________ 2 -9 ↓ 8 -36__ 16__ 2 -9 4 0 2 -9 4 |0| 2x2 – 9x + 4 (2x – 1)(x – 4) x = ½, 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

P 38 EX 2:  2 x 4 – x 3 + 5x + 3 x = -2 -2| 2 -1 0 5 3 ↓ | 2________________ ↓ -4 | 2 -5___________ ↓ -4 10 | 2 -5 10_______ ↓ -4 10 -20 | 2 -5 10 15  33  x = 3 3| 2 -1 0 5 3 ↓________________ | 2 ↓ 6___________ | 2 5 ↓ 6 15_______ | 2 5 15 ↓ 6 15 45_150_ 2 5 15 50  153

P 38 EX 2:  2 x 4 – x 3 + 5x + 3 x = -2 -2| 2 -1 0 5 3 ↓ -4 10 -20 30 | 2 -5 10 15 33 x = 3 3| 2 -1 0 5 3 ↓ 6 15 45 150 | 2 5 15 50 153

EX 1:  If P(x) = 2x4 + 5x3 – 11x2 – 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 ↓ 4 18 14 -12 2 9 7 -6 |0| -3| 2 9 7 -6 ↓ -6 -9 6 2 3 -2 |0| 2x2 + 3x – 2 (2x – 1)(x + 2)