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2.3 Synthetic Substitution
OBJ: To evaluate a polynomial for given values of its variables using synthetic substitution
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Top P 38 EX : P ( 2 ) P = 3 x 3 + 10 x 2 – 5x – 4 P ( 2 ) = 50
2| ↓_____________ 3 ↓ _________ ↓ ____ ↓ 3 x x2 – 5x – 4 3( )3+ 10( )2–5( )–4 3(2)3+ 10(2)2–5(2)–4 3(8) + 10(4) – 10 – 4 – 10 – 4 = 50 P ( 2 ) = 50
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EX 3: 3 x 4 – 2x 2 – 6x + 10 ↓ ______________ ↓ -6_____________
2| _ ↓ _____________ 3 2| _ ↓ 6_____________ 3 6 ↓ _________ 2| ↓ |38| -2| ↓ ______________ 3 ↓ -6_____________ 3 -6 ↓ ________ ↓ |62|
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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.
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EX 4:P(x)=x5 – 3x4 + 3x3– 5x2 +12 -1| 1 -1 1 -3 -6 |0|
2| ↓________________ 1________________ ↓ 2_____________ 1 -1_____________ ↓ __________ __________ ↓ __| 0 |__ -1| |0| ↓ _________________ 1__________________ ↓ -1_______________ 1 -2_______________ ↓ ___________ ___________ ↓ ____ __|0| ___ 1x3–2x2+3x – 6 = 0
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DEF: Factor Theorem If a polynomial is synthetically divided
by a # and the remainder is 0, then x– # is a factor.
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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
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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”
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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 ↓ 3_____________ 1 3 ↓ _________ ↓ ↓ _________ 1 ↓ -3_______ 1 0 ↓ __12__ x2 + 4 = 0 x = ± 2i
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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________________ 3______________ 1 2______________ __________ __________ ___-12 ___0_ -2| ↓_________________ 1_________________ ↓ -2_____________ _____________ ↓ _________ _________ ↓ _____ _____ x2 – 2 = 0 x =± √2
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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________________ ↓ 4 2 9_____________ 2| ↓ _________ ↓ ____ ↓ |0| -3| |0| ↓ 2_________________ -3| ↓ -6 ______________ ↓ |0|____ 2x2 + 3x – 2 (2x – 1)(x + 2)
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EX 4: P(x) = 2x3 –17x2 + 40x –16 ±1, 2, 4, 8, 16 Use calculator table to find the zeros. 4| ↓______________ 2 ↓ 8__________ ↓ __ 16__ |0| 2x2 – 9x + 4 (2x – 1)(x – 4) x = ½, 4
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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
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P 38 EX 2: 2 x 4 – x 3 + 5x + 3 x = -2 -2| ↓ | 2________________ ↓ -4 | ___________ ↓ | _______ ↓ | 33 x = 3 3| ↓________________ | 2 ↓ ___________ | ↓ _______ | ↓ _150_ 153
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P 38 EX 2: 2 x 4 – x 3 + 5x + 3 x = -2 -2| ↓ | x = 3 3| ↓ |
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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| ↓ |0| -3| ↓ |0| 2x2 + 3x – 2 (2x – 1)(x + 2)
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