2.3 Synthetic Substitution

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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

• 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 x3 10 x2 5x 4
• 3(2)3 10(2)25(2)4
• 3(8) 10(4) 10 4
• 24 40 10 4 50
• P ( 2 ) 50

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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

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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

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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 .

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EX 4?P(x)x5 3x4 3x3 5x2 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 ____
• 1x32x23x 6 0

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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.

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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 gt2

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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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EX 1 ? If P(x) 2x4 5x3 11x2 20x 12has
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____
• 2x2 3x 2
• (2x 1)(x 2)

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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 -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_____
• x2 2 0
• x v2

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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 0 -5 0 -36
• ? 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 3 4 12
• ? -3
• 1 0
• -3 1 3 4 12
• ? -3 0
• 1 0 4 0
• x2 4 0
• x 2i

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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 -17 40 -16
• ?
• 2
• 4 2 -17 40 -16
• ? 8
• 2 -9
• 4 2 -17 40 -16
• ? 8 -36
• 2 -9 4

• 2 -9 4 0
• 2x2 9x 4
• (2x 1)(x 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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15.9 Rational Zero Theorem

• OBJ ? To find the zeros of an integral
  polynomial using the rational zero theorem

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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.

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P 413 HW 3 P 415 (5-8) EX
1P(x) 6x4 x3 14x2 2x 4

• 1, 2, 4, ½, ?, ?, 1 ? , 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
• 6x2 12
• x v2

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EX 2 ? 6x 4 7x 3 3x 2 7x 3

• 1, 3, ½, ?, 1/6, 1½
• 3/2 6 -7 3 -7 -3
• 9 3 9 3
  
• -? 6 2 6 2 0
• -2 0 -2
• 6 0 6 0
• 6x2 6 0
• x i

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P414 HW 4 P415(1016, 2226e)EX 3 ? 3x4 5x3
10x2 20x 8 0

• 1, 2, 4, 8, ?, ?, 1
  ?, 2 ?
• 2 3 -5 10 -20 -8
• 6 2 24 8
• -? 3 1 12 4 0
• -1 0 -4
• 3 0 12 0
• 3x2 12 0
• x 2i