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

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( )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 + 102| 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|

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

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| ___ 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 allintegral coefficients

A # that if evaluated in apolynomial wouldresult in a 0 as theremainder

The integral zeros of apolynomial are theintegral factors of theconstant term, called “p”

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 36

1 3 4 12 0

-3 1 3 4 12

↓ _________

1

-3 1 3 4 12

↓ -3_______

1 0

-3 1 3 4 12

↓ -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 -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___-12 1 2 -2 -4___0_

-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 =± √2

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 6

2 3 -2 |0|____

2x2 + 3x – 2

(2x – 1)(x + 2)

EX 4: P(x) = 2x3 –17x2 + 40x –16

±1, 2, 4, 8, 16Use calculator table tofind the zeros. 4| 2 -17 +40 -16 ↓______________ 2 4| 2 -17 +40 -16 ↓ 8__________ 2 -9 4| 2 -17 +40 -16 ↓ 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________________-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 = 33| 2 -1 0 5 3 ↓________________ | 23| 2 -1 0 5 3 ↓ 6___________ | 2 5 3| 2 -1 0 5 3 ↓ 6 15_______ | 2 5 153| 2 -1 0 5 3 ↓ 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 +12has 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)