1. 2.3 Synthetic Substitution
OBJ:  To evaluate a polynomial for given values of its variables using synthetic substitution

2. Top P 38 EX :  P ( 2 ) P = 3 x 3 + 10 x 2 – 5x – 4 P ( 2 ) = 502|
↓_____________ 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

3. EX 3:  3 x 4 – 2x 2 – 6x + 10 ↓ ______________ ↓ -6_____________
2| _
↓ _____________ 3
2| _
↓ 6_____________
3 6
↓ _________ 2| ↓
|38|
-2|
↓ ______________ 3
↓ -6_____________
3 -6
↓ ________ ↓
|62|

4. 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.

5. 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

6. DEF:  Factor Theorem If a polynomial is synthetically divided
by a # and the remainder is 0, then
x– # is a factor.

7. 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

8. 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”

9. 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

10. 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

11. 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)

12. 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

13. 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

14. P 38 EX 2:  2 x 4 – x 3 + 5x + 3
x = -2
-2| ↓
| 2________________
↓ -4
| ___________ ↓
| _______ ↓
|  33 
x = 3 3|
↓________________
| 2
↓ ___________ |
↓ _______ |
↓ _150_
 153

15. P 38 EX 2:  2 x 4 – x 3 + 5x + 3
x = -2
-2| ↓ |
x = 3 3| ↓ |

16. 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)