OBJECTIVE 2 Use the remainder theorem to evaluate a polynomial.We can
use synthetic division to evaluate polynomials. For example, in the synthetic division of
Example 2,where the polynomial was divided by the remainder was
Replacing xin the polynomial with 2 gives
Replace xwith 2.
Evaluate the powers.
Multiply.
Add.the same number as the remainder. Dividing by produced a remainder equal to
the result when xis replaced with 2. This always happens, as the following remainder
theoremstates. This result is proved in more advanced courses.
x- 2
=-6,
=- 128 + 16 + 48 + 8 + 50
=- 4 # 32 + 16 + 6 # 8 + 2 # 4 + 50
=- 4 # 25 + 24 + 6 # 23 + 2 # 22 + 50
- 4 x^5 +x^4 + 6 x^3 + 2 x^2 + 50
x- 2 , -6.
APPENDIX B Synthetic Division 725
Remainder TheoremIf the polynomial P 1 x 2 is divided by x- k,then the remainder is equal to P 1 k 2.
Using the Remainder TheoremLet Use synthetic division to evaluate
Use the remainder theorem, and divide by
Value of kRemainderThus, NOW TRY
OBJECTIVE 3 Decide whether a given number is a solution of an equation.
We can alsouse the remainder theorem to do this.
Using the Remainder TheoremUse synthetic division to decide whether is a solution of the equation.
If synthetic division gives a remainder of 0, then is a solution. Otherwise, it is not.
Proposed solutionRemainderSince the remainder is 0, the polynomial has value 0 when So is a solu-
tion of the given equation. NOW TRY
The synthetic division in Example 4shows that divides the polyno-
mial with 0 remainder. Thus is a factorof the polynomial and
The second factor is the quotient polynomial found in the last row of the synthetic
division.
2 x^4 + 12 x^3 + 6 x^2 - 5 x+ 75 factors as 1 x+ 5212 x^3 + 2 x^2 - 4 x+ 152.
x- 1 - 52 =x+ 5
x- 1 - 52
k=-5. - 5
22 - 415 0
- 10 - 10 20 - 75
- 5 212 6- 575
- 5
2 x^4 + 12 x^3 + 6 x^2 - 5 x+ 75 = 0
- 5
EXAMPLE 4
P 1 - 22 =-19.
2 - 915 - 19
- 418 - 30
- 2 2 - 5 - 311
P 1 x 2 x- 1 - 22.
P 1 x 2 = 2 x^3 - 5 x^2 - 3 x+11. P 1 - 22.
EXAMPLE 3
NOW TRY ANSWERS
- 12 4.yes
NOW TRY
EXERCISE 3
Let
. Use synthetic
division to evaluate P 1 - 22.
5 x+ 30
P 1 x 2 = 3 x^3 - 2 x^2 +NOW TRY
EXERCISE 4
Use synthetic division to
decide whether is a solu-
tion of the equation.
5 x^3 + 19 x^2 - 2 x+ 8 = 0