Notes 8.3 - The Remainder and Factor Theorems

nostrilswelderElectronics - Devices

Oct 10, 2013 (4 years and 1 day ago)

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Notes 8.3
-

The Remainder and Factor Theorems


The Remainder Theorem
:




Here are three ways to find a remainder…


For the polynomial: f(x) = x
4

-

3
x
3

+
x
2

-

3x + 1, divide it by
x
-

2
.

Long Division:



Synthetic Division:


Finding f(x):

(you may

just want to watch this)












f(2
) =
2
4



3(2)
3

+ 2
2



3(2) + 1



x
3


-

x
2


-

x

-

5





2 1
-
3 1
-
3 1 16


3(8) + 4
-

6 + 1






2
-
2


-
2
-
10


16


24 + 4


6 + 1


-

(
x
4

-

2x
3
)






1

-
1
-
1


-
5
-
9



-
8 + 4


6 + 1
















-
4


6 + 1


-
x
3

+ x
2










-
10 + 1


-
(
-
x
3

+ 2x
2
)









-
9




-
x
2

-

3x


-
(
-
x
2

+ 2x)



-
5x + 1


Look how lucky

you are that





-
(
-
5x + 10)


someone discovered this!



-
9







What would be the remainder if x


8 was a factor of

x
4

+ 2x
3



10x
2

+ 5x


7
?

ZERO!



Quotient

If a polynomial f(x) is divided by x


a, the remainder is the
constant f(a).

Coefficients of the
Quotient

Remainder

Remainder


The Factor Theorem

The binomial x


a is a fa
ctor of the
polynomial f(x) if and only if f(a) = 0.


Is x


4 a factor of the polynomial
f(x) = x
4

+ x
3



13x
2



25x


12
?


f(4) = 4
4

+ 4
3



13(4)
2



25(4)


12

OR

4 1 1
-
13
-
25
-
12



= 256 + 64


208


100


12






4 20 28 12



= 0

YES








1 5 7 3 0


Ex 1) Use synthetic division to find f(4) for




f(x) = x
4



6x
3

+ 8x
2

+ 5x + 13

(Put the divisor in the box, list the coefficient
s in order,

bring down the first term, multiply each answer by the divisor, then add down)


4


1


-
6


8


5


13




+

4

-
8

0

20



1


-
2


0


5


33




f(4) = 33


Ex 2) Use synthetic divisi
on to determine if x+2 is a factor of the function,





f(x) = 3x
5



5x
3

+ 57


-
2


3


0


-
5


0


0


57



+

-
6

12

-
14


28

-
56


3

-
6

7


-
14

28 1




f(
-
2) = 1, so x+2 is not a factor.



KEY

The remainder

Remainder




Divide using synthetic division. Is the binomial a factor of the
polynomial?

Ex 3)
(x
3



64)

(x


4)


4


1


0


0


-
64






+ 4

16 64



1


4


16


0






yes







Ex 4)
(x
4



x
3

+ 2x


2)
(x


2)


2


1


-
1


0


2


-
2




+ 2 2 4 12



1


1


2


6


10




no





KEY

Ex 5) Show that x


2

is a factor of
x
3

+ 7x
2

+ 2x


40
.


Then find the remaining factors.


2


1


7


2


-
40






+ 2 18 40



1


9


20


0






(x


2)(x
2

+ 9x + 20)



(x


2)(x + 5)(x + 4)


so, x
3

+ 7x
2

+ 2x


40 =

(x


2)(x + 5)(x + 4)



Ex 6) Show that 2x + 7 is a factor of
2x
3

+ 17x
2

+ 23x


42.



Then find the remainin
g
factors.



2


17


2

3
-
42






+
-
7
-
35 42



2


10


-
12


0



(Since you used 2
as the denominator, you must divide 2 out of
the resulting expression)

…instead of 2x
2

+ 10x


12 it is x
2

+ 5x


6



(2x + 7)(x
2

+ 5x


6)



(2x + 7)(x


1)(x + 6)


so, 2x
3

+ 17x
2

+ 23x


42 = (2x + 7)(x


1)(x + 6)

KEY