Unit 1	Second Chance Exam II		Name:_____________
Multiple Choice
Identify the choice that best completes the statement or answers the question.

____	1.	What is the remainder when  is divided by ?
A	1070	C	962	
B	–1070	D	–962	

____	2.	If  is divided by  to give a quotient of  and
a remainder of –952, then which of the following is true?
A	 = –952	
B	 =  – 952	
C	 = 952	
D	=  + 952	

____	3.	When  is divided by , the remainder is
A	x2 x 	C		
B	P(–2) = –34	D		

____	4.	For a polynomial P(x), if  = 0, then which of the following must be a factor of P(x)?
A		C		
B		D		

____	5.	A factor of  is	
A	 	C	 	
B	x – 5	D	x – 8	

____	6.	Which set of values for x should be tested to determine the possible zeros of ?
A	±5, ±7, and ±35	C	1, 5, 7, 12, and 35	
B	±1, ±5, ±7, and ±35	D	1, 5, 7, and 35	

____	7.	Which of the following is the fully factored form of ?
A		C		
B		D		

____	8.	What is the maximum number of real distinct roots that a cubic equation can have?
A	infinitely many	C	3	
B	4	D	2	

____	9.	Which of the following graphs of polynomial functions corresponds to a cubic polynomial equation with roots 4, 1, and 3?
A		C		
B		D		

10	 If f(x) =  is divided by (x+2), then what is the restriction on x?

A)	x  0
B)	x 2
C)	x-2
D)	x-7

	1.	Algebraically determine what intervals is the function  positive?   6 marks

	2.	Factor 2x4 – 7x3 – 41x2 – 53x – 21 fully.						8 marks

	3.	Determine all roots to  2x4 – x3 = 7x2 – 9x + 3 algebraically.					8 marks

4.		Determine the equation of y=f(x) in factored form.    				5 marks

	

5.		Bill is preparing to make an ice sculpture.  He has a block of ice that measures 3 ft. wide , 4ft. high and 5 ft. long.  He wants to reduce the size of the block of ice by removing the same amount from each dimension such that the new block is 40% of the original volume and will fit in a freezer for transportation purposes.  Algebraically set up a polynomial function to determine how much has to be removed from each side.  
		7 marks

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Answer Section

MULTIPLE CHOICE

	1.	ANS:	C	PTS:	1	DIF:	Average	OBJ:	Section 3.2
NAT:	RF11	TOP:	The Remainder Theorem	KEY:	remainder theorem | remainder

	2.	ANS:	B	PTS:	1	DIF:	Average	OBJ:	Section 3.2
NAT:	RF11	TOP:	The Remainder Theorem	KEY:	quotient | remainder

	3.	ANS:	D	PTS:	1	DIF:	Difficult +	OBJ:	Section 3.2
NAT:	RF11	TOP:	The Remainder Theorem	KEY:	remainder theorem | remainder

	4.	ANS:	D	PTS:	1	DIF:	Easy	OBJ:	Section 3.3
NAT:	RF11	TOP:	The Factor Theorem		KEY:	factor theorem | factor

	5.	ANS:	A	PTS:	1	DIF:	Average	OBJ:	Section 3.3
NAT:	RF11	TOP:	The Factor Theorem		
KEY:	factor theorem | integral zero theorem | factor

	6.	ANS:	B	PTS:	1	DIF:	Average	OBJ:	Section 3.3
NAT:	RF11	TOP:	The Factor Theorem		KEY:	integral zero theorem

	7.	ANS:	D	PTS:	1	DIF:	Average	OBJ:	Section 3.3
NAT:	RF11	TOP:	The Factor Theorem		
KEY:	factored form | factor theorem | factor	

	8.	ANS:	C	PTS:	1	DIF:	Easy	OBJ:	Section 3.4
NAT:	RF12	TOP:	Equations and Graphs of Polynomial Functions
KEY:	polynomial equation | roots

	9.	ANS:	B	PTS:	1	DIF:	Average	OBJ:	Section 3.4
NAT:	RF12	TOP:	Equations and Graphs of Polynomial Functions
KEY:	polynomial equation | roots | graph

COMPLETION

	1.	ANS:	–2 < x < 6 and x < –9

PTS:	1	DIF:	Difficult +	OBJ:	Section 3.4	NAT:	RF12
TOP:	Equations and Graphs of Polynomial Functions	
KEY:	characteristics of polynomials | inequality	

SHORT ANSWER

	1.	ANS:	
Use the factor theorem. Find P(-1).

Since P(–1) = 0, x + 1 is a factor of P(x).
Long division or synthetic division can also be used.
	
Since the remainder is 0, x + 1 is a factor of P(x).

PTS:	1	DIF:	Easy	OBJ:	Section 3.3	NAT:	RF11
TOP:	The Factor Theorem		KEY:	factor theorem | factor | remainder

	2.	ANS:	
Possible values of x in the factor theorem are ±1, , ±2, ±4, and ±8.
Try x = 2.

Thus, x - 2 is a factor of P(x).
Divide.

Thus,

PTS:	1	DIF:	Difficult +	OBJ:	Section 3.3	NAT:	RF11
TOP:	The Factor Theorem		
KEY:	factor theorem | factor | integral zero theorem | grouping | rational zero theorem
NOT:	A variety of factoring techniques is required.

PROBLEM

	1.	ANS:	
Let P(x) = 2x4 – 7x3 – 41x2 – 53x – 21.
Test x = –1 in the factor theorem.

Thus, x + 1 is a factor. Divide to determine another factor.

Thus,

Now factor the cubic. Test x = –1 in the factor theorem. Let .

Thus, x + 1 is a factor. Divide to determine another factor.

Thus,

PTS:	1	DIF:	Average	OBJ:	Section 3.3	NAT:	RF11
TOP:	The Factor Theorem		KEY:	factor theorem | integral zero theorem | factor

	2.	ANS:	
Rewrite in the form P(x) = 0.
2x4 – x3 – 7x2 + 9x – 3 = 0
Try x = 1 in the factor theorem.

Since x = 1 is a zero of P(x), x – 1 is a factor. Divide to determine another factor.

Now factor the cubic. Try x = 1 in the factor theorem. Let .

Thus, x - 1 is a factor of Q(x). Use division to find another factor.

Thus, x = 1 or 2x2 + 3x - 3 = 0. Use the quadratic formula to find the other roots.

The solution is x = 1, , or .

PTS:	1	DIF:	Average	OBJ:	Section 3.3	NAT:	RF11
TOP:	The Remainder Theorem	KEY:	factor theorem | integral zero theorem | root