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Mathematics 141 Practice quiz 4
factoring

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[1] Factor 5355 completely :
 
(a) (5)(5)(3)(3)(13)
(b) (5)(7)(9)(17)
(c) (5)(63)(17)
(d) (5)(3)(3)(7)(17)
(e) none of the above

[2] Factor the following completely : 33x3y3 + 21x2y5
 
(a) 3x2y3(11x + 7y2)
(b) 3x2y3(11x + 21y2)
(c) 33x3y3(1 + 21x2y5)
(d) (33x3y3 + 1)(21x2y5)
(e) (33)(21)x2y3(x + y2)
(f) none of the above

[3] Factor the following completely : 27x2 - 12y4
 
(a) 3(9x2 - 4y4)
(b) 33x2 - 22(3)y4
(c) 3(3x - 2y)(3x + 2y)
(d) 3(3x - 2y)2
(e) (27 - 12)(x2 - y4)
(f) none of the above
[4] Factor the following completely : 54x3y6 - 16z9
 
(a) 54(xy2)3 - 16(x3)3
(b) 2(27x3y6 - 8z9)
(c) (2x2 - z9)(27y6 + 16)
(d) 2(3xy2 - 2z3)(9x2y4+ 6xy2z3 + 4z6)
(e) 2(3xy2 - 2z3)3
(f) none of the above
[5] Enter the two factors in the spaces provided :
 
2x2 - 3x - 2 = ( )( )  
[6] Enter the two factors in the spaces provided :
 
9x2y2 - 121 = ( )( )
[7] Enter the two factors in the spaces provided :
 
9x2y2 - 12xy + 4 = ( )( )
[8] Enter the two factors in the spaces provided :
 
9x2 - 3x - 3xy + y = ( )( )
[9]
Laura's dorm room is a rectangle 2 feet longer than wide, and has a floor area of 195 square feet. How long is Laura's room?
  Which of the following defines a variable and an equation correctly translating the boxed problem above?
 
(a) Let x = Laura's dorm room ; x(x - 2) = 195
(b) Let x = the area of Laura's dorm room ; x(x + 2) = 195
(c) Let x = the length of Laura's dorm room ; x(x + 2) = 195
(d) Let x = the length of Laura's dorm room ; x(x - 2) = 195
(e) Let x = the width of Laura's dorm room ; x(x - 2) = 195
(f) Let x = the area of Laura's dorm room ; x(x - 2) = 195
(g) none of the above
[10]
The perimeter of a rectangular field is 68 meters, and its diagonal measures 26 feet. What are the dimensions of the field?
  Which of the following defines a variable and an equation correctly translating the boxed problem above?
 
(a) Let x = the length of the field ; x2 + (34 - x)2 = 262
(b) Let x = the dimensions of the field ; x2 + (68 - x)2 = 262
(c) Let x = the width of the field ; x2 + (68 - 2x)2 = 262
(d) Let x = the dimensions of the field ; x2 + (34 - x)2 = 262
(e) Let x = the length of the field ; 2x + 2(26 - x)2 = 68
(f) Let x = the perimeter of the field ; x2 + (68 - 2x)2 = 262
(g) none of the above

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