# 159 - MA 15900 1. Solve the equation. EXAM 1 FALL 2010 2x 9...

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MA 15900 EXAM 1 FALL 2010 1 1. Solve the equation. 2 x 9 4 = 2 + x 12 1) x = 11 3 2) x = 11 3 3) x = 29 5 4) x = 29 5 5) x = 51 7 6) x = 51 7 7) x = 132 23 8) x = 132 23 9) None of the above 2. Factor the polynomial. 27 x 6 19 x 3 8 1) (3 x 2 + 2)(9 x 4 6 x 2 + 4)( x 1) 2) (3 x 2 2)(9 x 4 + 6 x 2 + 4)( x + 3) (27 x + 8)( x + 1)( x x 4 + x 2 + 4) (27 x 8)( x 2 + x 4 x 2 + 5) (3 x + 2)(9 x 2 6 x + 4)( x x 2 + x + 6) (3 x + 2)(9 x 2 6 x + 4)( x + x 2 x + 7) (3 x 2)(9 x 2 + 6 x + 4)( x x 2 + x + 8) (3 x 2)(9 x 2 + 6 x + 4)( x + x 2 x + 9) None of the above 3. Express as a polynomial. (3 x 2 2 y ) 2 (3 x 2 + 2 y ) 1) 27 x 6 18 x 4 y 12 x 2 y 2 + 8 y 3 2) 27 x 6 + 18 x 4 y 12 x 2 y 2 8 y 3 3) 27 x 6 + 18 x 4 y + 12 x 2 y 2 + 8 y 3 4) 9 x 4 6 x 2 y 2 + 6 x 2 y 4 y 3 5) 9 x 4 12 x 2 y 2 + 6 x 2 y 8 y 3 6) 9 x 4 + 12 x 2 y + 4 y 2 7) 9 x 4 12 x 2 y + 4 y 2 8) 9 x 4 4 y 2 9) None of the above

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MA 15900 EXAM 1 FALL 2010 2 4. Solve the equation. 2 x x 3 2 x + 3 = 36 x 2 9 1) x = 19 2) x = 7 3) x = 3 4) x = 3 5) x = 5 6) x = 7 7) x = 7, 3 8) x = 5,3 9) x = 7,3 5. Simplify the expression. 9 x 2 4 3 x 2 5 x + 2 9 x 2 6 x + 1 27 x 2 6 x 1 1) (3 x 2)(3 x 1) (3 x + 1)(9 x + 2) (3 x 2)(3 x + 2) ( x 2)(9 x + 3) (3 x 2)(3 x + ( x x + 4) (3 x + 2)(3 x ( x x + 5) (3 x + 2)(3 x ( x x 6) (3 x + 2)(3 x 2 ( x x 1)(3 x + 7) x 2)(3 x 2 ( x x 1)(3 x + 8) x 2) 2 x (3 x + 1)( x 2)(9 x + 9) (3 x 2)(3 x + 2)(3 x x + x 2)(9 x +
MA 15900 EXAM 1 FALL 2010 3 6. Simplify the expression and rationalize the denominator.

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## This note was uploaded on 09/14/2011 for the course MATH 159 taught by Professor Staff during the Spring '09 term at Purdue University-West Lafayette.

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159 - MA 15900 1. Solve the equation. EXAM 1 FALL 2010 2x 9...

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