Math 140 Placement Verification Test Solution Review of Similar Type Problems

Problem 1 Simplify: (4 -2 )

Problem 2 Simplify: (3x -6 )(-x 4 ) (9x) -2 [3x -6+4  3 (-1) 3 ](  x 2 ) = x 6+2 = -x 8 /27

Problem 3 Add and simplify:. Note: 625 = 5 4 & (-1) = (-1)   3 = -5·5 1   3 = -4·5 1  3 =

Problem 4 Perform these operations and simplify: (3x - 4) 2 - (x - 3)(8x + 1). (9x 2 -2 ·3 ·4x + 16) - (8x 2 +(1- 3 ·8)x - 3) = (9-8)x 2 + (-24+23)x = x 2 - x + 19

Problem 5 Simplify:.

Problem 6 Perform these operations and simplify:

Problem 7 Perform the indicated operations and simplify: (x - 2b 2 ) 3. (x - 2b 2 ) 3 = x 3 - 3(2b 2 ) x 2 + 3(2b 2 ) 2 x - (2b 2 ) 3 = x 3 - 6b 2 x b 4 x - 8b 6

Problem 8 Simplify: __________ 25 __________ 25 __________ 5

Problem 9 Solve: 7 - 5x < 2(x -21) x < 2x x < - 49 x > - 49/(-7) x > 7

Problem 10 Solve for x in the equation:

Problem 10 Continued Solve for x in the equation: Alternate approach: Multiply by LCD = acx. Then, cx + 2acx – ac + ax = 0 (c + 2ac + a)x = ac

Problem 11 Solve: Solve: |3 - 2x| < 5. a) 3 - 2x < 5 -2x < 2 x > -1 or b) –(3 - 2x) < 5 2x < 8 x < 4 Thus, -1< x < 4.

Problem 12 Solve this equation. Write the sum of the answers; that is, write the result when all possible answers are added together: (x + 5)(x - 1) = 16. x 2 + (5-1)x - 5 = 16 x 2 + 4x - 21 = 0 (x - 3)(x + 7) = 0 x – 3 = 0 or x + 7 = 0 x = {3, -7} is solution set. Sum is –4. Or from quadratic formula: x = [-b ± (b 2 – 4ac) 1/2 ]/(2a). Sum is –2b/(2a) = -4.

Problem 13 Perform the indicated operations on the expression: (16/81) -3/4. (81/16) 3/4 = (3 4 /2 4 ) 3/4 = [(3/2) 4 ] 3/4 = (3/2) 4·3/4 = (3/2) 3 = 27/8

Problem 14 Perform the indicated operations on the expression:.

Problem 15 Assuming a > 0 and b > 0, simplify the indicated expression equivalently so that all exponents are positive: (3ab 4 ) -3. (9ab 6 ) ·2 a 2 b 6·2 3 3 a 3 b 4·3 3/a Note two negative exponents => flipping fraction avoids later trouble with signs.

Problem 16 Assuming a > 0 and b > 0, simplify the indicated expression equivalently so that all exponents are positive: (144a -2/3 b -4/3 ) 3/ ·3/2 a (-2/3)(3/2) b (-4/3)·(3/2) 12 3 a -1 b /(ab 2 )

Problem 17 Simplify to standard form by performing the indicated operations on the expression: (x + 1)(3x - 4) 2 - (x - 3)(x + 3)(8x + 1). (x + 1)(9x 2 –24x + 16) - (x 2 - 9)(8x + 1) = 9x 3 + (-24+9) x 2 + (16-24)x +16 – (8x 3 + x 2 -72x - 9) = x x x + 25

Problem 18 Factor the expression over the integers or state it is prime if it cannot be factored: x 2 - 6x - 16 Work systematically. There are 5 possibilities: Two with 1 & 16 => (x - 1) (x + 16) & (x + 1) (x - 16), Two with 2 & 8 => (x - 2) (x + 8) & (x + 2) (x - 8), One with 4 & 4 => (x - 4) (x + 4) = (x + 4) (x - 4). But to get the first degree middle term (- 6x ) only the combination of factors (x + 2) (x - 8) works. x 2 - 6x - 16 = (x + 2) (x - 8).

__________ 2 2 — (5 1/2 ) 2 Problem 19 Simplify by rationalizing the denominator of the expression: 25. ____________

Problem 20 Simplify in factored form: