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Math 140 Placement Verification Test Solution Review of Similar Type Problems

Math 140 Placement Verification Test Solution Review of Similar Type Problems. Problem 1. Simplify: 12 -1 + 3(4 -2 ) . 12 -1 + 4 -2. Problem 2. [3 x -6+4  3 (-1) 3 ](3 -8+2 2 x 2 ) = -3 1-4 x 6+2 = - x 8 /27.

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Math 140 Placement Verification Test Solution Review of Similar Type Problems

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  1. Math 140Placement Verification Test Solution Review ofSimilar Type Problems

  2. Problem 1 Simplify: 12-1 + 3(4-2) . 12-1 + 4-2

  3. Problem 2 [3x-6+43(-1)3](3-8+22 x2) = -31-4 x6+2 = -x8/27 Simplify: (3x-6)(-x4)3. 38(9x)-2

  4. Problem 3 Note: 625 = 54 & (-1) = (-1)3. -54/3 + 51/3 = -5·51/3 + 51/3 = -4·51/3 = Add and simplify: .

  5. Problem 4 (9x2 -2 ·3 ·4x + 16) - (8x2 +(1- 3 ·8)x - 3) = (9-8)x2 + (-24+23)x + 16 + 3 = x2 - x + 19 Perform these operations and simplify: (3x - 4)2 - (x - 3)(8x + 1).

  6. Problem 5 Simplify: .

  7. Problem 6 Perform these operations and simplify:

  8. Problem 7 (x - 2b2)3 = x3 - 3(2b2) x2 + 3(2b2)2x - (2b2)3 = x3 - 6b2x2 + 12b4x - 8b6 Perform the indicated operations and simplify: (x - 2b2)3 .

  9. Problem 8 __________ 25 Simplify: __________ 25 __________ 5

  10. Problem 9 • 7 - 5x < 2x - 42 • 7x < - 49 • x > - 49/(-7) • x > 7 Solve: 7 - 5x < 2(x -21).

  11. Problem 10 Solve for x in the equation:

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

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

  14. Problem 12 x2 + (5-1)x - 5 = 16 x2 + 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 ± (b2 – 4ac)1/2]/(2a). Sum is –2b/(2a) = -4. 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.

  15. Problem 13 (81/16)3/4 = (34/24)3/4 = [(3/2)4]3/4 = (3/2)4·3/4 = (3/2)3 = 27/8 Perform the indicated operations on the expression: (16/81)-3/4.

  16. Problem 14 Perform the indicated operations on the expression: .

  17. Problem 15 Note two negative exponents => flipping fraction avoids later trouble with signs. Assuming a > 0 and b > 0, simplify the indicated expression equivalently so that all exponents are positive: (3ab4)-3. (9ab6)-2 32·2a2b6·2 33a3b4·3 3/a

  18. Problem 16 122·3/2a(-2/3)(3/2)b(-4/3)·(3/2) 123a-1b-2 1728/(ab2) Assuming a > 0 and b > 0, simplify the indicated expression equivalently so that all exponents are positive: (144a-2/3b-4/3)3/2 .

  19. Problem 17 (x + 1)(9x2 –24x + 16) - (x2 - 9)(8x + 1) = 9x3 + (-24+9) x2 + (16-24)x +16 – (8x3 + x2 -72x - 9) = x3 - 16x2 + 64x + 25 Simplify to standard form by performing the indicated operations on the expression: (x + 1)(3x - 4)2 - (x - 3)(x + 3)(8x + 1).

  20. Problem 18 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. x2 - 6x - 16 = (x + 2) (x - 8). Factor the expression over the integers or state it is prime if it cannot be factored: x2 - 6x - 16

  21. ____________ Problem 19 Simplify by rationalizing the denominator of the expression: 25 . __________ 22 — (51/2)2

  22. Problem 20 Simplify in factored form:

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