College Level Math Practice Test TSI, 49% OFF - Free Printable

Problem: Solve the given math practice test questions step by step.

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#### Section 1: Factor the Following Polynomials

1. a. \( 6x^5 - 8x^4 + 2x^3 \)
- Factor out the greatest common factor (GCF):
\[
6x^5 - 8x^4 + 2x^3 = 2x^3(3x^2 - 4x + 1)
\]
- The quadratic \( 3x^2 - 4x + 1 \) can be factored further:
\[
3x^2 - 4x + 1 = (3x - 1)(x - 1)
\]
- Final answer:
\[
6x^5 - 8x^4 + 2x^3 = 2x^3(3x - 1)(x - 1)
\]

2. b. \( x^2 - 25 \)
- This is a difference of squares:
\[
x^2 - 25 = (x - 5)(x + 5)
\]

3. c. \( x^3 + 27 \)
- This is a sum of cubes:
\[
x^3 + 27 = (x + 3)(x^2 - 3x + 9)
\]

4. d. \( ab + 3a + 2b + 6 \)
- Group terms to factor by grouping:
\[
ab + 3a + 2b + 6 = a(b + 3) + 2(b + 3)
\]
- Factor out the common binomial factor:
\[
ab + 3a + 2b + 6 = (a + 2)(b + 3)
\]

5. e. \( 2x^2 + 9x - 5 \)
- Use the AC method or trial and error to factor:
\[
2x^2 + 9x - 5 = (2x - 1)(x + 5)
\]

6. f. \( x^2 + 6x + 9 \)
- This is a perfect square trinomial:
\[
x^2 + 6x + 9 = (x + 3)^2
\]

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#### Section 2: Perform the Indicated Operation

1. a. \( (3x + 7y) + (4x^2 - 3x + 7) + (y - 1) \)
- Combine like terms:
\[
(3x + 7y) + (4x^2 - 3x + 7) + (y - 1) = 4x^2 + (3x - 3x) + (7y + y) + (7 - 1)
\]
\[
= 4x^2 + 0x + 8y + 6
\]
- Final answer:
\[
4x^2 + 8y + 6
\]

2. b. \( (x - 4)(2x + 9) \)
- Use the distributive property (FOIL method):
\[
(x - 4)(2x + 9) = x(2x) + x(9) - 4(2x) - 4(9)
\]
\[
= 2x^2 + 9x - 8x - 36
\]
\[
= 2x^2 + x - 36
\]

3. c. \( (-3xa + 4b)^2 \)
- Expand using the square of a binomial formula \((A - B)^2 = A^2 - 2AB + B^2\):
\[
(-3xa + 4b)^2 = (-3xa)^2 + 2(-3xa)(4b) + (4b)^2
\]
\[
= 9x^2a^2 - 24xab + 16b^2
\]

4. d. \( (4x^2 - 6xy + 9y^2) - (8x^2 - 6xy - y^2) \)
- Distribute the negative sign and combine like terms:
\[
(4x^2 - 6xy + 9y^2) - (8x^2 - 6xy - y^2) = 4x^2 - 6xy + 9y^2 - 8x^2 + 6xy + y^2
\]
\[
= (4x^2 - 8x^2) + (-6xy + 6xy) + (9y^2 + y^2)
\]
\[
= -4x^2 + 0xy + 10y^2
\]
- Final answer:
\[
-4x^2 + 10y^2
\]

5. e. \( \frac{3x^4y^9}{9xy^7} \)
- Simplify by canceling common factors in the numerator and denominator:
\[
\frac{3x^4y^9}{9xy^7} = \frac{3}{9} \cdot \frac{x^4}{x} \cdot \frac{y^9}{y^7}
\]
\[
= \frac{1}{3} \cdot x^{4-1} \cdot y^{9-7}
\]
\[
= \frac{1}{3} \cdot x^3 \cdot y^2
\]
- Final answer:
\[
\frac{x^3y^2}{3}
\]

6. f. \( \frac{3x^4 - 25x^2 - 20}{x - 3} \)
- Use polynomial long division or synthetic division to divide \( 3x^4 - 25x^2 - 20 \) by \( x - 3 \):
- Divide \( 3x^4 \) by \( x \) to get \( 3x^3 \).
- Multiply \( 3x^3 \) by \( x - 3 \) to get \( 3x^4 - 9x^3 \).
- Subtract: \( (3x^4 - 25x^2 - 20) - (3x^4 - 9x^3) = 9x^3 - 25x^2 - 20 \).
- Repeat the process until the remainder is found.
- After performing the division, the quotient is \( 3x^3 + 9x^2 + 2x + 6 \) with a remainder of \( 2 \).
- Final answer:
\[
3x^3 + 9x^2 + 2x + 6 + \frac{2}{x - 3}
\]

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#### Section 3: Simplify the Following Expressions

1. a. \( 3(5 - 7)^4 \)
- Simplify inside the parentheses first:
\[
5 - 7 = -2
\]
- Raise to the power of 4:
\[
(-2)^4 = 16
\]
- Multiply by 3:
\[
3 \cdot 16 = 48
\]
- Final answer:
\[
48
\]

2. b. \( \frac{-6 + |3 - 5|}{2} \)
- Simplify inside the absolute value:
\[
3 - 5 = -2 \quad \text{and} \quad |-2| = 2
\]
- Substitute back:
\[
\frac{-6 + 2}{2} = \frac{-4}{2} = -2
\]
- Final answer:
\[
-2
\]

3. c. \( 45 - 6^2 + 3^2 + \sqrt{1} \)
- Simplify each term:
\[
6^2 = 36, \quad 3^2 = 9, \quad \sqrt{1} = 1
\]
- Substitute back:
\[
45 - 36 + 9 + 1 = 19
\]
- Final answer:
\[
19
\]

4. d. \( \sqrt{\frac{1}{4}} + \sqrt{\frac{9}{4}} \)
- Simplify each square root:
\[
\sqrt{\frac{1}{4}} = \frac{1}{2}, \quad \sqrt{\frac{9}{4}} = \frac{3}{2}
\]
- Add the results:
\[
\frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2
\]
- Final answer:
\[
2
\]

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#### Section 4: Solve Each Linear Equation

1. a. \( 2(x + 3) = x + 5 \)
- Distribute and simplify:
\[
2x + 6 = x + 5
\]
- Subtract \( x \) from both sides:
\[
x + 6 = 5
\]
- Subtract 6 from both sides:
\[
x = -1
\]
- Final answer:
\[
x = -1
\]

2. b. \( \frac{x}{2} + \frac{2}{3} = \frac{3}{4} \)
- Find a common denominator (12) and rewrite the equation:
\[
\frac{6x}{12} + \frac{8}{12} = \frac{9}{12}
\]
- Combine terms:
\[
\frac{6x + 8}{12} = \frac{9}{12}
\]
- Eliminate the denominator:
\[
6x + 8 = 9
\]
- Subtract 8 from both sides:
\[
6x = 1
\]
- Divide by 6:
\[
x = \frac{1}{6}
\]
- Final answer:
\[
x = \frac{1}{6}
\]

3. c. \( P = 2L + 2W \); solve for \( W \)
- Isolate \( W \):
\[
P = 2L + 2W
\]
\[
P - 2L = 2W
\]
\[
W = \frac{P - 2L}{2}
\]
- Final answer:
\[
W = \frac{P - 2L}{2}
\]

4. d. \( \frac{m - 4}{3} - \frac{3m - 1}{5} = 1 \)
- Find a common denominator (15) and rewrite the equation:
\[
\frac{5(m - 4)}{15} - \frac{3(3m - 1)}{15} = 1
\]
\[
\frac{5m - 20 - 9m + 3}{15} = 1
\]
\[
\frac{-4m - 17}{15} = 1
\]
- Eliminate the denominator:
\[
-4m - 17 = 15
\]
- Add 17 to both sides:
\[
-4m = 32
\]
- Divide by -4:
\[
m = -8
\]
- Final answer:
\[
m = -8
\]

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#### Section 5: Solve Each Word Problem

1. a. Maria's Bicycle Trip
- Let \( t \) be the time spent on level roads and hilly roads (since they are equal).
- Distance on level roads: \( d_1 = 18t \)
- Distance on hilly roads: \( d_2 = 10t \)
- Total distance: \( d_1 + d_2 = 98 \)
\[
18t + 10t = 98
\]
\[
28t = 98
\]
\[
t = \frac{98}{28} = 3.5 \text{ hours}
\]
- Total time for the trip:
\[
2t = 2 \cdot 3.5 = 7 \text{ hours}
\]
- Final answer:
\[
\boxed{7}
\]

2. b. Jim's Backpack Coins
- Let \( n \) be the number of nickels and \( d \) be the number of dimes.
- Total coins: \( n + d = 20 \)
- Total value: \( 0.05n + 0.10d = 1.85 \)
- Solve the system of equations:
\[
n + d = 20 \quad \text{(1)}
\]
\[
0.05n + 0.10d = 1.85 \quad \text{(2)}
\]
- Multiply equation (2) by 100 to eliminate decimals:
\[
5n + 10d = 185 \quad \text{(3)}
\]
- Solve equation (1) for \( n \):
\[
n = 20 - d
\]
- Substitute into equation (3):
\[
5(20 - d) + 10d = 185
\]
\[
100 - 5d + 10d = 185
\]
\[
100 + 5d = 185
\]
\[
5d = 85
\]
\[
d = 17
\]
- Substitute \( d = 17 \) back into \( n = 20 - d \):
\[
n = 20 - 17 = 3
\]
- Final answer:
\[
\boxed{3 \text{ nickels, } 17 \text{ dimes}}
\]

3. c. Consecutive Integers
- Let the three consecutive integers be \( x \), \( x + 1 \), and \( x + 2 \).
- Sum of the integers: \( x + (x + 1) + (x + 2) = 3x + 3 \)
- Twice the smallest integer plus 13: \( 2x + 13 \)
- Set up the equation:
\[
3x + 3 = 2x + 13
\]
- Solve for \( x \):
\[
3x - 2x = 13 - 3
\]
\[
x = 10
\]
- The integers are:
\[
x = 10, \quad x + 1 = 11, \quad x + 2 = 12
\]
- Final answer:
\[
\boxed{10, 11, 12}
\]

4. d. Laura's Investments
- Let \( x \) be the amount invested in the mutual fund (8% interest).
- Let \( 24000 - x \) be the amount invested in the CD (9% interest).
- Annual interest from the mutual fund: \( 0.08x \)
- Annual interest from the CD: \( 0.09(24000 - x) \)
- Total annual interest: \( 2020 \)
- Set up the equation:
\[
0.08x + 0.09(24000 - x) = 2020
\]
- Distribute and simplify:
\[
0.08x + 2160 - 0.09x = 2020
\]
\[
-0.01x + 2160 = 2020
\]
\[
-0.01x = 2020 - 2160
\]
\[
-0.01x = -140
\]
\[
x = \frac{-140}{-0.01} = 14000
\]
- Amount invested in the mutual fund: \( x = 14000 \)
- Amount invested in the CD: \( 24000 - x = 24000 - 14000 = 10000 \)
- Final answer:
\[
\boxed{14000 \text{ in mutual fund, } 10000 \text{ in CD}}
\]

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Final Answers:

1. Factoring Polynomials:
- a. \( 2x^3(3x - 1)(x - 1) \)
- b. \( (x - 5)(x + 5) \)
- c. \( (x + 3)(x^2 - 3x + 9) \)
- d. \( (a + 2)(b + 3) \)
- e. \( (2x - 1)(x + 5) \)
- f. \( (x + 3)^2 \)

2. Perform Operations:
- a. \( 4x^2 + 8y + 6 \)
- b. \( 2x^2 + x - 36 \)
- c. \( 9x^2a^2 - 24xab + 16b^2 \)
- d. \( -4x^2 + 10y^2 \)
- e. \( \frac{x^3y^2}{3} \)
- f. \( 3x^3 + 9x^2 + 2x + 6 + \frac{2}{x - 3} \)

3. Simplify Expressions:
- a. \( 48 \)
- b. \( -2 \)
- c. \( 19 \)
- d. \( 2 \)

4. Solve Linear Equations:
- a. \( x = -1 \)
- b. \( x = \frac{1}{6} \)
- c. \( W = \frac{P - 2L}{2} \)
- d. \( m = -8 \)

5. Word Problems:
- a. \( \boxed{7} \)
- b. \( \boxed{3 \text{ nickels, } 17 \text{ dimes}} \)
- c. \( \boxed{10, 11, 12} \)
- d. \( \boxed{14000 \text{ in mutual fund, } 10000 \text{ in CD}} \)