Algebra Review for Math 108 This review targets those aspects of Grade 10 and 11 algebra that are particularly important for doing calculations in Math 108 (Applied Calculus). For most students, the following exercises will be relatively easy. However, if you have not done a Mathematics course recently, you may find these problems challenging. If you feel overwhelmed by the difficulty of these questions, you may want to consider taking a preparatory course such as Math 107, before attempting Math 108. Math 107 is University transferable and is eligible for student loans. If after trying these questions you are unclear regarding the best decision for you, talk to your instructor. Section 1 β Expanding and Simplifying Use proper order of operations to expand and simplify the following expressions: 1. 2. 3. 4. 5. 6. 3(5 β 4π₯) β 4(3π₯ + 1) 6 β 4(1 β 3π₯) + 2π₯(π₯ β 6) (3π₯ + 2)2 (π₯ + β β 4)2 (3π₯ + 3β β 2)2 β (3π₯ β 2)2 (π₯ β 5)3 Section 2 β Factoring. Use any appropriate factoring technique to factor completely: 1. 2. 3. 4. 5. 3π₯ 4 β 15π₯ 3 β 6π₯ 2 π₯ 2 β 7π₯ β 18 5π₯ 3 β 25π₯ 2 + 30π₯ 2π₯ 2 + 5π₯ β 12 π₯ 3 β 4π₯ 2 β 9π₯ + 36 Factor out the greatest common factor and simplify: 6. 8π₯(2π₯ 2 β 5)3 + 48π₯ 3 (2π₯ 2 β 5)2 7. 8π₯(π₯ 2 + 1)3 (4 β 3π₯)3 β 9(4 β 3π₯)2 (π₯ 2 + 1)4 Section 3 β Rational Expressions 1. For what values of x is the expression π₯ 2 +3π₯β18 2. Simplify: 3. Combine 4. Simplify: 5. Simplify: π₯β3 3 3 β π₯+2 π₯ 3 β4π₯ not defined? into a single expression. π₯+ββ7 π₯β7 8π₯(4π₯β1)2 β32π₯ 2 (4π₯β1) (4π₯β1)4 1 1 β 2π₯+2β+7 2π₯+7 Hint: factor the top to start. β Section 4 β Solving Equations 1. Solve the linear equation: 2 3 β 2 π₯ = 4 + 2(3π₯ β 5) 3 2. Solve the following polynomial equations: a) π₯ 2 β 10π₯ β 24 = 0 b) 3π₯ 2 β 8π₯ + 2 = 0 (Use the Quadratic Formula) c) 4π₯ 3 β 12π₯ 2 β 9π₯ + 27 = 0 (Hint: factor) d) (π₯ β 3)(π₯ + 6) = 10 3. Solve the following rational equations: a) b) c) d) 3π₯ 2 β8π₯+5 (π₯β4)2 π₯ 4π₯+3 β =0 1 π₯+2 π₯ 2 β25 π₯ 2 β10π₯+25 2 π₯ 1 + = 2 =0 =0 11 12 4. Solve the following radical equations: a) β7π₯ + 1 β 8 = 0 b) β2π₯ 2 β 4π₯ β 21 β π₯ = 0 Section 5 β Solving Inequalities 1. Solve the linear inequality: 9 β 4(2π₯ β 3) β₯ 3(5 β 2π₯) 2. Solve the following inequalities by simplifying to make one side zero (if needed), finding all values for which the expression is zero or undefined and then using test points on a number line. Give your answers using interval notation. a) π₯ 2 β 8π₯ + 14 < 2 b) (4 β π₯)(π₯ β 1)2 (π₯ + 2)3 > 0 c) d) π₯ 2 +2π₯β3 π₯β5 2π₯β1 π₯+2 β₯0 β€1 Section 6 β Exponents and Radicals 1. Convert the following radical expressions to ones involving rational exponents: 3 a) βπ₯ 5 4 b) οΏ½(2π₯ 2 + 1)3 c) 2βπ₯ 3 + 6βπ₯ 8 2. Convert the following expressions involving rational exponents to radicals: 3 a) π₯ 4 2 2 1 c) (π₯ 2 + π¦ 2 )2 b) π₯ 3 + π¦ 3 4 3. Use your calculator to estimate (β9)3 4. Convert (4β1 + 3β1 )β2 to a fraction. 5. Simplify the following expression: βπ₯ + β3 + βπ₯ 2 + 9 6. Simplify the following expression and give your answer with only positive exponents: π₯ β2 (3π₯ β1 )β4 π₯ β4 7. Factor out the greatest common factor and simplify the following expression: 3 1 6π₯ 2 (4π₯ + 1)2 + 12π₯ 3 (4π₯ + 1)2 8. Rationalize the denominator in the following expression: 3 βπ₯ β β3 9. Rationalize the numerator and simplify the following expression: β2π₯ β β10 π₯β5 Answers Section 1: 1. 11 β 24π₯ 2. 2π₯ 2 + 2 5. 9β2 + 18π₯β β 12β 6. π₯ 3 β 15π₯ 2 + 75π₯ β 125 3. 9π₯ 2 + 12π₯ + 4 4. π₯ 2 + β2 + 16 + 2π₯β β 8π₯ β 8β Section 2: 1. 3π₯ 2 (π₯ 2 β 5π₯ β 2) 2. (π₯ β 9)(π₯ + 2) 5. (π₯ β 3)(π₯ + 3)(π₯ β 4) 6. 8π₯(2π₯ 2 β 5)2 (8π₯ 2 β 5) 3. 5π₯(π₯ β 2)(π₯ β 3) 4. (2π₯ β 3)(π₯ + 4) 7. (π₯ 2 + 1)3 (4 β 3π₯)2 (β33π₯ 2 + 32π₯ β 9) Section 3: 1. π₯ = 0, 2, β2 3. 5. 2. π₯ + 6, π₯ β 3 β3β (π₯+ββ7)(π₯β7) β2 (2π₯+2β+7)(2π₯+7) Section 4: 1. π₯ = 8 9 2. a) π₯ = 12, β2 c) π₯ = 3, 3. a) π₯ = 1, 3 , 2 c) π₯ = β5 4. a) π₯ = 9 5 3 β 3 2 4. , ββ 0 β8π₯ (4π₯β1)3 b) π₯ = 2.39, 0.28 d) π₯ = 4, β7 b) π₯ = 3, β1 d) π₯ = 24 5 b) π₯ = 7 Section 5: 1. π₯ β€ 3 2. a) (2, 6) c) [β3, 1] βͺ (5, β) Section 6: 5 1. a) π₯ 3 4 2. a) βπ₯ 3 3. 18.7208 4. 144 6. π₯6 8. 49 81 3(βπ₯+β3 ) π₯β3 b) (β2, 1) βͺ (1, 4) d) (β2, 3] 3 3 b) (2π₯ 2 + 1)2 3 c) 2π₯ 4 + 6π₯ 4 3 b) βπ₯ 2 + οΏ½π¦ 2 c) οΏ½π₯ 2 + π¦ 2 5. It cannot be simplified any further. 1 7. 6π₯ 2 (6π₯ + 1)(4π₯ + 1)2 9. 2 β2π₯+β10
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