# Algebra Practice Problems for Precalculus and Calculus

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1 Algebra Practice Problems for Precalculus and Calculus Solve the following equations for the unknown x: 1. 5 = 7x x 3 = 5 x (x 3) + x = (4 x) 5 x = 2 x 3 Multiply the indicated polynomials and simplify. 5. (4x 1)( 3x + 2) 6. (x 1)(x 2 + x + 1) 7. (x + 1)(x 2 x + 1) 8. (x 2)(x + 2) 9. (x 2)(x 2) 10. (x 3 + 2x 1)(x 3 5x 2 + 4) Find the domain of each of the following functions in f (x) = 1 + x 12. f (x) = 1 1+x 13. f (x) = 1 x 14. f (x) = 1 1+x 15. f (x) = 1 1+x Given that f (x) = x 2 3x + 4, find and simplify f (3), f (a), f ( t), and f (x 2 + 1). Factor the following quadratics 17. x 2 x x 2 10x x x x 2 + 8x x x x 2 + 7x x 2 2 1

2 Solve the following quadratic equations in three ways: 1) factor, 2) quadratic formula, 3) complete the square 24. x 2 + 6x 16 = x 2 3x 2 = x 2 + 2x 4 = 0 Solve the following smorgasbord of equations and inequalities x = 2x 1 x 2 3 = 2x 29. x 5 = x x x+4 x 3 = x 2 x 2 > 0 Add/Subtract the following rational expressions: x x+2 + x 4 3 x 2 +1 (x 1)(x 2) x3 x 3 Simplify the following rational expressions (if possible): x 2 +x 2 x 2 1 x 2 +5x+6 x 2 3x x x x+1 x 1 Solutions 1. Given that 5 = 7x 16, add 16 to both sides to get 7x = 21. Now divide both sides by 7 to get x = 3. Checking, we see that 7(3) 16 = = Given that 2x 3 = 5 x, add x to both sides and then add 3 to both sides to get 3x = 8. Now divide both sides by 3 to get x = 8/3 = Checking, we see that 2(8/3) 3 = = = 7 3 and 5 (8/3) = = Given that 2 1 (x 3) + x = (4 x), we first simplify the left and right hand sides using the distributive property to get 1 2 x x = x. Combining like terms on both sides gives 2 3 x 2 3 = 29 3x. Now we add 3x and 2 3 to both sides, obtaining 2 9 x = Dividing both sides by 2 9 (or multiplying both sides by 2 9 ) gives x = = Checking we see ( ) that L H S = 2 1 ( ) = = = 78 9 and RH S = ( ) = = = =

3 4. Given that x 5 = x 3 2, we cross-multiply to obtain 5(x 3) = 2x. Distributing the 5 gives 5x 15 = 2x. Subtracting 2x and adding 15 to both sides gives 3x = 15. Dividing both sides by 3 gives x = 5. Checking, we see that 5 5 = 1 and = 2 2 = 1. NOTE: Checking is very important in this kind of problem. When there are x s in the denominator of fractions in equations, it is possible that your final solution doesn t satisfy the original equation (because you would divide by zero) so it is really not a solution to the original problem. In the multiply and simplify problems, we must multiply each term in the left-hand factor with each term in the right-hand factor, and then simplify by combining like terms. In the case of binomial binomial, we can use the so-called FOIL method. 5. (4x 1)( 3x + 2) = 12x 2 + 8x + 3x 2 = 12x x (x 1)(x 2 + x + 1) = x 3 + x 2 + x x 2 x 1 = x (x + 1)(x 2 x + 1) = x 3 x 2 + x + x 2 x + 1 = x (x 2)(x + 2) = x 2 + 2x 2x 4 = x (x 2)(x 2) = x 2 2x 2x + 4 = x 2 4x (x 3 + 2x 1)(x 3 5x 2 + 4) = x 6 5x 5 + 4x 3 + 2x 4 10x 3 + 8x x 3 + 5x 2 4 = x 6 5x 5 + 2x 4 7x 3 + 5x 2 + 8x For a number x to be in the domain of the function f (x) = 1 + x, we require 1 + x 0 (so we don t take the square root of a negative number). Subtracting 1 from both sides of this inequality yields x 1. Thus, in interval notation, the domain is the set [ 1, ). 12. For a number x to be in the domain of the function f (x) = 1+x 1, we require 1 + x = 0 (so we don t divide by zero). Thus, we require x = 1. In interval notation, the domain is the set (, 1) ( 1, ). 13. For a number x to be in the domain of the function f (x) = 1 x, we require x > 0 (so we don t divide by zero or take the square root of a negative number). In interval notation, the domain is the set (0, ). 14. For a number x to be in the domain of the function f (x) = 1 1+x, we require 1 + x > 0 (so we don t divide by zero or take the square root of a negative number). Subtracting 1 from both sides of this inequality yields x > 1. Thus, in interval notation, the domain is the set ( 1, ). 15. For a number x to be in the domain of the function f (x) = 1 1+x 2, we require that 1 + x2 = 0 (so we don t divide by zero). But no matter what x is, 1 + x 2 > 0. Therefore, the domain is R, the set of all real numbers. 16. f (3) = (3) 2 3(3) + 4 = = 4, f (a) = a 2 3a + 4, f ( t) = ( t) 2 3( t) + 4 = t 2 + 3t + 4, and f (x 2 + 1) = (x 2 + 1) 2 3(x 2 + 1) + 4 = x 4 + 2x x = x 4 x By trial & error, we see that: 17. x 2 x 20 = (x 5)(x + 4) 18. x 2 10x + 21 = (x 3)(x 7) 19. x x + 16 = (x + 8)(x + 2) 20. x 2 + 8x 105 = (x + 15)(x 7) 21. 4x x 3 = (4x 1)(x + 3) 22. 2x 2 + 7x + 15 = (2x 2 7x 15) = (2x + 3)(x 5) 23. x 2 2 = (x 2)(x + 2) tricky, tricky, tricky!!! :) 3

4 24. x 2 + 6x 16 = 0 (a) By factoring: x 2 + 6x 16 = 0 implies that (x + 8)(x 2) = 0. Thus, x = 8 or x = 2. (b) By the quadratic formula: x = b ± b 2 4ac 2a = 6 ± 36 4(1)( 16) 2(1) = 6 ± = 6 ± 10 2 = 2 or 8 (c) By completing the square: by adding 16 to both sides of x 2 + 6x 16 = 0, we get x 2 + 6x = 16. Now, adding 9 = 3 2 = (6/2) 2 to both sides makes the left-hand side a perfect square: x 2 + 6x + 9 = 25. We can factor the left hand side to get (x + 3) 2 = 25. Now take the square root of both sides, allowing for the two square roots of 25 on the right hand side to obtain x + 3 = ±5. Subtracting 3 from both sides gives x = 3 ± 5. In other words, x = 8 or x = 2. Checking: ( 8) 2 + 6( 8) 16 = = 0 and (2) 2 + 6(2) 16 = = x 2 3x 2 = 0 (a) By factoring: x 2 3x 2 = 0 implies that x 2 + 3x + 2 = 0 (multiply both sides by -1). (x + 1)(x + 2) = 0. Thus, x = 1 or x = 2. (b) By the quadratic formula: Factoring gives x = b ± b 2 4ac 2a = 3 ± 9 4( 1)( 2) 2( 1) = 3 ± 1 2 = 3 ± 1 2 = 2 or 1 (c) By completing the square: by adding 2 to both sides of x 2 3x 2 = 0, we get x 2 3x = 2. Now, multiply both sides by 1 to get x 2 + 3x = 2. Now, adding 9/4 = (3/2) 2 to both sides makes the left-hand side a perfect square: x 2 + 3x = = 1 4. We can factor the left hand side to get (x )2 = 4 1. Now take the square root of both sides, allowing for the two square roots of 1/4 on the right hand side to obtain x = ± 2 1. Subtracting 3/2 from both sides gives x = 2 3 ± 2 1. In other words, x = 1 or x = 2. Checking: ( 1) 2 3( 1) 2 = = 0 and ( 2) 2 3( 2) 2 = = x 2 + 2x 4 = 0 (a) By factoring: 2x 2 +2x 4 = 0 implies that x 2 + x 2 = 0 (divide both sides by 2). Factoring gives (x +2)(x 1) = 0. Thus, x = 2 or x = 1. (b) By the quadratic formula: x = b ± b 2 4ac 2a = 2 ± 4 4(2)( 4) 2(2) = 2 ± 36 4 = 2 ± 6 4 = 1 or 2 (c) By completing the square: by adding 4 to both sides of 2x 2 + 2x 4 = 0, we get 2x 2 + 2x = 4. Now divide both sides by 2 to give us x 2 + x = 2. Now, adding 1 4 = (1/2)2 to both sides makes the left-hand side a perfect square: x 2 + x = = 4 9. We can factor the left hand side to get (x )2 = 9 4. Now take the square root of both sides, allowing for the two square roots of 4 9 on the right hand side to obtain x = ± 2 3. Subtracting 1/2 from both sides gives x = 2 1 ± 2 3. In other words, x = 1 or x = 2. Checking: 2(1) 2 + 2(1) 4 = = 0 and 2( 2) 2 + 2( 2) 4 = = Squaring both sides of x = 2x 1 gives x = 2x 1. Solving this equation for x gives us x = 1. Checking in the original equation: L H S = 1 = 1, RH S = 2(1) 1 = 1 = Squaring both sides of x 2 3 = 2x gives x 2 3 = 2x. Subtract 2x from both sides to get x 2 2x 3 = 0. The left-hand side can now be factored to give (x 3)(x + 1) = 0, so x = 3 or x = 1. CHECKING in the ORIGINAL equation: L H S = (3) 2 3 = 9 3 = 6 = 2(3) = RH S, however, if you plug x = 1 into either side of this equation, you get the square root of a negative number. Therefore, for us, x = 1 is not a solution (even though the LHS and RHS are equal the imaginary number 2 = i 2). 4

5 29. x 5 = 4 implies that either x 5 = 4 or x 5 = 4. Thus, either x = 9 or x = 1. Checking shows that both of these numbers are solutions: 9 4 = 4 = 4 and 1 5 = 4 = Subtracting 4 from both sides of 2x gives 2x 1. Now divide both sides by 2 to get x 2 1. The logic also works in the other direction: if x 2 1, this will imply that 2x Thus, the solution set is the interval [ 2 1, ). 31. Subtracting 4 from both sides of 2x gives 2x 1. Now divide both sides by -2 and switch the direction of the inequality to get x 2 1. The logic also works in the other direction: if x 2 1, this will imply that 2x Thus, the solution set is the interval (, 2 1 ]. 32. Take the equation x+4 x 3 = 2 and multiply both sides by x 3 to get x + 4 = 2(x 3). Now solve this equation: x + 4 = 2x 6 = x = 10. Checking: = 14/7 = x 2 x 2 > 0 implies that (x 2)(x + 1) > 0. Thus, either x 2 > 0 and x + 1 > 0 OR x 2 < 0 and x + 1 < 0. Thus, either x > 2 and x > 1 OR x < 2 and x < 1. Thus, either x > 2 OR x < 1. The logic also works in the other direction: if x > 2 OR x < 1, then (x 2)(x + 1) > 0 so x 2 x 2 > 0. Thus, the solution set is (, 1) (2, ). 34. We need to get a common denominator. The simplest one to choose is (x + 2)(x 4). Multiplying the top and bottom of the first fraction by x 4 and multiplying the top and bottom of the second fraction by x + 2 and combining the fractions produces: Simplifying the top and bottom gives us: x x x 4 = x(x 4) (x + 2)(x 4) + 3(x + 2) (x + 2)(x 4) = (x2 4x) + (3x + 6) (x + 2)(x 4) x 2 x + 6 x 2 2x 8 If there were a common factor on the top and bottom, we would cancel it out. However, there are no common factors. Therefore, this is our final answer. 35. We need to get a common denominator. The simplest one to choose is (x 1)(x 2)(x 3). Multiplying the top and bottom of the first fraction by x 3 and multiplying the top and bottom of the second fraction by (x 1)(x 2) and combining the fractions produces the following expressions (which are equal to the original): (x 2 + 1)(x 3) (x 1)(x 2)(x 3) x3 (x 1)(x 2) (x 1)(x 2)(x 3) = (x3 3x 2 + x 3) (x 5 3x 4 + 2x 3 ) (x 1)(x 2)(x 3) Simplifying the top and bottom gives us: x 5 + 3x 4 x 3 3x 2 + x 3 (x 1)(x 2 5x + 6) = x5 + 3x 4 x 3 3x 2 + x 3 x 3 6x x 6 Again, we would technically need to check for common factors to simplify this completely. To do this, it is enough to determine whether 1, 2, or 3 are zeros of the polynomial in the numerator (since they are the zeros of the polynomial in the denominator). Let s call the numerator P, so P(x) = x 5 +3x 4 x 3 3x 2 + x 3. P(1) = = 4 = 0, P(2) = = 5 = 0, and P(3) = = 54 = 0. Therefore, there a no common factors, so the answer above is as simple as possible. 36. We can factor the top and bottom to obtain x 2 + x 2 (x + 2)(x 1) x 2 = 1 (x + 1)(x 1) There is a common factor of x 1, so we can cancel this to obtain our final answer: x + 2 x + 1 It should be pointed out that this expression is equal to the first as long as x = 1. 5

6 37. We factor the numerator and denominator to obtain: x 2 + 5x + 6 x 2 3x + 2 = (x + 3)(x + 2) (x 1)(x 2) The numerator and denominator here have no common factors. Thus, this expression cannot be simplified. 38. The quickest approach here is to immediately multiply the top and bottom of this double-decker fraction by (x + 2)(x 1). Doing this, and canceling things, we obtain: Simplifying: ( ) x x x+1 x 1 (x + 2)(x 1) x(x 1) + 3(x + 2)(x 1) = (x + 2)(x 1) (x + 1)(x + 2) x 2 x + 3x 2 + 3x 6 x 2 + 3x + 2 = 4x2 + 2x 6 x 2 + 3x + 2 The top of this can be factored as 2(2x + 3)(x 1), and so has no common factors with the bottom. Therefore, we are done. 6

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