A/ Factor Completely if Possible

1) 6a2b2 – 54b4 6b2(a2 – 9b2) 6b2(a – 3b)(a + 3b)	2) c2 – a2 – 10ab – 25b2 c2 – (a2 + 10ab + 25b2) c2 – (a + 5b)2 [c – (a + 5b)][c + (a + 5b)] [c – a – 5b][c + a + 5b]	3) 81 – m4 (9 – m2)(9 + m2) (3 – m)(3 + m)(9 + m2)
4) 3cx2 + 6cx – 9c 3cx2 + 6cx – 9c 3c(x2 + 2x – 3) 3c(x + 3)(x – 1)	5) 64 – x6 (8 – x3)(8 + x3) (2 – x)(4 + 2x + x2)(2 + x)(4 – 2x + x2)	6) a2b3 – b3 + a2 – 1 b3(a2 – 1) + (a2 – 1) (b3 + 1)(a2 – 1) (b + 1)(b2 – b + 1)(a – 1)(a + 1)
7) 3x2 + 11x + 10 (3x + 5)(x + 2)	8) 25 – x2 – 6x – 9 25 – (x2 + 6x + 9) 25 – (x + 3)2 [5 – (x + 3)][5 + (x + 3)] [5 – x – 3][5 + x + 3] = (2 – x) (8 + x)	9) 7x2 – 27x – 4 (7x + 1)(x – 4)
	10) 3x3 + 24 3(x3 + 8) 3(x + 2)(x2 – 2x + 4)

.

B/ Perform the Indicated Operations. Reduce to Lowest Terms.

1)

.

(2)

.

(3)

.

(4)

.

(5)

.

C/ Solve These Equations and Check Your Answer(s).

1) 12 + 3(m + 4) = 5m – 2 12 + 3m + 12 = 5m – 2 24 + 3m = 5m – 2 – 2m = – 26 becomes m = 13	2) 8(n + 3) + (6 + 2n) = 2(n – 2) + 5(5 + n) 8n + 24 + 6 + 2n = 2n – 4 + 25 + 5n 10n + 30 = 7n + 21 3n = – 9 becomes n = –3
3) 4(x + 2) – 2(5 – x) = x + 8 4x + 8 – 10 + 2x = x + 8 6x – 2 = x + 8 becomes 5x = 10 so x = 2	4) 11 – t2 = 2t2 – t – 6 – 3t2 + 8t + 3 14 = 7t becomes t = 2 but t = 2 makes the last fraction's denominator = 0. There are no solutions.
5) 2x + 1 = 15 or 2x + 1 = – 15 x = 7 or x = –8	6) x2 = 2x + 15 x2 – 2x – 15 = 0 (x – 5)(x + 3) = 0 becomes x = 5 or x = –3
7) 3x2 + 2x = 7 3x2 + 2x – 7 = 0 using the quadratic formula	8) . x2 – 6x + 9 = 4(x – 3) x2 – 10x + 21 = 0 which becomes (x – 7)(x – 3) = 0 so x = 7 or x = 3

9) 3x + 2y = 16 (eq. 1) 7x + y = 19 (eq. 2) becomes y = 19 – 7x 3x + 2(19 – 7x) = 16 3x + 38 – 14x = 16 becomes 11x = 22 x = 2 becomes y = 19 – 7x so y = 5

10) 4r2 + t2 = 25 (eq. 1) 2r + t = 7 (eq. 2) becomes t = 7 – 2r 4r2 + (7 – 2r)2 = 25 4r2 + 49 – 28r + 4r2 = 25 8r2 – 28r + 24 = 0 2r2 – 7r + 6 = 0 (2r – 3)(r – 2) = 0 so r = 3/2 or r = 2 both solutions work.

D/ Solve These Inequalities and Check Your Answers.

1) 3x + 2 < 5x – 6 –2x < –8 so x > 4	2) x2 + 4x > 12 x2 + 4x – 12 > 0 (x + 6)(x – 2) > 0 x < – 6 or x > 2
3) 3(3x + 2) < 2(5x – 4) 9x + 6 < 10x – 8 – x < – 14 becomes x > 14	4) –1 < x < 13/5 and x > 5
5) | 3x – 5| > 22 3x – 5 < – 22 becomes 3x < – 17 so x < –17/3 3x – 5 > 22 becomes 3x > 27 so x > 9

E/ Simplify these Radical Expressions:

1) =	2) = = =
3) =	d)
e)	f)
g)	h)

.

F/ Rationalize these denominators

1)

2)

3)

4)

.

G/ Simplify. Leave no negative exponents

1)

2)

3)

4)

.

H/ Solve these word problems:

(remember to make a "let" statement and a table or diagram when helpful.)

1) The larger of two numbers exceeds the smaller by 4. Eight times the smaller decreased by four times the larger is 12. What are the two numbers?

solution:

let x = the smaller #

x + 4 = the larger #

8x – 4(x + 4) = 12

8x – 4x – 16 = 12

4x = 28

x = 7 and x + 4 = 11

The numbers are 7 and 11.

2) The perimeter of a rectangle is 72 dm. Three times the width exceeds twice the length by 3 cm. Find the length and width of the rectangle.

Solution:

Since the perimeter = 72, half the perimeter = 36

So length + width = 36

Therefore, length = 36 – width.

Let x = the width of the rectangle

so, 36 – x = the length

3x = 2(36 – x) + 3

3x = 72 – 2x + 3

5x = 75 becomes x = 15

so 36 – x = 21.

The rectangle is 15 dm. wide and 21 dm. long.

3) Sam is presently 7 years older than Fred. One year ago, Sam was twice as old as Fred was then. How old is each now?

Let x = Fred's age now.

so x + 7 = Sam's age now.

x + 6 = 2(x – 1)

x = 2x – 8

–x = – 8, so x = 8

x + 7 = 15

Fred is 8 years old and Sam is 15.

.

4) The sum of the squares of two consecutive integers is 221. Find 2 solutions for these integers.

Let x = the 1st integer

x + 1 = the 2nd integer

x ² + (x + 1) ² = 221

x ² + x ² + 2x + 1 = 221

2x ² + 2x – 220 = 0

x ² + x – 110 = 0

(x + 11)(x – 10) = 0

so, x = – 11 or x = 10

The integers are –11 and –10, or 10 and 11.

.

5) A grocer wishes to make a mixture of 80 Kg. of coffee to sell at $4.00 per Kg. He has some coffee that sells for $5.50/kg and cheaper coffee that sells for $3.00/kg. How many kilograms of each type of coffee must he put in the mixture?

Let x = the amount of $5.50 coffee in the mixture.

80 – x = the amount of $3.00 coffee in the mixture.

5.50 x + 3(80 – x) = 320x

5.50 x + 240 – 3x = 320

2.50 x = 80 becomes x = 32

The mixture should be made up of 32 kg. of $5.50 coffee and 48 kg. of $3.00 coffee.

.

6) A bicycle party travels at 25 km/h. At what speed must a second bicycle party travel in order to overtake the first party in 3 hours if they start from the same place 1 hour later than the first party?

If the 2nd party overtakes the 1st, the 2 parties must have covered the same distance. So, since we have values for rate and time for both parties, and distance = rate × time we can set the distances for the 2 parties equal to each other.

Let x = the late party's rate

so 3x = the late party's distance.

3x = 100 becomes x =

The 2nd party must travel at 33 km/hour to overtake the 1st party in 3 hours.

( Algebra MathRoom Index )

(all content © MathRoom Learning Service; 2004 - ).