Sims Grade 8 Mathematics.docx

FACTORING AND SPECIAL PRODUCTS There are certain types of products that must be memorized. 

Difference of two squares

The square of the first term minus the square of the second term is equal to the product of the sum and difference of the two terms. (first term)2 – (second term)2 = (first term + second term)(first term – second term) a2 – b2 = (a + b)(a – b) a 2  a and

Note that

b2  b .

If the original problem does not have an explicit base with an exponent of 2, use the positive square root of the expression to find the first and second terms. For example if 4x 2 is given, the positive square root of 4x2, ( 4x 2 ) is 2x, so that we write (2x)2. Example 1 Factor each polynomial. (a) 9x2 – 25 (b) 16x4 – 81

(c) 2v2 - 72u2

Solution: (a) a

b 25 =5

9x 2 = 3x

(b)

a 16 x  4 x 4

2

b 81  9

9x2 – 25

= (3x)2 – (5)2 = (3x – 5)(3x + 5)

16x4 – 81

= (4x2)2 – (9)2 = (4x2 + 9)(4x2 – 9)

But 4x2 – 9 is also difference of two squares so that a 4x 2 = 2x

Substituting we have,

(c)

b 9 3 16x4 – 81

4x2 – 9 = (2x)2 – (3)2 = (2x + 3)(2x – 3)

= (4x2 + 9) (2x + 3)(2x – 3)

2v2 - 72u2 = 2(v2 – 36u2) = 2[(v)2 – (6u)2] = 2(v + 6u)(v – 6u)

Problem 1

a) x2 – 1

b) x2 – y2

c) 4x2 – y2

d) m4 –16

e) 9 – y6

f) 25x2 – 4a2

g) 100 – 49a6

h) s2 - 36

i) y2 – 49

j) 3a2 – 12x2

k) (2x + 3)2 – y2

l) (a+ b)2 – (a – b)2

m) x3 – x

n) 3ax2 – 3ay2

o) 100a2 – 64b2

p) s64 –t22

q) a4 – b2

r) x4 – 4y4

s) 9x2 – 64y2

t) 2x2 – 2y2

u) 44x4 – 99y4

v) 3a3 – 12a

w) a4 – b4

x) 16y12 – 1



Factoring perfect squares

If the middle term of a trinomial is equal to twice the product of the positive square root of the first term and the last terms, then that trinomial is a perfect square, and it is equal to the square of the sum or difference of the positive square root of the first and last terms. a2 + 2ab + b2 = (a + b)2 a2 - 2ab + b2 = (a - b)2 Note that middle term = twice the product of Example Factor each polynomial. (a) x2 – 10x + 25 (b) 16x2 + 24x + 9 Solution: (a)

x 2  x and

25 = 5 so that,

a 2  a and

b2  b = 2ab.

middle term = 2(x)(5) = 10x this match with the middle term of the original problem. x2 – 10x + 25 = (x – 5)2 (b)

9  3 so that,

16 x2  4 x and

middle term = 2(4x)(3) = 24x this match with the middle term of the original problem. 16x2 + 24x + 9 = (4x + 3)2

Problem Factor each polynomial. a) 9x2 – 30xy + 25y2

b) 4x2 + 36x + 81

c)

d) 16x2 – 40xy + 25y2

e) 16x2 + 56x + 49

f) x2 + 18x + 81

x2 + 24x + 144

Sum and Difference of two cubes The sum of two cubes is expressed as a3 + b3 = (first term)3 + (second term)3 Sum of two cubes = [first term + second term][(first term)2 – (first term)(second term) + (second term)2] In general, a3 + b3 = (a + b)(a2 – ab + b2) The difference of two cubes is expressed as a3 - b3 = (first term)3 - (second term)3 Difference of two cubes = [first term - second term][(first term)2 + (first term)(second term) + (second term)2] In general, a3 - b3 = (a - b)(a2 + ab + b2)

Note that

3

a3  a and

3

b3  b .

If the original problem does not have an explicit base with an exponent of 3, use the cube root of the expression to find the first and second terms. For example if 8x3 is given, the cube root of 8x3, ( 3 8x3 ) is 2x, so that we write (2x)3. Example Factor each polynomial. (a) 8x3 – 27 (b) 64x6 + 1 Solution: (a)

3

8x3 = 2x, and

3

27  3 so that,

a 2x

a2 (2x) = 4x2

b 3

ab (2x)(3) = 6x

2

b2 (3)2 = 9

8x3 – 27 = (2x)3 – (3)3 = (2x – 3)[(2x)2 + (2x)(3) + (3)2] = (2x – 3)(4x2 + 6x + 9)

(b)

3

64 x6  4 x2 , and 3 1  1 so that,

64x6 + 1

a 4x2

a2 (4x2)2 = 16x4

b 1

b2 (1)2 = 1

ab 2 (4x )(1) = 4x2

= (4x2)3 + (1)3 = (4x2 + 1)[(4x2)2 - (4x2)(1) + (1)2] = (4x2 + 1)(16x4 - 4x2 + 1)

Problem 2 a

b

a2

ab

Factor each polynomial with the aid of this chart. a) x3 – 1

b) x3 + y3

c) 27x3 – y3

d) m3 – 216

e) 27 – y3

f) 125x3 + 8a3

g) 1000 + 27a3

h) s3 - 64

i) y3 + 125

j) 3a3 – 81x2

k) (2x + 3)3 – y3

l) r3 + 8b3

m) x4 + x

n) 3ax3 – 3ay3

o) 54a3 – 128b3

b2

p) s6 – t3

q) 128a3 – 2b3

r) x3 + 64y6

s) 81x3 + 64y3

t) 2x3 – 2y3

u) 81x3 – 3y3

Strategic Intervention

Rational Expressions, Equations, and Functions Simplifying Rational Expressions  Rational Expressions

Rational Expressions Definition:

Simplifying:

Example:

Solution:

Additional Example 1:

Solution:

Additional Example 2:

Solution:

Additional Example 3:

Solution:

Additional Example 4:

Solution:

Simplify the following rational expressions. If the expression cannot be simplified any further, then simply rewrite the original expression. 1.

15 25

2.



30 36

3.



48 64

4.

26 39

5.

60 x 2 y 5 48 x 5 y 3

6.



7.



8.

49a 4b9 56a 7 b10 5 x3  x  y 

10 x5  x  y 

8c 6  c  d 

2

12c  c  d 

3

9.

x y yx

10.

cd d c

11.

7 3

2  a  b  c  d  6 b  a 

12. 

12  x  y  w  z  6  z  w x  y 

13.

4x  8 x2

14.

x3 5 x  15

15.

x5 x  25

16.

x3 x2  9

17.

a 2  b2 ab

18.

x 2  16 x4

2

19.

49  c 2 c 2  9c  18

20.

x 2  11x  10 100  x 2

21.

x 2  2 x  15 x 2  10 x  21

22.

m2  m  20 m2  m  30

23.

x2  5x  6 x 2  x  12

24.

x 2  7 x  12 x 2  7 x  30

25.

x 2  8 x  12 x 2  13x  42

26.

x 2  7 x  10 x 2  7 x  10

27.

x 2  36 x  12 x  36

28.

x2  8x  16 x2  16

29.

9 x  36 x2  4 x

30.

7 x 2  14 x x2

31.

10 x 2  30 x 5 x 2  10 x

32.

6 x2  8x 9 x3  12 x 2

33.

x2  7 x  6 8x2  8x

34.

4 x 2  20 x x2  4 x  5

35.

6 x 2  24 x  18 4 x 2  8 x  60

36.

5x2  10 x  40 10 x2  30 x  20

37.

4 x 2  17 x  15 5 x 2  13x  6

2

38.

4 x 2  8 x  21 8 x 2  24 x  14

39.

6 x2  5x  4 10 x 2  9 x  2

40.

15x2  4 x  4 5x2  22 x  8

41.

8x2  30 x  7 16 x2  1

42.

9 x 2  25 6 x 2  13x  5

43.

m3  m2  m  1 m3  m  m2 n  n

44.

ax  ay  bx  by ax  ay  2 x  2 y

45.

xy  3x  2 y  6 yz  3z  5 y  15

46.

ab  5a  2b  10 a 2 b  4b  5a 2  20

47.

x3  8 x2

48.

x5 x  125

49.

x3  27 x 2  3x  9

50.

x3  1 x2  x  1

3

Strategic Intervetion

____________________________________________ Ratio of a to b Simplifying Ratios:   

1.

12cm 4m

2.

a b

 a:b 1 m = 100 cm 16 oz= 1 lb 5, 280 ft = 1 mi 1,760 yd = 1 mi

denominators cannot be zero must have the same units must be simplified 6ft 18in

3.

6. The area of a rectangle is 108 cm2. The ratio of the width to the length is 3:4. Find the length and the width.

8. In the diagram,



BD BE . Find BA and BD.  DA EC

24oz 2lb

4.

14ft 6yd

12 in = 1 ft 3 ft = 1 yd

5.

440yd 2mi

7. If the measures of the angles in a triangle have the ratio of 4:5:6, classify the triangle as right, obtuse or acute.

9. In the diagram,

DE BE . Find AC.  AC BC

The ____________________ of two positive numbers a and b is the positive number x that satisfies:

Geometric Mean

so x 2  ab



11. Find the Geometric Mean of: a. 4 and 9 b. 16 and 18 c. 6 and 20 number?

d. 8 is the geometric mean of 4 and what

Notes: 6.2 Use Proportions to Solve Geometry Problems

________________________________________ ____ Proportion: equation that equates two ratios Properties: a. Cross Products a c If  , then b d

a c  , then b d

c. If

ab  b

a c  b d

b. Reciprocal a c If  , then b d

a  c

.

d. If

a c  , then b d

.

Practice: Complete each statement. 1. If

3. If

6 5 6  , then  x y 5 x 7 x4  , then  4 y 4

.

.

2. If

x y x  , then  12 26 y

.

4. If

9 x 11  , then  2 y 2

.

Decide whether the statement is True or False. x 8 y 3  , then  . y 3 x 8 x 8 x 3  , then  . y 3 8 y

5. If

8. If

6. If

x 8 x y  , then  . y 3 8 3

x 8 3 y  , then  . y 3 x 8

9. If

7. If

x 8 x8 y3  , then  . y 3 8 3

Solve for x. 10.

x 9  6 24

11.

4 3  y 3 y 3

12.

5 3  2y  7 y  3

Solve for the variable. 13. MN:MO is 3:4 M x O

14. PQR side lengths: STU side lengths is 1:3 P S 5

x

9 N

R U

36

T

Use the diagram and the given information to find the unknown length. AB AE AB AE 15. Given 16. Given   , find BC. , find BC. BC ED BC ED .

12

Q

Notes: 6.3 Use Similar Polygons

________________________________________ ____ Similar Polygons:

SYMBOL for SIMILAR: ________

Corresponding angles are _____________________________________________ Corresponding sides are _______________________________________________

Writing Similarity Statements: Corresponding <’s:

Proportional Sides:

A BC XYZ







If 2 polygons are _____________, then the ratio of the lengths of 2 corresponding sides is called the ___________________. What is the scale factor of ∆ABC to ∆XYZ? ________________ Practice: 1.) If polygon LMNO ~HIJK , completing proportions and congruence statements.

a. M  __?__

d.

MN  IJ

? JK

b. K  __?__

e.

HK HI  ? LM

c. N  __?__

f.

2.) In the diagram, polygon ABCD ~ GHIJ.

IJ HK  MN ?

Hint: Draw a diagram!!

A 11 D

8

x

x

B

G

y

H

5.5

11 C

J

8

I

a. Find the scale factor of polygon ABCD to polygon GHIJ.

b. Find the scale factor of polygon GHIJ to polygon ABCD.

c.

d. Find the perimeter of each polygon.

Find the values of x and y.

e. Find the ratio to the perimeter of ABCD to perimeter of GHIJ.

If 2 polygons are ___________, then the ratio of their perimeters is equal to the ratios of their ___________. 3.) The ratio of one side of ∆ABC to the corresponding side of similar ∆DEF is 3:5. The perimeter of ∆DEF is 48in. What is the perimeter of ∆ABC?

Strategic Intervention

Measures of Variability I. Range The range for a set of data items is the difference between the largest and smallest values. Although the range is the easiest of the numerical measures of variability to compute, it is not widely used because it is based on only two of the items in the data set and thus is influenced too much by extreme data values. II. Interquartile Range A form of the range that avoids the dependence on extreme values in the data set is the interquartile range (IQR), or Q-spread. This descriptive measure of variability is simply the difference between the third quartile (Q3 ) , or 75%-tile data item, and the first quartile (Q1 ) , or 25%-tile data item. In effect, it is showing the range for the middle 50% of the data and, as such, is not affected by the extreme values in the 3 data set. To calculate Q3 , let i  N where N is the number of data items. If i is 4 not an integer, then the next integer greater than i denotes the position of the 75%-tile; if i is an integer, then the 75%-tile is the average of the data values in positions i and 1 i + 1. Similarly, to calculate Q1 , let i  N and follow the same guidelines as above. 4 Example 1: Given the following data: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Find the IQR. N = 10  i 

i

3 (10)  7.5  Q3 is the 8th data item  Q3  19. Next, 4

1 (10) = 2.5  Q1 is the 3rd data item  Q1  5 . Therefore, IQR = 19-5 = 14. 4

Example 2: Given the following data: 2, 3, 5, 7, 11, 13, 17, 19. Find the IQR.

3 (8)  6  Q3 is the average of the data values in the 6th and 7th 4 13  17 1  15. Next, i  (8)  2  Q1 is the average of the positions  Q3  2 4 35  4. Therefore, IQR = 15-4 = 11. values in the 2nd and 3rd positions  Q1  2 N 8i 

1 III. Average Absolute Deviation from the Mean Obviously, there are limitations in using range or interquartile range as measures of variability. It would seem reasonable that any useful measure of variability should measure the spread around the mean since the mean is the “balance point” of a distribution. If you find the difference between each data item and the mean, you will get negative values for items that are less than the mean and positive values for items greater than the mean. If you then sum up all of these differences, you will get zero; this illustrates a special property of the mean. However, by taking the absolute

value of each difference, you will get the distance of each item from the mean, and the sum of these distances would measure the total spread around the mean. If you were to include more data items, equally spread around the mean, you would increase the total of the distances even though the new distribution might be less variable. Therefore, it is important to divide the total absolute deviation by the number of data items; this will give an average absolute deviation from the mean.

X X Average Absolute Deviation =

N

This average absolute deviation gives the average distance of any data item from the mean and thus is a good measure of spread. IV. Standard Deviation If you were to calculate the average absolute deviation of a distribution using a value other than the mean, you could possibly get a smaller average absolute deviation. This result is one of the reasons that the average absolute deviation is not the best measure of variability. Instead, calculate the average of the squared differences from the mean; this is the variance of a distribution. If you were to calculate the average of the squared differences of a distribution by using a value other than the mean, you would always get a larger value. The mean is the one number that minimizes the average of the squared differences in a distribution.

( X  X ) 2 Variance =   N 2

There are still two slight inconveniences in using variance as our measure of variability. First, variance does not give an estimate of the distance of a typical data from the mean; it is too big. Second, if the data items have a unit of measurement associated with them, then the variance would not have the same unit of measurement; it would have square units. By taking the square root of variance, we get standard deviation, which is the measure of variability that we want.

2 Standard Deviation =  

( X  X ) 2 N

The standard deviation can be calculated in an alternative way. Standard Deviation =  

2 X 2 X N

Example: Given the following histogram, estimate the standard deviation. %/cig 3 2

30% 40% (.5) 10%

20%

0 0

10

20 40 Number of cigarettes

80

Recall that the mean of a histogram can be determined by calculating a “weighted” average using the midpoints of the class intervals and the areas of the blocks. Thus, X  .1(5)  .3(15)  .4(30)  .2(60)  .5  4.5  12  12  29 cigarettes. The standard deviation of a histogram can also be calculated using the midpoints of the class interval, the area of the blocks, and the “weighted” average. Using the first formula, we get: .1(5  29) 2  .3(15  29) 2  .4(30  29) 2  .2(60  29) 2 SD     17.6 cig .1  .3  .4  .2

Using the second formula, we get: .1(5 2 )  .3(15 2 )  .4(30 2 )  .2(60 2 ) SD     29 2  17.6 cig .1  .3  .4  .2

3 Important Note: Some textbooks will give the following formulas for variance and standard deviation: ( X  X ) 2 X 2  N X  Variance = s  N 1 N 1

2

2

( X  X ) 2 X 2  N X Standard Deviation = s   N 1 N 1

2

These formulas should be used when N data items are taken as a sample from a larger population in which the variance and standard deviation of that population are unknown. These formulas give good approximations of the variance and standard deviation of the population.

Practice Sheet – Measures of Variability I. The following are 25 final averages in a math class: 46 49 53 60 61

64 66 66 67 71

(1) What is the range?

72 74 75 76 79

79 79 80 83 88

89 91 94 95 98

(2) What is the interquartile range? II. Given the following data: 5, 7, 11, 12, 13, 18. (1) (2) (3) (4) (5) (6) (7) (8) (9)

What is the mean? What is the average absolute deviation from the mean? What is the median? What is the average absolute deviation from the median? What is the standard deviation? Add 8 to each item. What is the new SD? Subtract 7 from each item. What is the new SD? Multiply each item by 7. What is the new SD? Divide each item by 5. What is the new SD?