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THIRD TERM
SUBJECT: MATHEMATICS CLASS: JSS3
SCHEME OF WORK
WEEKS TOPIC
1 | Geometric Construction |
2 | Tangent of an Angle |
3 | Measure of Central Tendency |
4 | Variation |
5 | Statistics |
6 | Factorization: Common Factors |
7 | Factorization of Quadratic Equations |
8 | Revision |
9 | Revision |
10 | Examination |
REFERENCE BOOKS
New General Mathematics by M. F Macrae et al bk 3
Essential Maths by AJS OluwasanmiBk 3
WEEK ONE
GEOMETRICAL CONSTRUCTION
Using ruler and compasses
Remember the following when making geometrical constructions.
1. Use a hard pencil with a sharp point. This gives thin lines which are more accurate.
2. Check that your ruler has good straight edge. A damaged ruler is useless for construction work.
3. Check that your compasses are not too loose. Tighten loose compasses with a small screw driver.
4. All construction lines must be seen. Do not rub out anything which leads to the final result.
5. Always take great care, especially when drawing a line through a point.
6. Where possible, arrange that the angles of intersection between lines and arcs are about 90^{0}.
Perpendicular bisector of a line segment
The locus of a point which moves so that it is an equal distance from two points, A and B, is the perpendicular bisector of the line joining A and B.
Perpendicular means at right angles to.
Bisector means cuts in half.
To construct this locus, you do the following (try this yourself on a piece of paper):
Draw the line segment XY.
Put your compass on X and set it to be over half way along the line. Draw an arc.
Without adjusting your compass put it on Y and draw another arc.
Label these points A and B.
Draw a straight line through A and B.
The point M where the lines cross is the midpoint of XY. And AB is perpendicular to XY.
Bisecting an angle
V is the vertex of the angle we want to bisect.
Place your compass on V and draw an arc that crosses both sides of the angle.
Label the crossing points A and B.
Place your compass on A and draw an arc between the two sides of the angle.
Without adjusting your compass place it on B and draw another arc that cuts the one you just drew. Label the point where they cross C.
Draw a straight line through V and C.
The line VC bisects the angle. Angles AVC and BVC are equal.
Constructing a 90^{0} Angle
We can construct a 90º angle either by bisecting a straight angle or using the following steps.
Step 1: Draw the arm PA.
Step 2: Place the point of the compass at P and draw an arc that cuts the arm at Q.
Step 3: Place the point of the compass at Q and draw an arc of radius PQ that cuts the arc drawn in Step 2 at R.
Step 4: With the point of the compass at R, draw an arc of radius PQ to cut the arc drawn in Step 2 at S.
Step 5: With the point of the compass still at R, draw another arc of radius PQ near T as shown.
Step 6: With the point of the compass at S, draw an arc of radius PQ to cut the arc drawn in step 5 at T.
Step 7: Join T to P. The angle APT is 90º.
Constructing a 30^{0} Angle
We know that: ½ of 60^{0 }= 30^{0}
So, to construct an angle of 30º, first construct a 60º angle and then bisect it. Often, we apply the following steps.
Step 1: Draw the arm PQ.
Step 2: Place the point of the compass at P and draw an arc that passes through Q.
Step 3: Place the point of the compass at Q and draw an arc that cuts the arc drawn in Step 2 at R.
Step 4: With the point of the compass still at Q, draw an arc near T as shown.
Step 5: With the point of the compass at R, draw an arc to cut the arc drawn in Step 4 at T.
Step 6: Join T to P. The angle QPT is 30º.
EVALUATION
Constructing a 60^{0} Angle
We know that the angles in an equilateral triangle are all 60º in size. This suggests that to construct a 60º angle we need to construct an equilateral triangle as described below.
Step 1: Draw the arm PQ.
Step 2: Place the point of the compass at P and draw an arc that passes through Q.
Step 3: Place the point of the compass at Q and draw an arc that passes through P. Let this arc cut the arc drawn in Step 2 at R.
Step 4: Join P to R. The angle QPR is 60^{0}, as the ∆PQR is an equilateral triangle
WEEK TWO
TANGENT OF AN ANGLE
The tangent is one of the three basic trigonometric functions, the other two being sine and cosine. These functions are essential to the study of triangles and relate the angles of the triangle to its sides. The simplest definition of the tangent uses the ratios of the sides of a right triangle, and modern methods express this function as the sum of an infinite series. Tangents can be calculated directly when the lengths of the sides of the right triangle are known and can also be derived from other trigonometric functions.
Step 1
Identify and label the parts of a right triangle. The right angle will be at vertex C, and the side opposite it will be the hypotenuse h. The angle θ will be at vertex A, and the remaining vertex will be B. The side adjacent to angle θ will be side b and the side opposite angle θ will be side a. The two sides of a triangle that are not the hypotenuse are known as the legs of the triangle.
Step 2
Define the tangent. The tangent of an angle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. In the case of the triangle in Step 1, tan θ = a/b.
Step 3
Determine the tangent for a simple right triangle. For example, the legs of an isosceles right triangle are equal, so a/b = tan θ = 1. The angles are also equal so θ = 45 degrees. Therefore, tan 45 degrees = 1.
Step 4
Derive the tangent from the other trigonometric functions. Since sine θ = a/h and cosine θ = b/h, then sine θ / cosine θ = (a/h) / (b/h) = a/b = tan θ. Therefore, tan θ = sine θ / cosine θ.
The ratio called tangent (tan) of an acute angle in a right angled triangle is defined as the ration between the side opposite the angle and the side adjacent to the angle.
Tan A = a/b Tan B =b/a
Example 1 Find the angle A
First
Tan A = ^{3}/_{4} = 0.75
We need to use the inverse function for tan, tan^{-1}, to find the angle. This function is on the same key on the calculator as the tan function (shift tan).
We use the following sequence of commands:
shift – tan^{-1} 0.75 = 37º
Try the following on your calculator to see the difference between tan and tan-1:
angle → ratio ratio → angle
tan 37^{0} = 0.75 tan^{–}1 0.75 = 37^{0}
EVALUATION
- Find the side b
tan 37^{0 }= ^{4}/_{b}
tan 37^{0} · b = 4
0.75· b = 4
b=5.3
WEEK THREE
MEASURE OF CENTRAL TENDENCY
A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data. As such, measures of central tendency are sometimes called measures of central location. They are also classed as summary statistics. The mean (often called the average) is most likely the measure of central tendency that you are most familiar with, but there are others, such as the median and the mode.
The mean, median and mode are all valid measures of central tendency, but under different conditions, some measures of central tendency become more appropriate to use than others. In the following sections, we will look at the mean, mode and median, and learn how to calculate them and under what conditions they are most appropriate to be used.
Mean (Arithmetic)
The mean (or average) is the most popular and well known measure of central tendency. It can be used with both discrete and continuous data, although its use is most often with continuous data (see our Types of Variable guide for data types). The mean is equal to the sum of all the values in the data set divided by the number of values in the data set. So, if we have n values in a data set and they have values x_{1}, x_{2}, …, x_{n}, the sample mean, usually denoted by (pronounced x bar), is:
This formula is usually written in a slightly different manner using the Greek capital letter, , pronounced “sigma”, which means “sum of…”:
You may have noticed that the above formula refers to the sample mean. So, why have we called it a sample mean? This is because, in statistics, samples and populations have very different meanings and these differences are very important, even if, in the case of the mean, they are calculated in the same way. To acknowledge that we are calculating the population mean and not the sample mean, we use the Greek lower case letter “mu”, denoted as µ:
The mean is essentially a model of your data set. It is the value that is most common. You will notice, however, that the mean is not often one of the actual values that you have observed in your data set. However, one of its important properties is that it minimizes error in the prediction of any one value in your data set. That is, it is the value that produces the lowest amount of error from all other values in the data set.
An important property of the mean is that it includes every value in your data set as part of the calculation. In addition, the mean is the only measure of central tendency where the sum of the deviations of each value from the mean is always zero.
When not to use the mean
The mean has one main disadvantage: it is particularly susceptible to the influence of outliers. These are values that are unusual compared to the rest of the data set by being especially small or large in numerical value. For example, consider the wages of staff at a factory below:
Staff | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Salary | 15k | 18k | 16k | 14k | 15k | 15k | 12k | 17k | 90k | 95k |
The mean salary for these ten staff is $30.7k. However, inspecting the raw data suggests that this mean value might not be the best way to accurately reflect the typical salary of a worker, as most workers have salaries in the $12k to 18k range. The mean is being skewed by the two large salaries. Therefore, in this situation, we would like to have a better measure of central tendency. As we will find out later, taking the median would be a better measure of central tendency in this situation.
Another time when we usually prefer the median over the mean (or mode) is when our data is skewed (i.e., the frequency distribution for our data is skewed). If we consider the normal distribution – as this is the most frequently assessed in statistics – when the data is perfectly normal, the mean, median and mode are identical. Moreover, they all represent the most typical value in the data set. However, as the data becomes skewed the mean loses its ability to provide the best central location for the data because the skewed data is dragging it away from the typical value. However, the median best retains this position and is not as strongly influenced by the skewed values. This is explained in more detail in the skewed distribution section later in this guide.
Median
The median is the middle score for a set of data that has been arranged in order of magnitude. The median is less affected by outliers and skewed data. In order to calculate the median, suppose we have the data below:
65 | 55 | 89 | 56 | 35 | 14 | 56 | 55 | 87 | 45 | 92 |
We first need to rearrange that data into order of magnitude (smallest first):
14 | 35 | 45 | 55 | 55 | 56 | 56 | 65 | 87 | 89 | 92 |
Our median mark is the middle mark – in this case, 56 (highlighted in bold). It is the middle mark because there are 5 scores before it and 5 scores after it. This works fine when you have an odd number of scores, but what happens when you have an even number of scores? What if you had only 10 scores? Well, you simply have to take the middle two scores and average the result. So, if we look at the example below:
65 | 55 | 89 | 56 | 35 | 14 | 56 | 55 | 87 | 45 |
We again rearrange that data into order of magnitude (smallest first):
14 | 35 | 45 | 55 | 55 | 56 | 56 | 65 | 87 | 89 | 92 |
Only now we have to take the 5th and 6th score in our data set and average them to get a median of 55.5.
Definition of Mode
A statistical term that refers to the most frequently occurring number found in a set of numbers. The mode is found by collecting and organizing the data in order to count the frequency of each result. The result with the highest occurrences is the mode of the set.
Other related terms include the mean, or the average of a set; and the median, or the middle value in a set.
For example, in the following list of numbers, 16 is the mode since it appears more times than any other number in the set:
3, 3, 6, 9, 16, 16, 16, 27, 27, 37, 48
A set of numbers can have more than one mode (this is known as bimodal) if there are multiple numbers that occur with equal frequency, and more times than the others in the set.
3, 3, 3, 9, 16, 16, 16, 27, 37, 48
In this example, both the number 3 and the number 16 are modes. If no number in a set of numbers occurs more than once, that set has no mode:
3, 6, 9, 16, 27, 37, 48
Range
The range of a set of data is the difference between the highest and lowest values in the set.
Cheryl took 7 math tests in one marking period. What is the range of her test scores?
89, 73, 84, 91, 87, 77, 94
Solution
Ordering the test scores from least to greatest, we get:
73, 77, 84, 87, 89, 91, 94
highest – lowest = 94 – 73 = 21
The range of these test scores is 21 points.
Example
The Jaeger family drove through 6 mid-western states on their summer vacation. Gasoline prices varied from state to state. What is the range of gasoline prices?
$1.79, $1.61, $1.96, $2.09, $1.84, $1.75
Solution
Ordering the data from least to greatest, we get:
$1.61, $1.75, $1.79, $1.84, $1.96, $2.09
highest – lowest = $2.09 – $1.61 = $0.48
The range of gasoline prices is $0.48.
Summary
The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.
EVALUATION
1. Find the average range of the following percentages
84%, 56%, 72%, 64%, 33%, 49%.
2. A marathon race was completed by 5 participants. What is the range of times given in hours given thus, 2.7 hr, 8.3 hr, 3.5 hr, 5.1 hr, 4.9 hr?
WEEK FOUR
VARIATION
Direct Variation
If a person buys some packets of sugar, the total cost is proportional to the number of packets bought.
The cost of 2 packets at Nx per packet isN2x.
The cost of 3 packets at Nx per packet isN3x.
The cost of n packets at Nx per packet is Nnx.
The ratio of total cost to number of packets is the same for any number of packets bought.
This is an example of direct variation, or direct proportion. The cost, C, varies directly with the number of packets, n. In the second example, the mass, M, varies directly with the length, L.
The symbol ∝ means ‘varies with’ or ‘is proportional to’. The statements in the previous paragraph are written:
C ∝ n
M ∝ L
M ∝ L really means that the ratio M/L is constant (i.e. stays the same).
Example
1 packet of sugar costs x naira. What will be the cost of 20 packets of sugar?
Cost varies directly with the number of packets bought.
Cost of 1 packet = x naira
Cost of 20 packets = 20 X x naira
= 20x naira
Example
C ∝ n and C = 5 when n = 20. Find the formula connecting C and n.
C ∝ n means C/n = k.
0r C = kn
C = 5 when n = 20
hence 5 = k V 20
k = ¼
Thus, C = ¼n is the formula which connects C and n.
A formula such as C = ¼n is often known as a relationship between the variables C and n.
Exercise
1. 1m of wire has a mass of x g. What is the mass of 25 m of the same wire?
2. A man cycles 15 Km in 1 hour. How many will he cycle in t hours in he keeps up the same rate?
3. Eggs cost N25 each, how many will n eggs cost?
Answers
1. 25x g 2. 15t km 3. N25n
Inverse Variation
Inverse variation is the relationship of two variables such that a variable increases in its value as the other variable decreases and vice-versa i.e the two variables are inversely proportional to each other. In other words, it is defined as the mathematical expression that shows the relationship between two variables whose product is a constant.
The Inverse Variation Formula is,
Y = k/x
Some solved problems on inverse variation are given below:
Examples
Question 1: If y varies inversely with x and when y = 100, x = 25. What is the value of y when x = 10 ?
Solution
Given,
y = 100
x = 25
The inverse variation formula is,
y = kx
100 = k25
k = 100 X 25
k = 2500
Now,
x = 10
k = 2500
y = kx
y = 250010
y = 250
Question
The time taken to reach the church is inversely proportional to the driving speed. If traveled at the speed of 30 miles per hour, it takes you 2 hours to reach the church. How long will it take to reach the church at the speed of 60 miles per hour?
Solution:
Given,
y = 2
x = 30
The direct variation formula is,
y = kx
2 = k30
k = 2 X 30
k = 60
Now,
x = 60
k = 60
y = kx
y = 6060
y = 1
Joint variation
The mass of a sheet of metal is proportional to both the area and the thickness of the metal. Therefore M ∝ At (where M, A and t are the mass, area and thickness). This is an example of joint variation. The mass varies jointly with the area and thickness.
At midday, the temperature, T^{0}C, inside a house is proportional to the outside temperature, S^{0}C, and inversely proportional to the thickness of the house walls, t cm. In theis case,
T ∝ S/t. This is another example of joint variation.
The inside temperature varies directly with the outside temperature and inversely with the wall thickness.
Example
x ∝ y/z. When y = 7 and z = 3, x = 42.
a. Find the relation between x, y and z.
b. Find x when y = 5 and z = 9.
Solution
a. x ∝ y/z
Thus, x = ky/z, where k is a constant.
When y = 7 and z = 3, x = 42.
Thus, 42 = k X 7/3
K = 3 X 42/7 = 18
Hence x = 18y/z
b. When y = 5 and z = 9,
x = 18 X 5/9 = 2 X 5
x = 10
Exercise
1. x ∝ yz. When y = 2 and z = 3, x = 30.
a. Find the relationship between x, y and z.
b. Find x when y = 4 and z = 6.
2. x ∝ y/z. x = 27 when y = 9 and z = 2.
a. Find the relationship between x, y and z.
b. Find x when y = 14 and z = 12.
Partial variation
When a tailor makes a dress, the total cost depends on two things: first the cost of the cloth; secondly the amount of time it takes to make the dress. The cost of the cloth is constant, but the time taken to make the dress can vary. A simple dress will take a short time to make; a dress with a difficult pattern will take a long time. This is an example of partial variation. The cost is partly constant and partly varies with the amount of time taken. In algebraic form, C = a + kt, where C is the cost, t is the time taken and a and k are the constants.
EVALUATION
R is partly constant and partly varies with E. When R = 530, E = 1 600 and when R = 730, E = 3 600.
a. Find the formula which connects R and E.
b. Find R when E = 1 300.
a. From the first sentence,
R = c + kE where c and k are both constants.
Substituting the given values given values gives two equations.
530 = c + 1 600k (1)
730 = c + 3 600k (2)
These are simultaneous equations.
Subtract (1) from (2).
200 = 2 000k
k = 200/2 000
=1/10
Substituting in (1),
530 = c + 1 600 X 1/10
530 = c + 160
Thus, c = 370
Thus, R = 370 + 1/10E is the required formula.
b. R = 370 + E/10
When E = 1 300,
R = 370 + 1 300/10
= 370 + 130
= 500
WEEK FIVE
STATISTICS
Types of Presentation
Good presentation can make statistical data easy to read, understand and interpret. Therefore it is important to present data clearly.
i. There are two main ways of presenting data: presentation of numbers or values in lists and tables;
ii. Presentation using graphs, i.e. picture. We use the following examples to show the various kinds of presentation.
An English teacher gave an essay to 15 students. |
She graded the essays from A (very good), through B, C.D, E to f (very poor). The grades of the students were:
B, C, A, B, A, D, F, E, C, C, A, B, B, E, B
Lists and tables
Rank and order list
Rank order means in order from highest to lowest. The 15 grades are given in rank order below:
A, A, A, B, B, B, B, C, C, C, E, E, F
Notice that all the grades are put in the list even though most of them appear more than once. The ordered list makes it easier to find the following: the highest and lowest grades; the number of students who got each grades; the most common grade; the number of students above and below each grade; and so on.
Frequency table
Frequency means the number of times something happens. For example, three students got grade A.
The frequency of grade A is three. A frequency table, gives the frequency of each grade.
Grade | A | B | C | D | E | F |
frequency | 3 | 5 | 3 | 1 | 2 | 1 |
Graphical presentation
In most cases, a picture will show the meaning of statistical data more clearly than a list of or table or numbers. The following methods of presentation give the data of the example in picture, or graph, form.
Pictogram
A pictogram uses pictures or drawings to give a quick and easy meaning to statistical data.
Bar chart
A bar chart represents the data as horizontal or vertical bars. The length of each bar is proportional to the amount that it represents.
There are 3 main types of bar charts.
Horizontal bar charts, vertical bar chart and double bar charts.
When constructing a bar chart it is important to choose a suitable scale to represent the frequency.
Pie Chart
Pie charts are useful to compare different parts of a whole amount. They are often used to present financial information. E.g. A Company’s expenditure can be shown to be the sum of its parts including different expense categories such as salaries, borrowing interest, taxation and general running costs (i.e. rent, electricity, heating etc.).
A pie chart is a circular chart in which the circle is divided into sectors. Each sector visually represents an item in a data set to match the amount of the item as a percentage or fraction of the total data set.
Example
A family’s weekly expenditure on its house mortgage, food and fuel is as follows:
Expenses | N |
Mortgage | 300 |
Food | 225 |
Fuel | 75 |
Draw a pie chart to display the information.
Solution
The total weekly expenditure = N300 + N225 + N75 = N600
We can find what percentage of the total expenditure each item equals.
Percentage of weekly expenditure on:
Mortgage = 300/600 X 100% = 50%
Food = 225/600 X 100% = 37.5%
Fuel = 75/600 X 100% = 12.5%
To draw a pie chart, divide the circle into 100 percentage parts. Then allocate the number of percentage parts required for each item.
Note
It is simple to read a pie chart. Just look at the required sector representing an item (or category) and read off the value. For example, the weekly expenditure of the family on food is 37.5% of the total expenditure measured.
A pie chart is used to compare the different parts that make up a whole amount.
EVALUATION
The following is a rank order list of an exam result: 87, 82, 78, 76, 75, 70, 66, 64, 59, 59, 59, 51, 49, 48, 41.
a. How many students took the exam?
b. What was the highest rank?
c. What was the lowest rank?
d. What is the mark of the student who came 6th?
e. What is the position of the student who got 76 marks?
f. Three students got 59 marks. What is their position?
g. How many students got less than 75 marks?
WEEK SIX
FACTORIZATION: COMMON FACTORS
a(b + c) = ab + ac
The reverse process, ab + ac = a(b + c), is called taking out the common factor.
Consider the factorisation of the expression 5x + 15.
Clearly, 5x = 5 X x
15 = 3 X 5
∴ HCF = 5
Thus 5x + 15 = 5 X x + 3 X 5
= 5(x + 3)
Note that the common factor 5 has been taken out and placed in front of the brackets. The expression inside the brackets is obtained by dividing each term by 5.
In general:
To factorise an algebraic expression, take out the highest common factor and place it in front of the brackets. Then the expression inside the brackets is obtained by dividing each term by the highest common factor.
Example
Factorise the following
a. 15x – 20
b. 3px + 12qx
c. x^{2 }+ x
d. 8x^{2}y + 6xy^{2}
Solution:
a. 15x – 20 {HCF = 5}
= 5(3x – 4)
b. 3px + 12qx {HCF = 3x}
= 3x(p + 4q)
c. x^{2 }+ x {HCF = x}
= x(x + 1)
d. 8x^{2}y + 6xy^{2 } {HCF = 2xy}
= 2xy(4x + 3y)
Note:
The process of taking out a common factor is of great importance in algebra. With practice you will be able to find the highest common factor (HCF) readily and hence factorise the given expression.
Example
Fcartorise the following:
a. -8x + 12
b. –x^{2 }– x + ax
c. -6x^{2 }+ 4x – 2px
Solution:
a. -8x
-8x + 12 {HCF = 4}
= 4(-2x + 3)
It is customary to have the first term in the brackets positive. So, we take out -4 as the common factor rather than 4.
So, -8x + 12 = -4(2x – 3)
- –x^{2 }– x + ax {Take out –x as the common factor}
= -x(x + 1 – a)
C. -6x^{2 }+ 4x – 2px {Take out the common factor -2x}
= -2x(3x – 2 + p)
Simplifying calculations by factorization
Example
By factorizing, simplify 79 X 37 + 21 X 37.
37 is a common factor of 79 X 37 and 21 X 37.
79 X 37 + 21 X 37 = 37(79 + 21)
= 37 X 100
= 3 700
Factorise the expression πr^{2 }+ 2πrh. Hence, find the value of πr^{2 }+ 2πrh when π = 22/7, r = 14 and h = 43.
π r^{2 }+2πrh = πr(r + 2h)
when π = 22/7, r = 14 and h = 43,
πr^{2 }+ 2πrh = πr(r + 2h)
= 22/7 X 14(14 + 2 X 43)
= 22 X 2(14 + 86)
= 44 X 100
= 4400
Factorization by grouping
This lesson introduces the technique of factoring by grouping, as well as factoring the sum and difference of cubes. Factoring by grouping builds on the ideas that were presented in the section on factoring the common factor. While there are many different types of grouping, as you will learn in higher algebra courses, all of the grouping problems in this book involve four terms, and they work by grouping the first two terms and the second two terms together. If you do it right, a common factor will always emerge! Remember, the method of grouping is one of trial and error. As always, there is no substitute for practice and experience. Remember, if you can learn this topic now, it will help you later.
Example
Factor x^{3 }+ 2x^{2 }+ 8x + 16
Solution: There are no common factors to all four terms. It is not a trinomial, and nothing discussed so far works to factor this. So, try grouping the first two terms together, and the last two terms together, and factor out the common factor within each grouping as follows:
(x^{3 }+ 2x^{2}) + (8x + 16) = x^{2 }(x + 2)+ 8(x + 2)
Notice that there is a common factor of (x+2) that can be factored out:
= (x + 2)( x^{2 }+ 8)
Example
Factor xy^{ }− 4y^{ }+ 3x − 12
Solution: Again, there are no common factors, and this is not a trinomial. Group the first two and the last two terms together, and factor out the common factor from each grouping:
(xy^{ }− 4y)^{ }+ (3x − 12) = y(x − 4) + 3(x − 4)
Now, take out the common factor, which is (x − 4):
= (x − 4)(y + 3)
Example
Factor xy^{ }− 4y^{ }− 3x + 12
Solution: Group the first two and the last two terms together, and factor out the common factor from each grouping:
(xy^{ }− 4y)^{ }+ (−3x + 12) = y(x − 4) + 3(−x + 4)
This time there is no common factor. Try again, this time factoring a −3 from the last grouping. This works!
xy^{ }− 4y^{ } − 3x + 12 = y(x − 4) − 3(x − 4)
= (x − 4)(y − 3)
Example
Factor xy^{ }− 4y^{ }+ 3x + 12
Solution: Group the first two and the last two terms:
xy^{ }− 4y^{ }+ 3x + 12 = y(x − 4) + 3(x + 4)
At this point, it is important to realize that no common factor resulted. Do not try to factor out something that is not common to both groupings. In fact, there is no way to group this problem to get a common factor. This one cannot be factored by grouping. In fact, it can not be factored by any method. Remember, not all problems can be factored. Remember that the entire process of grouping is one of trial and error, and, as you will see later, there are different types of grouping.
EVALUATION
- Factorise the following: 4xy + 6zy, kx – 6k – 9x + 24
- Factorise 12mn + 8kn and solve for k if m=5 and n=2
WEEK SEVEN
FACTORISATION OF QUADRATIC EXPRESSIONS
You can also factorise quadratic expressions. Remember that factorizing an expression simplifies it in some way. Factorizing is the reverse of expanding brackets.
Factorizing quadratic expressions
Multiply these brackets to remind yourself how to factorise.
( x + 2 ) ( x + 5 ) = x^{2} + 7x + 10
( x + 2 ) ( x + 3 ) = x^{2} + 5x + 6
( x – 3 ) ( x – 5 ) = x ^{2} – 8x + 15
( x + 6 ) ( x – 5 ) = x^{2} + x – 30
( x – 6 ) ( x + 5 ) = x^{2} – x – 30
Factorising
To factorise an expression such x^{2} + 5x + 6, you need to look for two numbers that add up to make 5 and multiply to give 6.
The factor pairs of 6 are:
1 and 6
2 and 3
2 and 3 add up to 5. So: (x +2) (x+3) = x^{2} + 5x + 6
Factorising expressions gets trickier with negative numbers.
Example
Factorise the expression: c^{2}– 3c – 10
Write down the expression: c^{2}– 3c – 10
Remember that to factorise an expression we need to look for common factor pairs. In this example we are looking for two numbers that:
multiply to give -10
add to give -3
Think of all the factor pairs of -10:
1 and -10
-1 and 10
2 and -5
-2 and 5
Which of these factor pairs can be added to get -3?
Only 2 + (-5) = -3
So the answer is:
c^{2} – 3c – 10 = (c + 2)(c – 5)
Factorising the difference of two squares
Some quadratic expressions have only a term in x^{2} and a number such as x^{2} – 25.
These quadratic expressions have no x term.
Using our method to factorise quadratics means we look for two numbers that multiply to make -25 and add to make 0.
The only factor pair that will work are 5 and -5. So:
(x + 5)(x – 5) = x² – 25
Not all quadratic expressions without an x term can be factorized.
Examples
Factorise:
x^{2} – 4 = (x + 2)(x – 2)
x^{2} – 81 = (x + 9)(x – 9)
x^{2} – 9 = (x + 3)(x – 3)
Solving quadratic equations by factorizing
To solve a quadratic equation, the first step is to write it in the form: ax^{2} + bx + c = 0. Then factorise the equation as you have revised in the previous section.
If we have two numbers, A and B, and we know that A × B = 0, then it must follow that either A = 0, or B = 0 (or both). When we multiply any number by 0, we get 0.
Example
Solve the equation: x^{2} – 9x + 20 = 0
Solution
First, factorise the quadratic equation x^{2}– 9x + 20 = 0
Find two numbers which add up to 9 and multiply to give 20. These numbers are 4 and 5.
(x – 4) (x – 5) = 0
Now find the value x so that when these brackets are multiplied together the answer is 0.
This means either (x – 4) = 0 or (x – 5) = 0
So x = 4 or x = 5.
You can check these answers by substituting 4 and 5 in to the equation:
x^{2}– 9x + 20
Substituting 4 gives:
4^{2} – 9 × 4 + 20 = 16 – 36 + 20 = 0
Substituting 5 gives:
5^{2} – 9 × 5 + 20 = 25 – 45 + 20 = 0
Remember these 3 simple steps and you will be able to solve quadratic equations.
Now try this question.
Completing the square
This is another way to solve a quadratic equation if the equation will not factorise.
It is often convenient to write an algebraic expression as a square plus another term. The other term is found by dividing the coefficient of x by 2, and squaring it.
Any quadratic equation can be rearranged so that it can be solved in this way.
Have a look at this example.
Example
Rewrite x^{2} + 6x as a square plus another term.
The coeffient of x is 6. Dividing 6 by 2 and squaring it gives 9.
x^{2} + 6x = (x^{2} + 6x + 9) – 9
= (x + 3)^{2} – 9
Example
We have seen in the previous example that x^{2} + 6x = (x + 3)^{2} – 9
So work out x^{2} + 6x – 2
x^{2} + 6x – 2 = ( x^{2} + 6x + 9 ) – 9 – 2 = (x + 3)^{2} – 11
Now try one for yourself.
Example
Solve x^{2} + 6x – 2 = 0
From the previous examples, we know that x^{2} + 6x – 2 = 0 can be written as (x + 3)^{2} – 11 = 0
So, to solve the equation, take the square root of both sides. So (x + 3)^{2} = 11
x + 3 = + √11
or x + 3 = – √11
x = – 3 + √11
or x = – 3 – √11
x = – 3 + 3.317 or x = – 3 – 3.317 (√11 is 3.317)
x = 0.317 (3 s.f) or x = – 6.317 (3 s.f)
Example
Rewrite 2x^{2} + 20x + 3
Rewrite to get x^{2} on its own.
2( x^{2} + 10x ) + 3
The coefficient of x is 10. Divide 10 by 2, and square to get 25.
= 2 ( ( x + 5)^{2} – 25) + 3
= 2 (x + 5)^{2} – 50 + 3
= 2 (x + 5)^{2} – 47
Now use the previous example to solve 2x^{2} + 20x + 3 = 0
From the previous example, we know that 2x^{2} + 20x + 3 can be rewritten as:
2 (x + 5)^{2} – 47
Therefore, we can rewrite the equation as:
2(x + 5 )^{2} – 47 = 0
2(x + 5 )^{2} = 47
(x + 5 )^{2} = 23.5 (dividing both sides by 2)
Take the square root of both sides.
x + 5 = √23.5
or x + 5 = – √23.5
x = – 5 + √23.5
or x = – 5 – √23.5
x = – 5 + √23.5
or x = – 5 – √23.5
x = – 0.152 (3 s.f) or x = – 9.85 (3 s.f)
EVALUATION
1. x^{2 }+ 4x – 5
2. x^{2 }+ 6x – 7
3. (a + 4)^{2}
4. (3x + y)^{2}
5. 49m^{2 }– n^{2 }
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