[Mathematic Form 3] Factorising Expressions

Factorising is the reverse process of expansion.

When you factorise an expression, you write it as a product of two or more common factors.

Tip: You may have to find the Highest Common Factor (HCF) for the terms first in order to arrive at an answer.

Eg:

Factorise each of the following:
i) 3p + 6
ii) 8a² - 6ab
iii) ab + ac + bd + cd

Solution:

i) 3p + 6 ( 3 is the HCF)
= 3(p + 2)

ii) 8a² - 6ab (2a is the HCF)
= 2a(4a - 3b)

iii) ab + ac + bd + cd
= a(b + c) + d(b + c)
= (b + c) (a + d)

Factorisation is also done by using the difference of two squares:

(a² - b²) = (a + b) (a - b)

or by your knowledge of perfect squares:

a² + 2ab + b² = (a + b)²
a² - 2ab + b² = (a - b)²

Eg:

Factorise 4p² - 25q²

Solution:

4p² - 25q²
= (2p + 5q) (2p - 5q)

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Factorise the following:

a. 10a + 15 = 5(2a + 3)

b. 12ab - 18b² = 6b(2a - 3b)

c. 4mn + 12mn² = 4mn(1 + 3n)

d. 9x² - 64y² = (3x + 8y) (3x - 8y)

e. 4p² - 100q² = 4(p + 5q) (p - 5q)

f. 12a² - 48b² = 12(a + 2b) (a - 2b)

g. x² + 6xy + 9y² = (x + 3y)²

h. 3p² + 6pq + 3q² = 3(p + q)²

i. a² + ab + 3a + 3b = (a + 3) (a + b)

j. mk - m² + 4k - 4m = (m + 4) (k - m)

k. (a - 7)² - 100 = (a + 3) (a - 17)

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[Mathematic Form 3] Common Factors

An algebraic term is made up of a numerical co-efficient and a variable. A divisor of an algebraic term is called the factor of the term.

Eg: 8p (8 is the numerical co-efficient and p is the variable)

Factors of an algebraic term can be multiplied together to produce the original term.

Eg: The factors of 4b are 1, 2, 4, b, 2b and 4b.

The common factors of algebraic terms can be determined by listing all the factors of the algebraic terms and then choosing the factors that are the same.

Example:

The factors of
6xy = 1, 2, 3, 6, x, y, xy, 2x, 2y, 2xy, 3x, 3y, 3xy, 6x, 6y, 6xy
3y² = 1, 3, y, y², 3y, 3y²

The common factors are 1, 3, y and 3y.

The Highest Common Factor or HCF is 3y because it has the highest value among the common factors.

** The number '1' is a factor and common factor for all algebraic terms. All algebraic terms are also factors of themselves.

[Mathematic Form 3] Pie Chart

A pie chart can be used to represent statistical data. It is easy to remember the pie chart because, just like the most pies, it is in the shape of circle.

A pie chart is divided into sectors, whose angles at the center of circle are proportional to the frequency of a certain category of the data. The angles can also be represented as percentages in a pie chart.

To interpret the data in a pie chart, you have to:

i) Determine the angles at the center of each sector. The angle of each sector can be calculated by using the following formula:

Angle of sector = (Quantity represented by sector X 360^o ) / Total quantity

ii) Note that the bigger an angle, the larger is the frequency or quantity represented
To convert the angle of a sector into a percentage, use the following formula:

Percentage = ( Angle of sector X 100% ) / 360^o

Eg:

The pie chart on the right represent the number of students who play basketball, football, tennis, hockey and netball in the third form of secondary school.

Basketball = 90^o
Football = 100^o
Tennis = 20^o
Netball = 30^o
Hockey = x ^o

a) Find the value of x.
Solution:
x = 360^o - 90^o - 100^o - 20^o - 30^o = 120^o [Sum of angles at the center = 360^o]

b) If the number of students who play netball is 45, calculate the total number of students in the third form.
Solution:
30^o (the number of students who play netball) represent 45 students.
1^o would represent (45/30) students

Hence the total number of students (represented by 360^o) will be
= (45/30) X 360^o
= 540 students

c) How much percentage of the total number of students in the third form play basketball?
Solution:
The percentage of students who play basketball
= (90/360) X 100%
= 25%

[Mathematic Form 3] Mean

The MEAN of a set of data is obtained by adding up all the values of the data and dividing them by the total number of data.

Mean = (Sum of values of data) / (Total number of data)

Eg:

Find the mean of the following sets of data:

a) -8, -5, -3, 0, 0, 1, 2, 8

Solution:

Mean = ( (-8) + (-5) + (-3) + 0 + 0 + 1 + 2 + 8 ) / 8

Mean = - (5/8) or -0.625