Page 72 - Mathematics_Form_Two

Algebra

P 2y 3 S Thus, 2y + 2 5y + 3 is the expanded form

y A B of (2y + 3)(y + 1).
Therefore, (2y + 3)(y + 1) 2y= 2 + 5y + 3.

C
D
1 Figure 4�2: Regions of rectangle PQRS R Example 4�13 - 3).
FOR ONLINE READING ONLY
Q
Mathematics for Secondary Schools When the rectangle PQRS is subdivided Solution 3) = zz 3z + 3) 2(2z+ 4z − 6 - 3)
Expand (z
2)(2z +
(2 -
(z +
2)(2z -
into four rectangles A, B, C and D, as
shown in Figure 4.2, the following can be
2
= 2z −
deduced:
z
2
6
Area of region A =
2
Therefore,
units. 2y y× = 2y square = 2z +− 6.
2
-
3)
2z=
(z
+
2)(2z
z + -
Area of region B
B3 y= ´ 3y = square units.
Area of region C Example 4�14
C 12y= ´ 2y = square units. Find the coefficient of n and n in the
2
Area of region D expansion of (n + 9 )(n + 3 ).
D = 31 3´ = square units.
Solution
The total Area of regions A, B, C, and D (n + 9 )(n + ) 3 = ( nn + ) 3 + ( 9 n + ) 3
= (2y + 2 3y + 2y + 3) square units .
= n 2 3n + + 9n + 27
= (2y + 2 5y + 3) square units. = n + 12n + 7 2
2
From Figure 4.1, the total area is Therefore, the coefficients of n and n 2
(2y + 3)(y + 1) square units. are 12 and 1, respectively.
Thus, (2y+3)(y+1) = 2y +5y+3. Example 4�15
2
2
Expand (6x + ) 5 .
The expression 2y + 2 5y + 3 is called
a quadratic expression expanded from Solution
(2y + 3)(y + 1). . This expansion can also (6x + 5) = 2 (6x + 5)(6x + 5)
be done by multiply each term of 2y+3 = 6 (6xx + 5) 5(6x+ + 5)
2
by y+1 as follows: = 36x + 30x + 30x + 25
=
1) 3(y +
(2y + 3)(y + 1) 2 ( y y ++ 1) = 36x + 2 60x + 25
= 2y + 2 2y + 3y + 3 Therefore, (6x + 5) = 2 36x + 2 60x + 25.
= 2y + 2 5y + 3

66
Student's Book Form Two

11/10/2024 20:11:52
MATHEMATIC F2 v5.indd 66 11/10/2024 20:11:52
MATHEMATIC F2 v5.indd 66