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In today’s Mathematics class, We will be looking at Factorisation. We hope you enjoy the class!
FACTORISATION
CONTENT
 Factorization of simple expression
 The difference of two squares
 Factorization of a quadratic expression
FACTORISATION OF SIMPLE EXPRESSION
To factorize an expression completely, take the HCF outside the bracket and then divide each term with the HCF.
Example:
Factorize the following completely.
 8xy + 4x^{2}y
 6ab – 8a^{2}b + 12ab
Solution:
 8xy + 4x^{2}y
8xy = 2 X 2 X 2 X x X y
4x2y = 2 X 2 X xXxX y
HCF = 4xy
8xy + 4x2y = 4xy ( 2 + x)
= 4xy ( 2 + x)
 9a^{2}bc^{3} – 12ab^{2}c^{2}
9a^{2}bc^{3} = 3 X 3 X a X a X c X c X c
12ab^{2}c^{2} = 2 X 2 X 3 X b X b X c X c
HCF = 3abc^{2}
= 3abc^{2 }(3ac – 4b)
EVALUATION
factorize the following expression
 9x^{2}yz^{2} – 12x^{3}z^{3}
 14cd + 35cd^{2}f
 20m^{2}n – 15mn^{2}
FACTORISATION BY GROUPING
To factorize an expression containing four terms, you need to group the terms into pairs. Then factorize each pair of terns.
Example:
factorize ab – 2cb + 2cf – af
Solution:
Group ab and af together and 2cb and 2cf together
i.e. ab – 2cb + 2cf –af = ab – af – 2cb + 2cf
= a( b – f ) 2c( b – f )
= (a – 2c)( b – f)
EVALUATION
factorize these expressions;
 16uv – 12vt + 20mu – 15mt
 ap +aq +bq + bp
 mn – pqpn +mq
FACTORISATION OF QUADRATIC EXPRESSIONS
A quadratic expression has two (2) as its highest power; hence this at times is called a polynomial of the second order. The general representation of quadratic expression is
ax^{ 2} + bx + c where a ≠ 0.
From the above expression, a, b, and c stands for a number and are called constants.
NOTE
 if ax^{ 2 }+bx + c= 0, this is known as quadratic equation
 a is coefficient of x^{2}, b is coefficient of x and c is a constant term.
 When an expression contains three terms, it is known as trinomial.
 To be able to factorize trinomial, we need to convert it to contain four terms.
Examples: factorization of trinomial of the form x^{2} +bx + c.
 Factorise x^{2} +7x +6
Steps:
 Multiply the 1^{st} and the last term (3rd term) of the expression.
 Find two factors of the above multiple such that if added gives the second term (middle) and when multiplied gives the result in step 1.
 Replace the middle term with these two numbers and factorize by grouping.
Solution to example:
X^{2} x 6 = 6x^{2}
Factors: 6 and 1
X^{2} + 6x + x + 6
X(x+6) +1(x+6)
(x+6)(x+1)
EVALUATION
 z^{2} – 2z + 1
 x^{2} +10x – 24
FACTORISATION OF QUADRATIC EQUATIONS OF THE FORM ax^{ 2} +bx +c
Example: 5x^{ 2 }9x +4
Solution:
Product: 5x^{ 2 }x 4 = 20x^{ 2 }
Factors: 5 and 4
Sum: 54 = 9
Hence, 5x^{ 2 }– 9x + 4
5x^{2 } 5x 4x +4
5x(x1)4(x1)
(5x4)(x1)
EVALUATION
 2x^{ 2 }+13x +6
 13d^{ 2 }– 11d – 2
FACTORISATION OF TWO SQUARES
To factorise two squares with a difference, we need to remember the law guiding difference of two squares i.e. x^{ 2 }– y^{2 } = (x + y) (x y).
Examples:
 P^{ 2} – Q^{2 }= (P+Q) (PQ)
 36y^{ 2 }– 1= 6^{ 2 }y^{ 2 }– 1^{ 2}
= (6y)^{2 } – 1^{ 2 } = ( 6y+1) (6y1).
EVALUATION
 121 y^{ 2}
 x^{2}y^{2} – 4^{2}
READING ASSIGNMENT
Essential Mathematics for J.S.S.3 Pg2936
Exam focus for J.S.S CE Pg101105
WEEKEND ASSIGNMENT
 The coefficient of x^{ 2 }in x^{ 2 } + 3x 5 is 3 B. 1 C. 5 D. 2
 Simplify

 A. e^{ 2} – f^{ 2 } (e+f)(ef)
 B. (e+f)(f+e)
 C. (ef)(fe) D. e+f

 Factorize x^{ 2 }+x 6 (x+3)(x+2) B. (x2)(x+3) C. (x+1)(x+5) D. x + 2
 Solve by grouping 5h^{ 2} 20h + h – 4 A. (h4)(5h+1) B. (h+4)(5h1) C. (h+2)(h5) D. h – 4
 49m^{ 2} – 64n^{ 2 }when factorized will be A. (7m+8n)(8m+7n) B. (8m7n)(8m+7n)
 (7m8n)(7m+8n) D. 7m – 8n
THEORY
Factorise the following
 4p^{2} – 12p +9q^{2}
 f^{ 2 }– 2f + 1
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
In our next class, we will be talking about Simple Fractional Expressions. We are very much eager to meet you there.
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