Square of a Binomial:
i) (a + b)2 = a2 + 2 * a * b + b2 = a2 + 2 ab + b2
= (a + b)( a + b) .
(or)
ii) (- a – b )2 = a2 + 2 * a * b + b2. = a2 + 2 ab + b2
= (a +
b) ( a + b).
iii) (a – b)2 = a2 – 2* a * b + b2 = a2 – 2ab + b2
= ( a – b )
( a – b ).
Note : The quadratic equation or binomial ax2 + bx + c is a perfect square ; if b2 = 4ac.
Factorize the expressions using Identities:
1) 4x2 +
12xy + 9y2
Answer:
In the given expression 4x2 + 12xy + 9y2
There are two variables: x , y and has three terms.1st , 3 rd terms are Perfect
squares.
The middle term has Positive sign.
1st term : a2 = 4x2 ;
a = 2x
3 rd term : b2 = 9y2 ;
b = 3y
Middle term (12xy) = 2 * 1st term * 3 rd term
= 2 * a * b
=
2 * 2x * 3y = 12xy
By using the Square of Binomial Identity :
(a + b )2 = a2 + 2 ab +
b2 = (a + b) ( a + b)
Therefore, factorization of : 4x2 + 12xy + 9y2
= (
2x + 3y ) 2
=
(2x + 3y) ( 2x + 3y).
2) 4x2 – 4x + 1
Answer:
In the given expression 4x2 –
4x + 1
There are two variables: x , y and has three terms.1st , 3 rd terms are Perfect
squares.
The middle term has Negative sign.
1st term, a2 =
4x2 ; a = 2x
3rd term, b2 = 1; b =
1.
Middle term, (- 4x ) = - 2 * 2x * 1 = -
4x = 2 ab
By using the Square of Binomial Identity :
(a - b )2 = a2 - 2
ab + b2 = ( a – b)( a – b)
Therefore, factorization of 4x2 –
4x + 1
= ( 2x – 1 )2
= ( 2x - 1 ) ( 2x - 1).
3
) w2 – 2w/x + 1/ x2
Answer:
The given expression is
written as : w2 – 2w/x + 1/ x2
;also 1st , 3 rd terms are Perfect
squares . The middle term has Negative sign. The expression is written as:
=
w2 – 2 * w * 1/x + (1/x)2
; it is in the form of :
(a - b )2 = a2 - 2
ab + b2 = ( a – b)( a – b)
Here, a = w , b = 1/x , ab = w/x ; 2 ab = 2w/x.
Therefore ; factorization of :w2 – 2w/x + 1/ x2
=
( w – 1/x)2
=
( w – 1/x)( w - 1/x) .
4)
1/t2 – 8/t + 16
Answer:
The given expression is
written as: 1/t2 – 8/t + 16
;also 1st , 3 rd terms are Perfect
squares . The middle term has Negative sign. The expression is written as:
= 1/t2 – 2 *
1/t * 4 + 42 ; it is in the
form of :
(a - b )2 = a2 - 2
ab + b2 = ( a – b)( a – b)
Here, a =
1/t ; b = 4 ; ab = 4/t ; 2 ab = 8/t .
Therefore ; factorization of : 1/t2 – 8/t + 16
= ( 1/t –
4 )2
= (1/t –
4)( 1/t – 4).
5) p2
+ 169 +26p
Answer:
The given expression is
written as: p2 + 169 +26p
;rearranged as : p2 + 26p + 169
In the given expression
:1st , 3 rd terms are Perfect
squares . The middle term has Positive sign. The expression is written as:
= p2
+ 2 *p * 13 + 132 ; it is in the form of :
(a + b )2 = a2 + 2 ab +
b2 = (a + b) ( a + b)
Here, a = p ; b = 13 ; ab = 13p ;
2ab = 26p.
Therefore; factorization of : p2
+ 169 +26p
= ( p + 13 )2
= ( p + 13 ) ( p + 13 ).
6) (3a – 5b)2 + 2 (3a – 5b)(2b –a) + ( 2b – a )2
Answer:
In this expression : (3a – 5b)2 + 2 (3a – 5b)(2b –a) + ( 2b – a )2
The number of terms are Three ,1st and 3rd terms are Perfect squares . The middle term has Positive sign.The expression is in the form of : a2 + 2 ab + b2
1st term, a2 = ( 3a – 5b )2 ; a = (3a – 5b)
3rd term, b2 = ( 2b – a )2 ; b = ( 2b – a )
Middle term : 2 ab = 2 (3a – 5b)(2b –a)
By using the Square of Binomial Identity :
(a + b )2 = a2 + 2 ab + b2 = ( a + b )( a + b )
Solving : ( a + b ) = (( 3a – 5b ) + ( 2b – a )) ,
= ( 3a – 5b + 2b – a ) = (2a – 3b )
Therefore, factorization of : (3a – 5b)2 + 2 (3a – 5b)(2b –a) + ( 2b – a )2
= ( ( 3a – 5b ) + ( 2b – a ) )2 ;it is in the form: ( a + b)2
= ( 2a – 3b )2 ; since : ( a + b = 2a - 3b )
= (2a – 3b ) ( 2a – 3b) .
No comments:
Post a Comment