CONJUGATE BINOMIALS
Binomials which have the same terms but opposite signs between the terms are called conjugate binomials.
Examples :
(a + b) and (a - b)
(p - q) and (p + q)
(x + 4) and (x - 4)
Multiplying Conjugate Binomials
We can multiply two conjugate binomials (a + b) and (a - b) using FOIL method.
(a + b)(a - b) = a2 - ab + ab - b2
(a + b)(a - b) = a2 - b2
The difference of two squares a2 and b2 is equal to the product of two conjugate binomials (a + b) and (a - b).
In other words, the factored form of (a2 - b2) is equal to
(a + b)(a - b)
And a2 - b2 = (a + b)(a - b) is considered as algebraic identity.
Hence, a2 - b2 = (a + b)(a - b) is true for any values of a and b.
Solving Problems on Conjugate Binomials
Problems 1-10 : Expand and simplify.
Problem 1 :
(x + 2)(x - 2)
Solution :
= (x + 2)(x - 2)
= x2 - 22
= x2 - 4
Problem 2 :
(2x + 1)(2x - 1)
Solution :
= (2x + 1)(2x - 1)
= (2x)2 - 12
= 22x2 - 1
= 4x2 - 1
Problem 3 :
(2 + √3)(2 - √3)
Solution :
= (2 + √3)(2 - √3)
= 22 - (√3)2
= 4 - 3
= 1
Problem 4 :
(√5 - 7)(√5 + 7)
Solution :
= (√5 - 7)(√5 + 7)
= (√5)2 - 72
= 5 - 49
= -44
Problem 5 :
(2√3 + 1)(2√3 - 1)
Solution :
= (2√3 + 1)(2√3 - 1)
= (2√3)2 - 12
= 22(√3)2 - 1
= 4(3) - 1
= 12 - 1
= 11
Problem 6 :
(√5 - √2)(√5 + √2)
Solution :
= (√5 - √2)(√5 + √2)
= (√5)2 - (√2)2
= 5 - 2
= 3
Problem 7 :
(2√a - 3)(2√a + 3)
Solution :
= (2√a - 3)(2√a + 3)
= (2√a)2 - 32
= 22(√a)2 - 9
= 4a - 9
Problem 8 :
(√x + 2√y)(√x - 2√y)
Solution :
= (√x + 2√y)(√x - 2√y)
= (√x)2 - (2√y)2
= x - 22(√y)2
= x - 4y
Problem 9 :
(2x + 3y)(2x - 3y)
Solution :
= (2x + 3y)(2x - 3y)
= (2x)2 - (3y)2
= 22x2 - 32y2
= 4x2 - 9y2
Problem 10 :
(x2 - 5y)(x2 + 5y)
Solution :
= (x2 - 5y)(x2 + 5y)
= (x2)2 - (5y)2
= x4 - 52y2
= x4 - 25y2
Problems 11-20 : Factor the given expression.
Problem 11 :
x2 - y2
Solution :
= x2 - y2
= (x + y)(x - y)
Problem 12 :
x2 - 4
Solution :
= x2 - 4
= x2 - 22
= (x + 2)(x - 2)
Problem 13 :
y2 - 9
Solution :
= y2 - 9
= y2 - 32
= (y + 3)(y - 3)
Problem 14 :
4x2 - 25
Solution :
= 4x2 - 25
= 22x2 - 52
= (2x)2 - 52
= (2x + 5)(2x - 5)
Problem 15 :
a2 - 25b2
Solution :
= a2 - 25b2
= a2 - 52b2
= a2 - (5b)2
= (a + 5b)(a - 5b)
Problem 16 :
4p2 - 9q2
Solution :
= 4p2 - 9q2
= 22p2 - 32q2
= (2p)2 - (3q)2
= (2p + 3q)(2p - 3q)
Problem 17 :
a2 - 3
Solution :
= a2 - 3
= a2 - (√3)2
= (a + √3)(a - √3)
Problem 18 :
x - 16
Solution :
= x - 16
= (√x)2 - 42
= (√x + 4)(√x - 4)
Problem 19 :
y - 5
Solution :
= y - 5
= (√y)2 - (√5)2
= (√y + √5)(√y - √5)
Problem 20 :
x4 - y4
Solution :
= x4 - y4
= (x2)2 - (y2)2
= (x2 + y2)(x2 - y2)
= (x2 + y2)(x + y)(x - y)
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Recent Articles
-
Simplifying Algebraic Expressions with Fractional Coefficients
May 17, 24 08:12 AM
Simplifying Algebraic Expressions with Fractional Coefficients -
The Mean Value Theorem Worksheet
May 14, 24 08:53 AM
The Mean Value Theorem Worksheet -
Mean Value Theorem
May 14, 24 02:48 AM
Mean Value Theorem
