Let’s resolve into factorsx2 + 4abx-(a2-b2)2

x2 + 4abx-(a2-b2)2 … (1)(a2-b2)2=[(a + b)2(a-b)2]2Given, x2 + (p + q)x + pq=(x + p)(x + q) … (2)Comparing 1 and 2p + q = 4abpq =-(a2-b2)2we know that (a + b)2=a2 + 2ab + b2 … (3)similarly, (a-b)2=a2-2ab + b2 … (4)subtracting 4 from 3(a + b)2-(a-b)2 =a2 + 2ab + b2-( a2-2ab + b2)=4ab∴ (1) can be written asx2 + (a + b)2x-(a-b)2x-[(a + b)2(a-b)2]2=[{x + (a + b)2}{x-(a-b)2}]Apply the formula (a+b)2 = a2 + b2 + 2ab(a-b)2 = a2 + b2 - 2ab=[{x + a2 + b2 + 2ab }{x-( a2 + b2 - 2ab )}]=[{x + a2 + b2 + 2ab }{x- a2 - b2 + 2ab }]∴ the resolved factors are [{x + a2 + b2 + 2ab }{x- a2 - b2 + 2ab }]

Q. 2 P

Let’s resolve into factors

x² + 4abx-(a²-b²)²

Class 8th West Bengal - Mathematics 13. Factorisation of Algebraic Expressions

Answer :

x² + 4abx-(a²-b²)² … (1)

(a²-b²)²=[(a + b)²(a-b)²]²

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = 4ab

pq =-(a²-b²)²

we know that (a + b)²=a² + 2ab + b² … (3)

similarly, (a-b)²=a²-2ab + b² … (4)

subtracting 4 from 3

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

=4ab

∴ (1) can be written as

x² + (a + b)²x-(a-b)²x-[(a + b)²(a-b)²]²

=[{x + (a + b)²}{x-(a-b)²}]

Apply the formula (a+b)² = a² + b² + 2ab

(a-b)² = a² + b² - 2ab

=[{x + a² + b² + 2ab }{x-( a² + b² - 2ab )}]

=[{x + a² + b² + 2ab }{x- a² - b² + 2ab }]

∴ the resolved factors are [{x + a² + b² + 2ab }{x- a² - b² + 2ab }]

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

PREVIOUSLet’s resolve into factors(a + b)2-5a-5b + 6 NEXTLet’s resolve into factorsx2-

Related Videos

Division of Algebraic Expressions36 mins

NCERT | Practice Qs on Addition and Subtraction of Algebraic Expressions42 mins

Division of Algebraic Expressions37 mins

Master Algebraic Identities43 mins

Operations on Algebraic Expressions51 mins

Algebraic Expressions and Identities43 mins

How to Use Algebraic Identities40 mins

Genius Quiz | Algebraic Expressions and Identities30 mins

Champ Quiz | Division of Polynomials33 mins

NCERT | Factorisation Problems38 mins

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation

view all courses