Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
4.8k views
in Expansion formulae by (49.5k points)
closed by

Expand : 

i. (a + 2)(a – 1) 

ii. (m – 4)(m + 6) 

iii. (p + 8) (p – 3) 

iv. (13 + x)(13 – x) 

v. (3x + 4y) (3x + 5y) 

vi. (9x – 5t) (9x + 3t)

vii. (m + (2/3)) (m - (7/3))

viii. (x + (1/x)) (x - (1/x))

ix. ((1/y) + 4) ((1/y) - 9)

1 Answer

+1 vote
by (48.8k points)
selected by
 
Best answer

i. (a + 2)(a – 1) 

= a2 + (2 – 1) a + 2 × (-1) ... [∵ (x + A) (x + B) = x2 + (A + B)x + AB] 

= a2 + a – 2

ii. (m – 4)(m + 6) 

= m2 + (- 4 + 6) m + (-4) × 6 …[∵ (x + a) (x + b) = x2 + (a + b)x + ab] 

= m2 + 2m – 24 

iii. (p + 8) (p – 3) 

= p2 + (8 – 3) p + 8 x (-3) …[∵ (x + a) (x + b) = x2 + (a + b)x + ab] 

= p2 + 5p – 24 

iv. (13 + x) (13 – x) 

= (13)2 + (x – x) 13 + x × (-x) …[∵ (x + a) (x + b) = x2 + (a + b)x + ab] = 169 + 0 × 13 – x2 

= 169 – x2 

v. (3x + 4y) (3x + 5y) 

= (3x)2 + (4y + 5y) 3x + 4y × 5y …[∵ (x + a) (x + b) = x2 + (a + b)x + ab] 

= 9x2 + 9y × 3x + 20y2 = 9x+ 27xy + 20y2

vi. (9x – 5t) (9x + 3t) 

= (9x)2 + [(-5t) + 3t] 9x + (-5t) × 3t …[∵ (x + a) (x + b) = x2 + (a + b)x + ab] 

= 81x+ (-2t) × 9x – 15t2 

= 81x2 – 18xt – 15t2

vii. (m + (2/3)) (m - (7/3))

viii. (x + (1/x)) (x - (1/x))

ix. ((1/y) + 4) ((1/y) - 9)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...