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
1.5k views
in Algebra by (44.5k points)
closed by

Factorize:

(i) 7(2x + 5) + 3 (2x + 5)

(ii) 12x3y4 + 16x2y5 – 4x5y2

2 Answers

+1 vote
by (55.5k points)
selected by
 
Best answer

(i) 7(2x + 5) + 3 (2x + 5)

= (2x + 5) (7 + 3)

(ii) 12x3y4 + 16x2y5 – 4x5y2

= 4x2y2 (3xy2 + 4y3 – x3)

+1 vote
by (50 points)

(I)  7(2x + 5) + 3(2x + 5)

(7 + 3) (2x + 5)

(II) 12 x3y4  + 16 x2y5 - 4 x5y2

= 4 x2y(4xy+ 4y3 - x3)

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.

Categories

...