wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Factorise 2a7128a


A

2a(a+2)(a2)(a22a+4)(a2+2a+4)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

(a+2)(a2)(a22a+4)(a2+2a+4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

2a(a+2)(a2)(a22a+4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

2a(a+2)(a22a+4)(a2+2a+4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

2a(a+2)(a2)(a22a+4)(a2+2a+4)


2a7128a = 2a(a664)

= 2a[(a3)2(8)2]

= 2a(a3+8)(a38)

= 2a(a3+23)(a323)

= 2a(a+2)(a22a+4)(a2)(a2+2a+4)
= 2a(a+2)(a2)(a22a+4)(a2+2a+4)


flag
Suggest Corrections
thumbs-up
12
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Common Factors/ HCF
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon