1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# There are x4+57x+15 pens to be distributed in a class of x2+4x+2 students. Each student should get the minimum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed (xϵN).

A
9x13
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
9x15
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
9x20
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
9x+13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 9x−13Number of pens to be distributed in a class =x4+57x+15number of students =z2+4x+2Divisor polynomial is x2+4a+2 & dividend is x4+57x+15Dividingx2+4x+2)x2−4x+14x4+57x+15x4+4x3+2x2−4x3−2x2+57x−4x3−16x2−8x14x2+65x+1514x2+56x+289x−13 ∴ each students gets x2−4x+14 pens & no. of pens left undistributed is 9x−13

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Second Derivative Test for Local Minimum
MATHEMATICS
Watch in App
Join BYJU'S Learning Program