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

H.C.F of (x33x+2) and (x24x+3) is:

A
(x1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(x1)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(x1)(x+2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(x1)(x3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A (x1)
Given p(x)=x33x+2
Let x=1
p(1)=13+2=0. So, (x1) is a factor of p(x)
x2(x1)+x(x1)2(x1)
p(x)=(x2+x2)(x1)
p(x)=(x2+2xx2)(x1)
p(x)=(x1)(x1)(x+2)
And, q(x)=x24x+3
q(x)=x23xx+3
q(x)=x(x3)1(x3)
q(x)=(x3)(x1)
q(x)=(x3)(x1)
Hence, HCF =(x1)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
GCD Long Division
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon