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

If x23x+2 is a factor of f(x)=x4px2+q ,then (p,q)=

A
(4,5)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(4,5)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(5,4)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(5,4)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D (5,4)
Let us first factorize x23x+2 as follows:

x23x+2=x2x2x+2=x(x1)2(x1)=(x1)(x2)

It is given that x23x+2 is a factor of the polynomial f(x)=x4px2+q that is (x1)(x2) are the factors of f(x)=x4px2+q and therefore, x=1 and x=2 are the zeroes of f(x).

Now, we substitute x=1 and x=2 in f(x)=x4px2+q as shown below:

f(1)=14p(1)2+q0=1p+qpq=1....(1)

f(2)=24p(2)2+q0=164p+q4pq=16....(2)

Subtract equation 1 from equation 2:

(4pp)+(q+q)=1613p=15p=153p=5

Substituting the value of p in equation 1, we get:

5q=1q=51q=4

Hence, (p,q)=(5,4).

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