Answer :

Given p(x) = x3 + x2 + x


Let the number to be added be “k”.


Then, the new polynomial q(x) = x3 + x2 + x + k


Now, (x – 1) is a factor of x3 + x2 + x + k.


i.e. x = 1 is the root of the polynomial.


Then, put the polynomial to zero we get,


x3 + x2 + x + k = 0


(1)3 + (1)2 + 1 + k = 0


1 + 1+1 + k = 0


k = – 3


Hence, “ – 3” should be added to the polynomial such that (x – 1) is a factor of q(x).


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A square ground hKerala Board Mathematics Part-1

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

By adding a numbeKerala Board Mathematics Part-2

By adding a numbeKerala Board Mathematics Part-2

Find a third degrKerala Board Mathematics Part-2

Find a second degKerala Board Mathematics Part-2