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

If the polynomials (2x3+ax2+3x5) and (x3+x24xa) leave the same reminds when divided by (x - 1), find the value of a.

Open in App
Solution

Case 2 lets divide
2x2+(a+2)x+(a+5)x1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯2x3+ax2+3x5(2x32x2______________0+x2(2+a)+3xx2(a+2)(a+2)x______________+x(3+a+2)5=x(a+5)5x(a+5)(a+5)____________0+a+55=a Remainder 1
Case 2 x2+2x2x1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x3+x24xa(x3x2______________02x24x2x22x______________02xa2x+2____________0a2 Remainder 2
Since Remainders are equal
a=a2
2a=2
a=1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
(a + b)^2 Expansion and Visualisation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon