Question

plzz guys help me.........

Answer 1

f(x)=x^4-2x^3+3x^2-ax+b
putting(x-1)=0
x=1
f(1)=1^4-2(1)^3+3(1)^2-a(1)+b
5=1-2+3-a+b
5=2-a+b
5-2=-a+b
3+a=b
b=3+a
now putting x+1=0
x=-1
f(-1)=(-1)^4-2(-1)^3+3(-1)^2-a(-1)+b
19=1+2+3+a+b
19-6=a+b
13-a=b
b=13-a
now putting equation 1=equation 2
13-a=3+a
13-3=a+a
10=2a
a=5
now putting value of a in equation 1
b=3+5
b=8
putting x-2=0
x=2
f(2)=2^4-2(2)^3+3(2)^2-(5)(2)+8
f(2)=16-16+12-10+8
f(2)=2+8
f(2)=10
so,10 is the remainder
I hope this solution will help you:-)