Doubt by Saksham
Solution :
Case I
p(x)=x⁴-2x³+3x²-ax+b
g(x)=(x-1)
r(x) = 5
Using Remainder Theorem
g(x) = 0
x-1=0
x=1
p(x)=x⁴-2x³+3x²-ax+b
p(1)=(1)⁴-2(1)³+3(1)²-a(1)+b
p(1) = 1-2+3-a+b
r(x) =2-a+b
5=2-a+b
5-2=-a+b
3=-a+b — (1)
Case II
p(x)=x⁴-2x³+3x²-ax+b
g(x)=(x+1)
r(x) = 19
Using Remainder Theorem
g(x) = 0
x+1=0
x=-1
p(x)=x⁴-2x³+3x²-ax+b
p(-1)=(-1)⁴-2(-1)³+3(-1)²-a(-1)+b
p(-1) = 1+2+3+a+b
r(x) =6+a+b
19=6+a+b
19-6=a+b
13=a+b — (2)
Adding equation (1) and (2)
3+13 = -a+b+a+b
16 = 2b
16/2 = b
8 =b
b=8
putting this value of b in equation (1)
3=-a+8
a=8-3
a=5
Hence, a=5 and b=8
So, p(x) = x⁴-2x³+3x²-5x+8
Now Remainder when p(x) is divided by (x-2)
Again, using remainder theorem
x-2=0
x=2
p(2)=(2)⁴-2(2)³+3(2)²-5(2)+8
p(2) = 16-16+12-10+8
p(2) = 10
∴ When p(x) is divided by x-2 then remainder is 10.