Question

answer question no 2 please

Answer 1

Let p(x)=x100 and q(x)=x2–3x+2=x2–2x–x+2=x(x–2)–1(x–2)=(x–2)(x–1)

Let Q(x)

and r(x) be the quotient and remainder when p(x) is divided by q(x). Hence by division algorithm, we have p(x)=q(x)⋅Q(x)+r(x)

That is, x100=(x–2)(x–1)Q(x)+r(x)

Let r(x)=(ax+b)

where 0<degr(x)<2

x100=(x–2)(x–1)Q(x)+(ax+b)(1)

Put x=1

in Equation (1), we get 1=0+(a+b)

Hence (a+b)=1(2)

Now put x = 2 in equation (1), we get

2100=0+(2a+b)

2a+b=2100(3)

Solving (2)

and (3), we get a=(2100–1) and b=2–2100

Therefore the remainder =(ax+b)=(2100–1)x+(2–2100)