Question

x power 100 divided by x square minus 3x +2​

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)

Answer 2

As per the reminder theory for any 

polynomial equation 

p(x)=g(x).q(x)+r(x)→(i)

where p(x),g(x)= polynomials 

q(x)= quotient 

r(x)= reminder 

Assuming r(x)=Ax+B

here p(x)=x100,g(x)=x2−3x+2

Simplifying g(x)

g(x)=x2−3x+2

=x2−2x−x+2

=(x−2)(x−1)

Substituting in eqn (i)

p(x)=(x−2)(x−1).q(x)+Ax+B

taking (x)=1

p(1)=(1−2)(1−1).q(1)+A(1)+B

1100=0+A+B

A+B=1→(i) 

taking x=2

p(2)=(2−2)(2−1).q(x)+A(2)+B

2100=0+2A+B

2A+B=2100→(ii)

from eqn (i) B

Hope it helps u

I just studied this yesterday

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