# The polynomial f is divided by g. Find the reminder of the division?f=(x-1)^300 + x-1 g=x-2

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f = (x-1)^300 +x-1

g = x-2

We need to find the remainder of dividing g from f

then f = gQ + r, where r is the remainder

then ,

r = f-gQ

We know that g(2) = 0 , since 2 is a root for g.

Then.

r = f(2) - g(2)Q(2) = f(2)

r = f(2) = 1+2-1 = 2

Then the remainder is 2

f(x) = (x-1)^300 +(x-1)

To find the remainder after dividing by g(x) = x-2.

Solution:

We knowthat

f(x) = g(x)* Q(x) +R, where Q(x) is the quotient after dividing f(x) by g(x) and R is the remainder. So

(x-1)^300 +(x-1)= (x-2)Q(x) +R.............(1)

Putting x=2, in eq(1), we get:

(2-1)^300 +(2-1) = (2-2)Q(2) +R. Or

1^300 +1 = 0*Q(2) +R. Or

1+1 = R. So R = 2 is the remainder.

First, let's recall the rule of division with reminder.

f=g*Q+R

According to the rule, the degree of the polynomial which represents the reminder has to be smaller than the degree of g. In our case, g is a linear function, so the reminder is a constant.

R=a

First, let's find out the roots of g.

x-2=0

x=2

Now, we'll substitute the root of g, into the rule of division with reminder:

f(2)=g(2)*Q+a*2+0, where f(2)=1+2-1=2

2=0+a

**a=2**

**The reminder is R=2.**