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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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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.


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

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

Then the remainder is 2


neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

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


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.


giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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


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.


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



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



The reminder is R=2.

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