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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
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
f(x) = (x-1)^300 +(x-1)
To find the remainder after dividing by g(x) = x-2.
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.
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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