How To Find Remainder in case of Polynomial

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Number = Divisor * Dividend + Remainder
Any polynomial of degree n when divided by any other polynomial of degree m then remainder will be polynomial of degree less than m.
when x^3 is divided by x^2+1 then remainder will of form ax + b here a and b are constant so remainder could be linear or constant also when a=0
Now We will Apply this in Questions
Let P(x) be a polynomial in x. When P(x) is divided by (x2) the remainder is 8. When P(x) is divided by (x+2) the remainder is 4. What is the remainder when P(x) is divided by x^2  4
When P(x) is divided by (x2) the remainder is 8 so remainder will be of form ax+b
so P(x)=Q(x) * (x^2  4 ) + (ax+b)
so P(x)=Q(x) (x  2)(x + 2) + (ax+b)
so put x = 2 in remainder that will be equal to 8
so 2a + b = 8
now put x = 2 then remainder will be 4
and 2a + b = 4
solve these two equations
so b=6 and a=1
so remainder = x + 6Find the remainder when x^2014 + 1 is divided by x^2  2x + 1
Method  1
x^2014 mod (x1)^2
let x  1 = a
so (1+a)^2014 +1 mod a^2
= 1 + (1 + 2004 * a + 2004 * a^2 + ... 2004 * a^2014) mod a^2
so 1 + (1 + 2004 * a)
2 + 2004 * a
2 + 2004 (x1)Method 2 (common method for all kind of problem of this type)
x^2014 + 1 = P * (x1)^2 + mx + n
where mx + n is remainder because divisor power is x^2 so remainder will be less than that so
now
x^2014 + 1 = P * (x1)^2 + mx + n
now put x=1
so m + n = 2
now differentiate it
2014 * x^(2013) = p(x) * 2(x1) + (x1)^2 * p'(x) + m
now again put x=1
2014 = m
so m + n = 2
so n = 2012
so remainder mx+n
2014x  2012