How To Find Remainder in case of Polynomial


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    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 (x-2) 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 (x-2) 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 + 6

    Find the remainder when x^2014 + 1 is divided by x^2 - 2x + 1

    Method - 1
    x^2014 mod (x-1)^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 (x-1)

    Method 2- (common method for all kind of problem of this type)
    x^2014 + 1 = P * (x-1)^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 * (x-1)^2 + mx + n
    now put x=1
    so m + n = 2
    now differentiate it
    2014 * x^(2013) = p(x) * 2(x-1) + (x-1)^2 * p'(x) + m
    now again put x=1
    2014 = m
    so m + n = 2
    so n = -2012
    so remainder mx+n
    2014x - 2012


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