What must be subtracted from 6x4+13x3+13x2+30x+20, so that the resulting polynomial is exactly divisible by 3x2+2x+5?
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Solution
p(x)=g(x)×q(x)+r(x) [Division algorithm for polynomials] ⇒p(x)−r(x)=g(x)×q(x) It is clear the RHS of the above equation is divisible by g(x). i.e., the divisor. ∴ LHS is also divisible by the divisor. Therefore, if we subtract remainder r(x) from dividend p(x), then it will be exactly divisible by the divisor g(x)] Let us divide, 6x4+13x3+13x2+30x+20 by 3x2+2x+5 we get, quotient q(x)=2x2+3x−1 remainder r(x)=17x+25 ∴ If we subtract 17x+25 from 6x4+13x3+13x2+30x+20, it will be exactly divisible by (3x2+2x+5).