Remainder theorem:let p(x)be any polynomial of degree greater than or equal to one and let 'a'be any real number .If p(x)is divided by the linear polynomial (x-a),then the remainder is p(a).
ex:by division algorithm, p(x)=g(x).q(x)+r(x) p(x)=(x-a).q(x)+r(x) g(x)=(x-a) since the degree of(x-a)is 1 and the degree of r(x)is less than the degree of (x-a) therefore,degree of r(x)=0,implies r(x)is a constant,say K so,for every real value of x,r(x)=K therefore, p(x)=(x-a)q(x)+K If x=a,then p(a)=(a-a)q(a)+K =0+K =K hence proved.
hope this helps u:))
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abhinav18802
can't understand . please solve with a non variable
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Remainder theorem:let p(x)be any polynomial of degree greater than or equal to one and let 'a'be any real number .If p(x)is divided by the linear polynomial (x-a),then the remainder is p(a).ex:by division algorithm,
p(x)=g(x).q(x)+r(x)
p(x)=(x-a).q(x)+r(x) g(x)=(x-a)
since the degree of(x-a)is 1 and the degree of r(x)is less than the degree of (x-a)
therefore,degree of r(x)=0,implies r(x)is a constant,say K
so,for every real value of x,r(x)=K
therefore,
p(x)=(x-a)q(x)+K
If x=a,then p(a)=(a-a)q(a)+K
=0+K
=K
hence proved.
hope this helps u:))