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Remainder Theorem : This theorem represents the relationship between the divisor of the first degree in the form (x - a) and the remainder r(x).
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).
Proof: Let,
q(x) = quotient, and r(x) = remainder
according to the theorem,
P(x) is divided by (x - a)
∴ P(x) = Dividend, and (x - a ) = Divisor
By using the formula,
Dividend = Divisor × Quotient + Remainder
or, P(x) = ( x - a ) q(x) + r (x)
But, r (x) is a constant polynomial, therefore, r (x) = r
∴ P(x) = (x - a ) q(x) + r
Substitute, x = a
∴ P(a) = ( a - a ) q (a) + r
or, P (a ) = 0 × q(a) + r
∴ P(a) = r
This shows that the remainder is P (a) when P(x) is divided by (x - a)
We can write, P(x)/ (x - a) = P(a) = r
Remark 1. If a polynomial P (x) is divided by ( x + a) , the remainder is the value of P(x) at x = -a is P(-a)
[ ∵ x + a = 0, ∴ x = -a]
Remark 2. If P(x) is divided by ( ax - b), then remainder is P(x) at x = b/a is P(b/a).[ ∵ ax - b = 0, ∴ x = b/a]
Factor Theorem :
Theorem: Let p(x) be a polynomial of degree greater than or equal to 1 and ‘a’ be a real number such that:
i) If p(a) = 0, then ( x - a) is a factor of p(x).
ii) If ( x - a) is a factor of p(x), then p(a) = 0.
Proof: We know,
By remainder theorem, when p(x) is divided by ( x - a) gives remainder p(a).
then, p(x) = ( x - a) q(x) + p(a)…..(1)
where, q(x) = quotient.
i) Given, p(a) = 0
Substitute, p(a) = 0 in (1)
then, p(x) = ( x - a)q(x) + 0
or, p(x) = ( x - a) q(x)
or, x - a is a factor of p(x) proved.
ii) Given, x - a is a factor of p(x).
We know, when x - a is a factor of p(x) then p(x) = (x - a) q(x) + 0 ……(2)
where, q(x) = quotient
Comparing (1) and (2)
p(a) = 0 . proved.
