Remainder Theorem | Factor Theorem

Remainder Theorem

Let p(x) be a polynomial of degree  ≥ 1  and let a be any real number. When  p(x) is divided by ( x – a), then the remainder is p(a).
Proof : Suppose that when p(x) is divided by (x – a), the quotient is q(x) and the remainder is r.
          ∴          Dividend = Divisor x Quotient + Remainder 
                               p(x) = (x – a) . q(x) + r        ……….. (i)
                                Putting x = a in (i),
                                 We get r = p(a)
Thus, when p(x) is divided by (x – a), then the remainder is p(a).

Example: Find the remainder when the polynomial
                 p(x) = x3 + 3x2 – 2x +1 is divided by (x – 4).
Solution:       p(x) = x3 + 3x2 – 2x +1
                           x – 4 = 0  ⇒ x = 4
              By the remainder theorem, we that when p(x) is divided by (x – 4), the remainder is p(4).
Now, remainder = p(4)
                              = (43 + 3 x 42 – 2 x 4 + 1)
                              = (64 + 3 x 16 – 8 +1 ) = (64 + 48 – 7) = 105
Hence, the required remainder is 105.

Factor Theorem

Let p(x) be a polynomial of degree n > 1 and a be any real number.
(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: (i). p(a) = 0.
                On dividing p(x) by (x – a ), let q(x) be the quotient.   
 By remainder theorem, when p(x) is divided by (x – a), then      remainder is p(a).
Therefore p(x) = (x – a) . q(x) + p(a)
  p(x) = (x – a) . q(x)                                           { ∴  p(a) = 0 }
 ⇒ (x – a) is a factor of p(x).

(ii) Let (x – a) be a factor of p(x).
 On dividing p(x) by (x – a), let q(x) be the quotient.
Then, p(x) = (x – a) . q(x)
 ⇒  p(a) = 0                                              {putting x = a}
Thus, (x – a) is a factor of p(x) imply f(a) = 0

Example: Show that (x – 2) is a factor of the polynomial
                     p(x) = x3 – 2x2 – 3x + 6

Solution: (x – 2) = 0  x = 2
By the factor theorem, (x – 2) will be a factor of p(x) if p(2) = 0
Now, p(x) = x3 – 2x2 – 3x + 6
 ⇒                 p(2) = 23 – 2 x 22 – 3 x 2 + 6
                             = 8 – 2 x 4 – 6 + 6
                             = 8 – 8
                             = 0
 ⇒ (x – 2) is a factor of p(x)

Hence, (x – 2) is a factor of the given polynomial p(x).

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