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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