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Remainder Theorem

Remainder theorem gives us a method for finding the remainder without actual division.
Let us understand this with the examples:
 
Example-1
Let p(x) = x4 − 3x2 + 2x + 5. Find the remainder when p(x) is divided by (x − 1).
Solution
Rather going for the actual division, Let us compute p(1), i.e., value of p(x) when x is replaced by 1, we have, p(1) = 1 − 3 + 2 + 5 = 5
We find that the remainder when p(x) is divided by (x − 1) is equal to p(1), i.e., the value of p(x) at x = 1.
 
 
Example-2
Find the remainder when p(y) = y3 + y2 + 2y + 3 is divided by y + 2
Solution
Compute p(−2) = Value of p(y) when y is replaced by −2, we have, P(−2) = (−2)3 + (2)2 + 2(−2) + 3 = −8 + 4 − 4 + 3 = −5
 
It follows from these two examples that the remainder obtained when p(x) is divided by (x − a) is equal to p(a), i.e., the value of p(x) at x a.
 
The above result is stated as remainder theorem (Look in chapter 2—Number system for more reference.)
 
 
Note:
  1. If a polynomial p(x) is divided by (x + a), the remainder is the value of p(x) at x = −a, i.e., p(−a).
  2. If a polynomial p(x) is divided by (ax − b), the remainder is the value of p(x) at x = b/a, i.e., P (b/a).
  3. If a polynomial p(x) is divided by (ax b), the remainder is the value of p(x) at x = −b/a, i.e., p(−b/a).
  4. If a polynomial p(x) is divided by (b − ax), the remainder is equal to the value of p(x) at x = b/a, i.e., p(b/a).
 
Example-3
Determine the remainder when the polynomial p(x) = x4 − 3x2 + 1 is divided by x − 1.
Solution
By remainder theorem, the required remainder is equal to p(1).
 
Now, p(x) = x4 − 3x2 + 2x + 1
 
⇒ p(1) = (1)4 − 3 × 12 + 2 × 1 + 1 = 1 − 3 + 2 + 1 = 1
 
Hence, required remainder = p(1) = 1.
 

Factor Theorem

If g(x) divides f(x), we say that f(x) is divisible by g(x) or g(x) is a factor of f(x).
 
Factor theorem gives us a method to determine whether a polynomial g(x) is a factor of a polynomial f(x) or not without actual division.
 
Let f(x) be a polynomial of degree greater than or equal to 1 and ‘a’ be a real number such that f(a) = 0, then (x − a) is a factor of f(x). Conversely, if (x − a) is a factor of f(x) then f(a) = 0.
 

Note:

  1. (x + a) is a factor of a polynomial f(x) if f(−a) = 0
  2. (ax − b) is a factor of a polynomial f(x) if f(b/a) = 0
  3. ax + b is a factor of polynomial f(x) if f(−b/a) = 0
  4. (x − a) (x − b) is a factor of a polynomial f(x) if f(a) = 0 and f(b) = 0.
 
Example-1
Show that (x − 3) is a factor of the polynomial: x3 − 3x2 + 4x − 12
Solution
Let f(x) = x3 − 3x2 + 4x − 12 be the given polynomial. By factor theorem, (x − a) is a factor of a polynomial f(x) if f(a) = 0. Therefore, in order to prove that (x − 3) is a factor of f(x), it is sufficient to show that f(3) = 0.
 
Now, f(x) = x3 − 3x2 + 4x − 12
 
⇒ f(3) = 33 − 3 × 32 + 4 × 3 − 12 = 27 − 27 + 12 − 12 = 0
 
Hence, (x − 3) is a factor of f(x).
 
 
Example-2
Find out if (x − 1) is a factor of (i) and (ii).
Solution
Let f(x) = x10 − 1 and g(x) = x11 − 1.
 
To find out that (x − 1) is a factor of both f(x) and g(x), it is sufficient to show that f(1) = 0 and g(1) = 0.
 
Now, f(x) = x10 − 1 and g(x) = x11 − 1
 
⇒ f(1) = 110 − 1 = 0 and g(1) = 111 − 1 = 0
 
⇒ (x − 1) is a factor of both f(x) and g(x)
 




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