The remainder theorem is applicable when a polynomial is divided by a linear polynomial. It states when a polynomial a(x) is divided by a linear polynomial b(x) whose zero is x=k, then the remainder of that division is given by r = a(k).
The general formula for the remainder theorem is p(x) = (x-c)·q(x) + r(x).
When p(x) is divided by (x-a), then the remainder is given by p(a).
When p(x) is divided by (ax+b), then the remainder is given by p(-b/a).
We know that dividend = (divisor x quotient) + remainder.
If r(x) is the constant then, p(x) = (x-c)·q(x) + r.
Let us put x=c
p(c) = (c-c)·q(c) + r
p(c) = (0)·q(c) + r
p(c) = r
Let us take an example of a polynomial a(x) = 6x⁴ - x³ + 2x² - 7x + 2 that needs to be divided by b(x) = 2x+3.
Step 1) Find the multiple for the first quotient that completely divides the dividend. In this case, it is 3x³.
Step 2) Multiply the whole divisor by 3x³ to get the remainder as -10x³.
Step 3) Copy down the next part of the dividend. In this case, it is 2x² - 7x + 2.
Step 4) Again find such a quotient that completely divides the dividend.
Step 5) Repeat the steps till you get a remainder in the form of zero or some numeric value that cannot be further divided. In this case, we get a remainder 203/4.
If one has found the zeroes for the polynomial equation 2x+3, they can place the value of this zero in the polynomial equation 6x⁴ - x³ + 2x² - 7x + 2 to get the remainder.
By equating 2x+3= 0, we get x = -3/2.
Put this value in the polynomial 6x⁴ - x³ + 2x² - 7x + 2 to get remainder as 203/4.
This proves that the remainder found out is correct.
This is the shortcut method to find the remainder for various polynomial equations.
NCERT Class 9 Maths Chapter 2 extensively talks about polynomials and its application along with Remainder Theorem.
Find the remainder when x³ - 2x² + x + 1 is divided by x-1. Solution
Equating x -1 = 0, we get x = 1.
Substitute the value of x in the equation x³ - 2x² + x + 1,
We get, (1)³ – 2(1)² + 1 + 1 = 1
This gives the remainder equal to 1.
Euler’s theorem states that if n and X are two coprime positive integers, then Xφ⁽ⁿ⁾ = 1 (mod n).
φ(n) is Euler’s function or Euler’s totient function = φ(n) = n (1-1/a).(1-1/b).(1-1/c), where n is a natural number, such that n = aᵖ.bᑫ. cʳ,
In this a, b, c are prime factors of n and p, q, r are positive integers.
Find the Euler totient function of 35.
Solution: The factors of 35 = 5×7.
Φ(35) = 35 (1- (1/5) . (1-(1/7) = 24
This gives a totient function of 35 as 24.