# 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*) =*x*^{4}âˆ’ 3*x*^{2}+ 2*x*+ 5. Find the remainder when*p*(*x*) is divided by (*x*âˆ’ 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 âˆ’ 3 + 2 + 5 = 5We find that the remainder when

*p*(*x*) is divided by (*x*âˆ’ 1) is equal to*p*(1), i.e., the value of*p*(*x*) at*x*= 1.Example-2

Find the remainder when

*p*(*y*) =*y*^{3}+*y*^{2}+ 2*y*+ 3 is divided by*y*+ 2Solution

Compute

*p*(âˆ’2) = Value of*p*(*y*) when*y*is replaced by âˆ’2, we have,*P*(âˆ’2) = (âˆ’2)^{3}+ (2)^{2}+ 2(âˆ’2) + 3 = âˆ’8 + 4 âˆ’ 4 + 3 = âˆ’5It follows from these two examples that the remainder obtained when

*p*(*x*) is divided by (*x*âˆ’*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 2â€”Number system for more reference.)

**Note:**

- If a polynomial
*p*(*x*) is divided by (*x*+*a*), the remainder is the value of*p*(*x*) at*x*= âˆ’*a*, i.e.,*p*(âˆ’*a*). - If a polynomial
*p*(*x*) is divided by (*ax*âˆ’*b*), the remainder is the value of*p*(*x*) at*x*=*b*/*a*, i.e.,*P*(*b*/*a*). - If a polynomial
*p*(*x*) is divided by (*ax*+*b*), the remainder is the value of*p*(*x*) at*x*= âˆ’*b*/*a*, i.e.,*p*(âˆ’*b*/*a*). - If a polynomial
*p*(*x*) is divided by (*b*âˆ’*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*) =*x*^{4 }âˆ’ 3*x*^{2 }+ 1 is divided by*x*âˆ’ 1.Solution

By remainder theorem, the required remainder is equal to

*p*(1).Now,
â‡’
Hence, required remainder =

*p*(*x*) =*x*^{4}âˆ’ 3*x*^{2}+ 2*x*+ 1*p*(1) = (1)^{4}âˆ’ 3 Ã— 1^{2}+ 2 Ã— 1 + 1 = 1 âˆ’ 3 + 2 + 1 = 1*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)*.

*f(x)*. Conversely, if (x âˆ’ a) is a factor of f(x) then f(a) = 0.

**Note:**

- (x + a) is a factor of a polynomial f(x) if f(âˆ’a) = 0
- (ax âˆ’ b) is a factor of a polynomial f(x) if f(b/a) = 0
- ax + b is a factor of polynomial f(x) if f(âˆ’b/a) = 0
- (x âˆ’ a) (x âˆ’ b) is a factor of a polynomial f(x) if f(a) = 0 and f(b) = 0.

Example-1

Show that (x âˆ’ 3) is a factor of the polynomial: x

^{3}âˆ’ 3x^{2}+ 4x âˆ’ 12Solution

Let f(x) = x
Now, f(x) = x
â‡’ f(3) = 3
Hence, (x âˆ’ 3) is a factor of f(x).

^{3}âˆ’ 3x^{2}+ 4x âˆ’ 12 be the given polynomial. By factor theorem, (x âˆ’ a) is a factor of a polynomial f(x) if f(a) = 0. Therefore, in order to prove that (x âˆ’ 3) is a factor of f(x), it is sufficient to show that f(3) = 0.^{3}âˆ’ 3x^{2}+ 4x âˆ’ 12^{3}âˆ’ 3 Ã— 3^{2}+ 4 Ã— 3 âˆ’ 12 = 27 âˆ’ 27 + 12 âˆ’ 12 = 0Example-2

Find out if (x âˆ’ 1) is a factor of (i) and (ii).

Solution

Let f(x) = x
To find out that (x âˆ’ 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) = x
â‡’ f(1) = 1
â‡’ (x âˆ’ 1) is a factor of both f(x) and g(x)

^{10}âˆ’ 1 and g(x) = x^{11}âˆ’ 1.^{10}âˆ’ 1 and g(x) = x^{11}âˆ’ 1^{10}âˆ’ 1 = 0 and g(1) = 1^{11}âˆ’ 1 = 0