**Introduction:**

Polynomials are algebraic expressions with no division by a variable.

For ex:The expression is 3x^{2}-2x+5 is an example of polynomial.

Consider the algebraic expression `(7x+3)/(y)` which is not a polynomial as the variable ‘y’ appears in the denominator or there is a division by a variable.

Wecan perform all mathematical operations like addition, subtraction, multiplication, division, exponentiation etc on polynomials. When one polynomial is divided by another polynomial we get like numeric division, we get a quotient and a remainder. Remainder theorem helps us to calculate the remainder easily in some special cases without actuallyperforming the Division of Polynomials.

Remaindertheorem says when an polynomial f(x) is divided by (x-a), the remainderis f(a). Let us illustrate this with an example

**Ex 1 :** Consider function f(x) = 3x^{2}+4 is divided by x-3, the remainder is f(3)

**Sol: Step 1: ** Plug in x = 3 in f(x) = 3x^{2}+4

**Step 2:** f(3)= 3x3^{2}+4 = 31

Nowlet us consider a situation where f(x) is divided by (x+a) then the remainder can be found using the same formula by putting (x+a) as (x- (-a)).

So,dividing f(x) by (x+a) is equivalent to dividing f(x) by (x- (-a)). So as per theorem the remainder is f(-a). Let us solve an example here:

**Ex 2:** Find the remainder when f(x) = 7x^{2}+4x+6 is divided by (x+3)

**Sol:** **Step 1:** f(x) = 7x^{2}+4x+6 is divided by (x+3) or (x- -3). So, the remainder is f(-3)

**Step 2:** Plug in x = (-3) in f(x) = 7x^{2}+4x+6

f(-3) = 7x(-3)^{2}+4x(-3)+6=63-12+6=57

NowConsider the case f(x) is divided by (ax+b). In this we need to make itan fashion equal to f(x) divided by (x-a). This can be done as

`(f(x))/(ax+b) = (f(x))/(a(x-(-b/a)))` . So the remainder is f ( `(-b)/(a)` )

**Ex3:** Find the remainder when f(x) = 3x^{2}-4x+3 is divided by (3x+2)

**Sol:** **Step 1: **The given division is (3x^{2}-4x+3) is divided by (3x+2).

This is equivalent to dividing`(3x^(2)-4x+3)/(3 )` by `(x -(-2/3))`. So, the remainder is f(`(-2)/(3)` )

**Step 2: **f(`(-2)/(3)`) = 3`(-2/3)^2`–4(`(-2)/(3)`)+3 = `(4)/(3)` -`(-8/3)`+3 = 4+3 =7

**Prob 1:** Find the remainder when 7x^{3}-4x^{2}+8x+9 is divided by 7x-5

**Ans: Step 1: **As seen above, when f(x) is divided by (ax+b), the remainder is f(-b/a).

**Step 2:** When 7x^{3}-4x^{2}+8x+9 is divided by 7x-5, the remainder is f(`(5)/(7)`)

**Step 3: **f((`(5)/(7)`)= 7((`(5)/(7)`)^{3}-4((`(5)/(7)`)^{2}+8((`(5)/(7)`)+9 = (`(746)/(49))`

**Prob 2**: For what value of a will f(x) = 5x^{2}-6x+a be completely divisible by (x-5)?

**Ans: Step 1: **If f(x) is completely divisible by (x-5) then the remainder is zero. I.e. f(5)=0

**Step 2:** f(5)= 5(5)^{2}-6(5)+a=0

125-30+a=0

**Step 3:** 95+a=0

a=-95

5x^{2}-6x-95 is completely divisible by 95

**Prob 3:** : For what value of a will f(x) = 5x^{2}-6x+a when divided by (x-5) will leave a remainder of 6?

**Ans: Step 1: **If there is a reminder of 6 then f(x)-6 will be completely divisible by (x-5)

So f(5)-6=0

**Step 2: **5x5^{2}-6x5+a-6=0

125-30+a-6=0

**Step 3: **89+a=0

a="-89

When 5x^{2}-6x-89 is divided by (x-5), the remainder is 6