Polynomial is an expression of the form: . Where n is a natural number and . There are 2 theorems which play a vital role in solving polynomial which is remainder theorem and factor theorem. Factor theorem is being derived from the remainder theorem which allows us to initially study remainder theorem first then the factor theorem. According to remainder theorem: Let p(x) be any polynomial of degree greater than or equal to one and is divided by the linear polynomial x-a where a be any number which would be the divisor and we get the value of x = a, then the remainder is p (a).

Factor theorem is a special case for the remainder theorem. It states that when the polynomial p(x) is equally divided by another polynomial g(x) with the divisor x-a and if the remainder is zero i.e. R(x) = 0. So x-a is said to be the factor of that polynomial p(x). We can write it in the form: P(x) = (x-a).q(x) where the remainder is said to be zero.

This theorem also shows us the relationship between factors of the polynomials as well as zero of polynomial and helps to find out the zero of a polynomial having two or higher degrees.

## Definition

If p(x) is a polynomial of degree n which is greater than or equal to one and a is any real number which will be the divisor, then there will be two conditions fulfilled:

- If p (a) =0, then x-a is a factor of that polynomial p(x).
- x-a would be the factor of the polynomial if the r(x) i.e. remainder is 0.

### Proof

By the Remainder Theorem, Dividend = Divisor * Quotient + Remainder where P(x) = (x – a). Q(x) + p (a)

- Consider a polynomial p(x) which is divided by (x-a), then p (a) =0. Therefore, p(x) = (x-a).q(x) +p (a) = (x-a).q(x) +0 = (x-a).q(x). Thus, x-a is a factor of p(x) when the remainder is zero.
- If the (x-a) is a factor of polynomial p(x), then the remainder must be zero. So, we can say that x-a exactly divides p(x). Thus p(x) =0. Hence, the theorem is being proved.

**Example 1:** Examine whether x – 2 is a factor of .

**Solution: **Let P(x) = . By the factor theorem, x – 2 is the factor of P(x) if P(2) = 0.

By putting the value of P(2) in the equation we get P(2) = = 4-14+10 = 0

Since P(2) = 0. Therefore, x – 2 is a factor of P(x).

**Example 2: **Examine whether x + 2 is a factor of and of 2x+4

**Solution: **The zero of x+2 is -2. Let p(x) = and s(x) = 2x+4

Then, p(-2) = = -8 + 12 – 10 + 6 = 0.

So, by the factor theorem, x+2 is a factor of .

Again, s(-2) = 2(-2) + 4 = 0. So, x+2 is a factor of 2x+4.

**Example 3: **Find the value of k, if x-1 is a factor of .

**Solution: **As x-1 is a factor of p(x) = , p(1) = 0

Now, p(1) = = 4 + 3 – 4 + 4 = 0

So, k = -3

## Exercise

Use the Factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

- p(x) = , g(x) = x+1
- p(x) = , g(x) = x+2
- p(x) = , g(x) = x-3
- p(x) = , g(x) = x+2

Find the value of k, if x-1 is a factor of p(x) in each of the following cases:

- p(x) =
- p(x) =
- p(x) =

R Spencer says

Can you help me with this?

If p(x) = cx, and q(x) = x-c is q(x) a divisor or p(x)?

more specifically, under what conditions is it a factor?

I tried factor theorem of polynomials, but, does does p(x) have to be a higher degree than q(x)?