Factor theorem of polynomialPolynomial is an expression consisting of variables and coefficients of the form: P_{n}(x)= a_{n}x^{n}+a_{n-1}x^{n-1}+…+a_{0} , where n is a natural number and a_{n}eq 0 .

There are 2 necessary theorems which play a significant role while handling polynomials that is remainder theorem and other is factor theorem. Factor theorem comes from the remainder theorem. Which allows us to initial study the remainder theorem, according to that: Let p(x) be any polynomial of degree greater than or equal to one and let a be any number.

If p(x) is divided by the linear polynomial x-a, then the remainder is p(a) and the quotient is Q(x) and the remainder is R(x). Thus, P(x) = (x – a).Q(x) + R(x).

Factor theorem could be a special case of the remainder theorem. Factor theorem states that if a polynomial P(x) is equally divided by another polynomial (x – a), then it leaves no remainder; i.e. R(x) = zero and therefore (x – a) is said to be the factor of polynomial P(x).

According to that: P(x) = (x – a) Q(x) [R(x)=0]. In other words, we can say that (x-a) is a factor of P(x), if P(a) = 0.This theorem also shows us the relationship between factors of the polynomials as well as zero of polynomial and also the theorem helps to find out the zero of a polynomial having with two or higher degrees.DefinitionIf P(x) is a polynomial of degree n geq 1 and a is any positive number, then (i) x – a is a factor of P(x), if P(a) = 0, and (ii) P(a) = 0, if x – a is a factor of P(x).ProofBy the Remainder Theorem, Dividend = Divisor * Quotient + Remainder where P(x) = (x – a) . Q(x) + p(a)If p(a) = 0, then p(x) = (x – a).q(x), which shows that x – a is a factor of p(x).

Since x – a is a factor of p(x), p(x) = (x – a) g(x) for same polynomial g(x). In this case, p(a) = (a – a) g(a) = 0. Thus, p(a) = 0 when divided by (x-a) by applying remainder theorem.Example 1: Examine whether x – 2 is a factor of x^{2}-7x+10 .

Solution: Let P(x) = x^{2}-7x+10 . 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 x^{2}-7x+10 we get P(2) = 2^{2}-7*2+10 = 4-14+10 = 0Since P(2) = 0.

Therefore, x – 2 is a factor of P(x).Example 2: Examine whether x + 2 is a factor of x^{3}+3x^{2}+5x+6 and of 2x+4Solution: The zero of x+2 is -2. Let p(x) = x^{3}+3x^{2}+5x+6 and s(x) = 2x+4Then, p(-2) = (-2)^{3}+3(-2)^{2}+5(-2)+6 = -8 + 12 – 10 + 6 = 0.So, by the factor theorem, x+2 is a factor of x^{3}+3x^{2}+5x+6 .

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 4x^{3}+3x^{2}-4x+k .Solution: As x-1 is a factor of p(x) = 4x^{3}+3x^{2}-4x+k , p(1) = 0Now, p(1) = 4(1)^{3}+3(1)^{2}-4(1)+k = 4 + 3 – 4 + 4 = 0So, k = -3ExerciseUse the Factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases:p(x) = 2x^{3}+x^{2}-2x-1 , g(x) = x+1p(x) = x^{3}+3x^{2}+3x+1 , g(x) = x+2p(x) = x^{3}-4x^{2}+x+6 , g(x) = x-3p(x) = x^{4}+2x^{3}-ax^{2}+x-2 , g(x) = x+2Find the value of k, if x-1 is a factor of p(x) in each of the following cases:p(x) = x^{2}+x+k p(x) = kx^{2}-sqrt{2}x+1 p(x) = 2x^{2}+kx+sqrt{2} kx^{2}-3x+k