- Books Name
- Kaysons Academy Maths Foundation Book
- Publication
- Kaysons Publication
- Course
- JEE
- Subject
- Maths
Factorization of Polynomials
Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, (2t + 1) is a factor of q(t), i.e., q(t) = (2t + 1) g(t) for some polynomial g(t). This is a particular case of the following theorem.
Factor Theorem: If p(x) is a polynomial of degree n > 1 and a is any real 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).
Proof: By the Remainder Theorem, p(x)=(x – a) q(x) + p(a).
(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).
(ii) 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.