Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, `q (-1/2) =0 , (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.
`text (Factor Theorem : )` If `p(x)` is a polynomial of degree `n ge 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)`.
`text (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`.
Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, `q (-1/2) =0 , (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.
`text (Factor Theorem : )` If `p(x)` is a polynomial of degree `n ge 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)`.
`text (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`.