Proof of this factor theorem
Let p(x) be a polynomial of degree greater than or equal to one and a be areal number such that p(a) = 0. Then, we have to show that (x – a) is a factor of p(x).Let q(x) be the quotient when p(x) is divided by (x – a). By remainder theorem Dividend = Divisor x Quotient + Remainder p(x) = (x – a) x q(x) + p(a) [Remainder theorem] ⇒ p(x) = (x – a) x q(x) [p(a) = 0]⇒ (x – a) is a factor of p(x) Conversely, let (x – a) be a factor of p(x). Then we have to prove that p(a) = 0 Now, (x – a) is a factor of p(x)⇒ p(x), when divided by (x – a) gives remainder zero. But, by the remainder theorem, p(x) when divided by (x – a) gives the remainder equal to p(a). ∴ p(a) = 0 Proof of remainder theorem. Let q(x) be the quotient and r(x) be the remainder obtained when the polynomial p(x) is divided by (x–a). Then, p(x) = (x–a) q(x) + r(x), where r(x) = 0 or some constant. Let r(x) = c, where c is some constant. Then p (x) = (x–a) q(x) + c Putting x = a in p(x) = (x–a) q(x) + c, we getp(a) = (a–a) q(a) + c ⇒ p(a) = 0 x q(a) + c ⇒ p(a) = c This shows that the remainder is p(a) when p(x) is divided by (x–a).1. [a is –8.] Download study material for polynomial
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srikANTH
8/9/2013 01:52:53 pm
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