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  Mathematics (NTSE/Olympiad)  

Polynomial

Remainder Theorem

Important Points of a Remainder Theorem

(i) Remainder can be obtained on dividing polynomial p(x) by x – a is equal to p(a).
(ii) If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = –a.
(iii) (x – a) is a factor of polynomial p(x) if p(a) = 0
(iv) (x + a) is a factor of polynomial p(x) if p(–a) = 0
(v) (x – a) (x – b) is a factor of polynomial p(x), if p(a) = 0 and p(b) = 0.

Example: Find the remainder when 4x3 – 3x2 + 2x – 4 is divided by
(a) x – 1 (b) x + 2 (c) x + 1/2  

Solution:

Let the given cubic equation is p(x) = 4x3 – 3x2 + 2x – 4
(a) When p(x) is divided by (x – 1), then by remainder theorem, the required remainder will be p(1)
p(1) = 4 (1)3 – 3(1)2 + 2(1) – 4
= 4 × 1 – 3 × 1 + 2 × 1 – 4
= 4 – 3 + 2 – 4 = – 1

(b) When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (–2).
p(–2) = 4 (–2)3 – 3 (–2)2 + 2(–2) – 4
= 4 × (–8) – 3 × 4 – 4 – 4
= – 32 – 12 – 8 = – 52

(c) When p(x) is divided by x + 1/2, then by remainder theorem, the required remainder will be
p(–1/2)= 4 (–1/2)3 – 3(–1/2)2 + 2(–1/2) – 4
= 4 × (–1/8) – 3 (1/4)× – 2 × (1/2) – 4
= –1/2 – 3/4 – 1– 4 = –1/2 –3/4  – 5
= (–2 – 3 – 20)/4 = –25/4


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