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REMAINDER THEOREMWe know that: Dividend = (Divisor x Quotient) + Remainder

If in general, we denote:

Dividend as p(x),

Divisor as g(x),

Quotient as q(x),

Remainder as r(x).

Such that:(i) p(x), g(x), q(x) and r(x) are polynomials, and(ii) g(x) ≠ 0

(iii) Degree of r(x) < Degree of g(x)

(iv ) Degree of g(x) ≤ Degree of p(x)

Then, we have p(x) = [g(x) . q(x)] + r(x)

**Note:** Theorem means a statement which is already proved.

**Theorem:** Let p(x) be any polynomial of degree greater than or equal to one and let ‘a’ be any real number.

If p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).

**Proof:** Let p(x) be divided by (x – a) such that:

Quotient = q(x), Remainder = r(x)

∴ p(x) = [(x – a) . q(x)] + r(x) …(1)

Since, remainder is always smaller than the divisor.

∵ Divisor (x – a) is having degree as 1.

∴ Degree of remainder must be zero.

We know that the degree of a constant polynomial is zero.

∴ r(x) is a constant, say ‘r’.

Now, from (1), we have p(x) = [(x – a) . q(x)] + r

If x = a then P(a) = [(a – a) . q(a)] + r

or p(a) = [0 x (q (a))] + r

or p(a) = 0 + r

or p(a) = r i.e. Remainder = p(a).

**Ques 1: Find the remainder when x ^{3} + 3x^{2} + 3x + 1 is divided by (i) x + 1 (ii) x – 1/2(iii) x (iv) x + p ( v ) 5 + 2x**

**Ans: **(i) ∵ The zero of x + 1 is –1 [∵ x + 1 = 0 ⇒ x = –1]

And by remainder theorem, when p(x) = x^{3} + 3x^{2} + 3x + 1 is divided by x + 1, then remainder is p(–1).

∴ p(–1) = (–1)^{3} + 3 (–1)^{2} + 3(–1) + 1 = –1 + (3 x 1) + (–3) + 1

= –1 + 3 – 3 + 1

= 0

Thus, the required remainder = 0

(ii) ∵ The zero of x –1/2 is 1/2 [∵ x –1/2 = 0 ﬁ x = 1/2]

and p(x) = x^{3} + 3x^{2} + 3x + 1

∴ For divisor = x – 1/2, remainder is given as

Thus, the required remainder = 27/8

(iii) We have p(x) = x^{3} + 3x^{2} + 3x + 1 and the zero of x is 0

∴ p(0) = (0)^{3} + 3(0)^{2} + 3(0) + 1 = 0 + 0 + 0 + 1 = 1

Thus, the required remainder = 1.

(iv) We have p(x) = x^{3} + 3x^{2} + 3x + 1 and zero of x + p = (–π) [∵ x + π = 0 ⇒ x = – π]

∴ p(–π)= (– 5)^{3} + 3

(–π)^{2} + 3(–π) + 1 . = –π^{3} + 3(π^{2}) + (–3π) + 1 = –p3 + 3π^{2} – 3π + 1

Thus, the required remainder is –π + 3π^{2} – 3π + 1.

(v) We have (p(x) = x^{3 }+ 3x^{2} + 3x + 1 and zero of 5 + 2x is 5/2 [∵ 5 + 2x = 0 ⇒ x = -5/2]

Thus, the required remainder is 27/8

**Ques 2: Find the remainder when x ^{3} – ax^{2 }+ 6x – a is divided by x – a.** We have p(x) = x

Ans:

∵ Zero of x – a is a. [∵ x – a = 0 ⇒ x = a]

∴ p(a) = (a)^{3} – a(a)^{2} + 6(a) – a = a^{3} – a^{3} + 6a – a = 0 + 5a = 5a

Thus, the required remainder = 5a

**Ques 3: Check whether 7 + 3x is a factor of 3x ^{3} + 7x. Ans: **We have p(x) = 3x

since

i.e. the remainder is not 0.

∴ 3x^{3 }– 7x is not divisible by 7 + 3x.

Thus, (7 + 3x) is not a factor of 3x^{3} – 7x.

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