1. According to Remainder Theorem: Let

**p(x)**be any polynomial and**a**be any real number. If p(x) is divided by the linear polynomial**x - a**, then the remainder is**p(a)**.
2. If

**p(x)**is divided by**(x + a)**, then remainder is**p(-a)**.
3. If

**p(x)**is divided by**(ax -b)**then remainder is**p(b/a)**.
4. If

**p(x)**is divided by**(ax + b)**, then remainder is**p(-b/a)**.
5. In above cases,

**-a**,**b/a**and**-b/a**are the__zeros of the divisors__**x + a**,**ax-b**and**ax+b**respectively.**Exercise 2.3**

**Q1: Find the remainder when x**

^{3}+ 3x^{2}+ 3x + 1 is divided by**(i) x + 1**

**(ii) x - 1/2**

**(iii) x**

**(iv) x + π**

**(v) 5 + 2x**

Answer: The remainder of the polynomial can be found by following methods:

- long division method
- applying remainder theorem.
- synthetic division (not covered here)

(i) x + 1

By long division method,

x^{2}+ 2x + 1 ____________________________ x+1) x^{3}+ 3x^{2}+ 3x + 1 x^{3}+ x^{2}- - ―――――――――― 2x^{2}+ 3x 2x^{2}+ 2x - - ―――――――――― x + 1 x + 1 - - ―――――――― 0

Therefore remainder is 0.

**IInd Method**: q(x) = x

^{3}+ 3x

^{2}+ 3x + 1

If t(x) = x+1 is the divisor of the polynomial, by remainder theorem, remainder is q(-1)

q(-1) = (-1)

^{3}+ 3(-1)^{2}+ 3(-1) + 1 = -1 + 1 -1 +1 = 0.
∴ remainder is

**0**.
(ii) x - 1/2

p(x) = x

^{3}+ 3x^{2}+ 3x + 1
Zero of x-1/2 = 1/2

By remainder theorem, when p(x) is divided by x-1/2, the remainder is p(1/2).

∴ p(1/2) = (1/2)

^{3}+ 3(1/2)^{2}+ 3(1/2) + 1
= 1/8 + 3/4 + 3/2 + 1 =

**27/8**(iii) x

Zero of x is 0

∴ p(0) = (0)

^{3}+ 3(0)^{2}+ 3(0) + 1 = 1
So remainder is 1.

(iv) x + π

Root of x + π is (x + π = 0 ⇒ x = -π) = -π

By remainder theorem, if -π divides p(x)then remainder is p(-π) i.e.

p(-π) = (-π)^{3}+ 3(-π)

^{2}+ 3(-π) + 1 = -π

^{3}+ 3π

^{2}- 3π + 1 ... (answer)

(v) 5 + 2x

Root of 5 + 2x is -5/2

If -5/2 divides p(x) = x

^{3}+ 3x^{2}+ 3x + 1, by remainder theorem the remainder is p(-5/2)
∴ p(-5/2) = (-5/2)

^{3}+ 3(-5/2)^{2}+ 3(-5/2 + 1
= -125/8 + 3(25/4) -15/2 + 1

= -125/8 + 75/4 - 15/2 + 1

= **-27/8**

**Q2: Find the remainder when x**

^{3}– ax^{2}+ 6x – a is divided by x – a.
Answer: By remainder theorem, if (x = a) divides p(x) = x

^{3}– ax^{2}+ 6x – a then remainder of polynomial is p(a).
∴p(a) = a

^{3}– a(a)^{2}+ 6a – a = a^{3}– a^{3}+ 6a – a =**5a ...(answer)**

**Q3: Check whether 7 + 3x is a factor of 3x**

^{3}+ 7x.
Answer: Zero of 7 + 3x is:

7+ 3x = 0 ⇒ 3x = -7 ⇒ x = -7/3

If p(x) = 3x

^{3}+ 7x. is divisible by (7 + 3x) then its remainder is zero i.e. ⇔ p(-7/3) = 0
p(-7/3) = 3(-7/3)

^{3}+ 7(-7/3) = 3(-343/27) - 49/3
= -343/9 -49/3 = -490/9 ≠ 0

⇒ 7 + 3x is

**not**a factor of 3x^{3}+ 7x**Q4: Prove Remainder theorem.**

Answer:

*Remainder theorem*: Let p(x) be any polynomial of degree ≥ 1, and '**a'**is any real number. If p(x) is divided by (x-a) then remainder is p(a).
Proof: Suppose q(x) is the quotient and r(x) is the remainder, when (x-a) divides p(x).

Since Dividend = Divisor ✕ Quotient + Remainder.

Similarly we can have,

p(x) = (x-a) ✕ q(x) + r(x) ...(I)

Where r = 0 or degree of r(x) < degree of (x -a).

Since (x-a) is a linear equation. The degree of (x-a) is 1 and degree of r(x) is less than the degree of x-a, ⇒ the degree of r(x) = 0.

It means r(x) is a constant, say r. ∴ Eqn. I will become:

p(x) = (x-a) ✕ q(x) + c

Replacing x by a i.e. x =a , we have

p(a) = (a-a) ✕ q(x) + c = 0 + c = c

⇒p(a) = c which proves the remainder theorem.

**Q5: 6x**

^{2}+ ax + 7 when divided by x-2 gives remainder 13. Find the value of a.
Answer: Let p(x) = 6x

^{2}+ ax + 7.
According to remainder theorem, if (x-2) divides p(x) and remainder is 13 then, p(2) = 13

p(2) = 6(2)

^{2}+ a(2) + 7 = 13
⇒ 24 + 2a + 7 = 13

⇒ 2a = -24-7+13 = -18

⇒ a =

**-9****Q6: The polynomial x**

^{4}- 2x^{3}+ 3x^{2}- ax + b when divided by (x+1) and (x-1) gives remainders 19 and 5 respectively. Find the remainder when the polynomial is divided by (x-3).
Answer: Let p(x) = x

^{4}- 2x^{3}+ 3x^{2}- ax + b
Writing the divisors in (x -a) form i.e. (x-(-1)) and (x-1).

According to remainder theorem, if (x-a) divides the polynomial p(x) and c is the remainder, then p(a) =c.

∴ p(-1) = 19 and p(1) = 5

⇒ p(-1) = (-1)

^{4}- 2(-1)^{3}+ 3(-1)^{2}- a(-1) + b = 19
⇒ 1 - 2(-1) + 3 + a + b = 19

⇒ 1 + 2 + 3 + a + b = 19

⇒ a + b = 13 ...(I)

and p(1) = (1)

^{4}- 2(1)^{3}+ 3(1)^{2}- a(1) + b =5
⇒ 1 - 2 + 3 -a + b = 5

⇒ -a + b = 3 ...(II)

Adding equation I and II, gives

2b = 16 ⇒ b = 8

and a = 5

∴ p(x) = x

^{4}- 2x^{3}+ 3x^{2}- 5x + 8
The divisor is (x-3). Applying remainder theorem, remainder = p(3).

i.e. p(3) = (3)

^{4}- 2(3)^{3}+ 3(3)^{2}- 5(3) + 8 = 81 - 54 + 27 - 15 + 8 =**47**...(answer)
Watch KhanAcademy's (Wonderful Elearning site) Video on Long Division:

Class IX Maths chapter 3,9,10,11,12 are missing please add them as fast as u can

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