CBSE Class 9 - Maths - Polynomials - Exercise 2.3

Remainder Theorem

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 x³ + 3x² + 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,

        x2 + 2x + 1
       ____________________________
   x+1) x3 + 3x2 + 3x + 1
        x3 + x2 
       -   -   
        ――――――――――
            2x2 + 3x 
            2x2 + 2x
           -    -
            ――――――――――
                x + 1
                x + 1
               -  -
               ――――――――
               0 

Therefore remainder is 0. 

IInd Method:  q(x) = x³ + 3x² + 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) =  x³ + 3x² + 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 = -π³ + 3π² - 3π + 1  ... (answer)

(v)  5 + 2x

Root of 5 + 2x is -5/2

If -5/2 divides p(x) =  x³ + 3x² + 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 x³ – ax²  + 6x – a is divided by x – a.

Answer: By remainder theorem, if (x = a) divides  p(x) = x³ – ax²  + 6x – a then remainder of polynomial is p(a).

∴p(a) = a³ – a(a)²  + 6a – a  = a³ – a³  + 6a – a  = 5a  ...(answer)

As an alternate method you may find remainder by long division method.

Q3:  Check whether 7 + 3x is a factor of 3x³ + 7x.

Answer: Zero of 7 + 3x is:

   7+ 3x = 0  ⇒ 3x = -7   ⇒ x = -7/3

If  p(x) = 3x³ + 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 3x³ + 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: 6x² + ax + 7 when divided by x-2 gives remainder 13. Find the value of a.

Answer: Let p(x) = 6x² + 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 x⁴ - 2x³ + 3x² - 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) =  x⁴ - 2x³ + 3x² - 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) = x⁴ - 2x³ + 3x² - 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)

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