IX Std Term I - Chapter 3- Ex 3.3

Exercise 3.3

 

1.      Find the quotient and remainder of the following division.

(1)   (5x³ - 8x² + 5x - 7) ÷ (x - 1)

Solution:

     

                       5x² – 3x + 2

________________

  x – 1| 5x³ - 8x² + 5x – 7                           

                    |5x³ -  5x²
                    |-     +
                    |------------------------
                    |       - 3x² + 5x
                    |       - 3x² + 3x
                    |       +       -
                    |------------------------
                    |                  2x – 7
                    |                  2x – 2
                    |                 -     +
                    |------------------------
                    |                       - 5      

 

Quotient = 5x² – 3x + 2

Remainder = - 5

----------------------------------------------------------------------------------------------------------

(2)   (2x² -  3x - 14) ÷ (x + 2)

Solution:

(i)                  2x²/x = 2x

(ii)               2x(x + 2) = 2x² +4x

(iii)             -7x/x = -7

(iv)             -7(x+2) = -7x – 14

     

2x - 7             ___________

 

        x+2| 2x² - 3x – 14           
              | 2x² +4x
              |-      -
              |--------------------
              |        -7x – 14
              |        -7x – 14
              |        +     +
              |--------------------
              |                 0

Quotient = 2x - 7

Remainder = 0

----------------------------------------------------------------------------------------------------------

(3)   (9 + 4x + 5x² +  3x³ ) ÷ (x + 1)

Solution:

Writing the given polynomial in descending order, we get

(3x³ + 5x² + 4x + 9) ÷ (x + 1)

(i)                 3x3/x = 3x2 (ii)               3x2(x+ 1) = 3x3 + 3x2 (iii)             2 x2/x = 2x (iv)             2x(x + 1) = 2 x2 + 2x (v)               2x/x = 2 (vi)             2(x + 1) = 2x + 2                       3x2 + 2x + 2                    ______________

            x + 1| 3x³ + 5x² + 4x + 9
                      | 3x³ + 3x²
                     | -     -
                     |-------------------------
                     |           2x² + 4x
                     |           2 x² + 2x
                     |           -       -
                     |-------------------------
                     |                      2x + 9
                     |                      2x + 2
                     |                     -     -
                     |-------------------------
                     |                             7
 
  
           Quotient = 3x² + 2x + 2

Remainder = 7

        ---------------------------------------------------------------------------------------------------------

     (4)   (4x³ -  2x² +  6x +  7) ÷ (3 + 2x)

Solution:

Writing the given polynomials in descending order, we get

(4x³ - 2x² +  6x +  7) ÷ (2x + 3)

(i) 4x³/2x = 2x²

(ii) 2x²(2x + 3) = 4x³ + 6x²

(iii) - 8x²/2x = - 4x

(iv) – 4x(2x + 3) = - 8x² – 12x

(v) 18x/2x = 9

(vi) 9(2x + 3) = 18x +27

 

          

2x2 – 4x + 9                     ________________

            2x + 3| 4x³ - 2x² +  6x +  7
                      | 4x³ + 6x²
                      |-      -
                      |-----------------------------
                      |        - 8x² + 6x
                      |        - 8x² – 12x
                      |        +      +    
                      |-----------------------------
                      |           18x + 7            
                      |                     18x +27
                      |           -       -  
                      |-----------------------------
                      |                            - 20
                         

Quotient = 2x² – 4x + 9

Remainder = - 20

           -----------------------------------------------------------------------------------------------------------

(5)   ( -18 - 9x + 7x² ) ÷ (x - 2)

Solution:

Writing the given polynomial in descending order, we get

             (7x² – 9x – 18) ÷ (x – 2)
 

(i) 7x²/x = 7x

(ii) 7x(x – 2) = 7x² – 14x

(iii) 5x/x = 5

(iv) 5(x – 2) = 5x - 10

 

7x + 5

                    ____________
            x – 2| 7x² – 9x – 18
                    | 7x² – 14x
                    |-      +
                    |---------------------
                    |         5x - 18  
                    |         5x - 10     
                    |        -     +
                    |----------------------
                    |              - 8
            

Quotient = 7x + 5

Remainder = - 8

------------------------------------------------------------------------------------------------------------

IX Std Term I - Chapter 3 - Ex 3.2

Exercise 3.2

 1. Find the zeros of the following polynomials.

 (i)                 p(x) = 4x – 1

Solution:

Given that,

p(x) = 4x – 1 = 4(x -1/4)  

p(1/4) = 4(1/4 - 1/4) = 0

Hence, 1/4  is the zero of p(x).

-----------------------------------------------------------------------------------------------------

(ii)               p(x) = 3x + 5

Solution:

Given that,

p(x) = 3x + 5 = 3(x + 5/3)

p(-5/3) = 3(-5/3 + 5/3) = 0

Hence, - 5/3 is the zero of p(x).

-------------------------------------------------------------------------------------------------------

(iii)             p(x) = 2x

Solution:

Given that,

p(x) = 2x

p(0) = 2 × 0 = 0

Hence, 0 is the zero of p(x).

---------------------------------------------------------------------------------------------------------

(iv)             p(x) = x + 9

Solution:

Given that,

p(x) = x + 9

p(-9) = -9 + 9 = 0

Hence, -9 is the zero of p(x).

        --------------------------------------------------------------------------------------------

2. Find the roots of the following polynomial equations.

 

(i)                 x - 3 = 0

Solution:

x – 3 = 0

ð  x = 3

 ------------------------------------------------------------------------------------------------------

(ii)               5x - 6 = 0

Solution:

5x – 6 = 0

ð  5x = 6

ð  x = 6/5 

--------------------------------------------------------------------------------------------------------

(iii)             11x + 1 = 0

Solution:

11x + 1 = 0

ð  11x = -1

ð     x  = - 1/11

 

-----------------------------------------------------------------------------------------

(iv)             -9x = 0

Solution:

-9x = 0

ð  x = 0/-9    

ð  x = 0

----------------------------------------------------------------------------------------

3. Verify whether the following are roots of the polynomial equations indicated against them.

(i)                 x² -  5x + 6 = 0; x = 2, 3

Solution:

Let p(x) = x² -  5x + 6 = 0

P(2) = 2² – (5 × 2) + 6 = 4 – 10 +6 = 0

Hence, x = 2 is the root of p(x).

P(3) = 3² – (5× 3) + 6 = 9 – 15 + 6 = 0

Hence, x = 3 is the root of p(x).  

 
-------------------------------------------------------------------------------------------------------

(ii)               x² + 4x + 3 = 0; x = -1, 2

Solution:

Let p(x) = x² + 4x + 3 = 0

P(-1) = (-1)² + (4× (-1)) + 3 = 1 – 4 + 3 = 0

Hence, x = -1 is the root of p(x).

P(2) = 2² + (4 × 2) + 3 = 4 +8 + 3 = 15 ≠ 0

Hence, x = 2 is not the root of p(x).

------------------------------------------------------------------------------------------------------

(iii)             x³ - 2x² - 5x + 6 = 0; x = 1, -2, 3

Solution:

Let p(x) = x³ - 2x² - 5x + 6 = 0

P(1) = 1³ – (2 × 1²) – (5 × 1) + 6 = 1 – 2 – 5 + 6 = 0

Hence, x = 1 is the root of p(x).

P(-2) = (-2)³ – (2 × (-2)²) – (5 × (-2)) + 6 = -8 – 8 + 10 + 6 = 0

Hence, x = -2 is the root of p(x).

P(3) = 3³ – (2× (3)²) – (5 × 3) + 6 = 27 – 18 – 15 + 6 = 0

Hence, x = 3 is the root of p(x).

--------------------------------------------------------------------------------------------------------

(iv)             x³ -  2x² – x + 2 = 0; x = -1, 2, 3

Solution:

Let p(x) = x³ - 2x² – x + 2 = 0

 

P(-1) = (-1)³ – (2 × (-1)²) – (-1) + 2 = -1 - 2 + 1 + 2 = 0

Hence, x = -1 is the root of p(x).

P(2) = (2)³ – (2 × 2²) – 2 + 2 = 8 – 8 - 2 + 2 = 0

Hence, x = 2 is the root of p(x).

P(3) = (3)³ – (2 × 3²) – 3 + 2 = 27 – 18 – 3 + 2 = 8 ≠ 0

Hence, x = 3 is not the root of p(x).

----------------------------------------------------------------------------------------