9th Std - I Term Maths - Chapter 3 - Ex 3.4

Exercise 3.4

1. Find the remainder using remainder theorem, when

 

(i)                 3x³ +  4x² -  5x +  8 is divided by x – 1

Solution:

Let p(x) = 3x³ + 4x² - 5x + 8

The zero of (x – 1) is 1.

When p(x) is divided by (x – 1) the remainder is p(1). Now,

p(1) = 3(1)³ +  4(1)² -  5(1) +  8

        = 3 + 4 – 5 + 8

       = 10

؞ The remainder is 10.

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

(ii)               5x³ +  2x² -  6x +  12 is divided by x + 2

Solution:

Let p(x) = 5x³ +  2x² -  6x +  12

The zero of (x + 2) is -2.

When p(x) is divided by (x + 2) the remainder is p(-2). Now,

p(-2) = 5(-2)³ +  2(-2)² -  6(-2) +  12

         = -40 + 8 + 12 + 12

         = -8

؞ The remainder is -8.

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

(iii)             2x³ -  4x² + 7x + 6 is divided by x – 2

Solution:

Let p(x) = 2x³ -  4x² + 7x + 6

The zero of (x - 2) is 2.

When p(x) is divided by (x - 2) the remainder is p(2). Now,

p(2) =2(2)³ -  4(2)² + 7(2) + 6

         = 16 - 16 + 14 + 6

         = 20

؞ The remainder is 20.

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

 

(iv)             4x³ -  3x²+  2x -  4 is divided by x + 3

Solution:

Let p(x) = 4x³ -  3x²+  2x -  4

The zero of (x + 3) is -3.

When p(x) is divided by (x+ 3) the remainder is p(-3). Now,

p(-3) = 4(-3)³ – 3(-3)²+ 2(-3) -  4

         = -108 – 27 – 6 – 4

         = -145 

؞ The remainder is -145.

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

 

(v)               4x³ -  12x² +  11x -  5 is divided by 2x – 1

Solution:

Let p(x) = 4x³ -  12x² +  11x -  5

The zero of (2x - 1) is ½ .

When p(x) is divided by (2x- 1) the remainder is p(1/2). Now,

p(1/2) = 4(1/2)³ –  12(1/2)² +  11(1/2) -  5

          = 4/8 – 12/4 + 11/2 - 5

           = ½  - 3 + 11/2 - 5

           = 12/2 – 8 = 6 – 8

          = -2 

 ؞ The remainder is -2.

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

(vi)             8x⁴+ 12x³- 2x²- 18x+14 is divided by x + 1

Solution:

Let p(x) = 8x⁴+ 12x³- 2x²- 18x+14

The zero of (x + 1) is -1.

When p(x) is divided by (x+1) the remainder is p(-1). Now,

p(-1) = 8(-1)⁴+ 12(-1)³- 2(-1)²- 18(-1)+14

         = 8 – 12 – 2 + 18 + 14

         = 26

؞ The remainder is 26.

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

(vii)            x³- ax²- 5x+ 2a is divided by x – a

Solution:

Let p(x) = x³- ax²- 5x+ 2a  

The zero of (x - a) is a.

When p(x) is divided by (x- a) the remainder is p(a). Now,

p(a) = (a)³- a(a)²- 5(a)+ 2a

         = a³ – a³ – 5a + 2a

         = -3a

؞ The remainder is -3a.

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

2.  When the polynomial 2x³ – ax² + 9x - 8 is divided by x - 3 the remainder is 28.

     Find the value of a.

Solution:

                  Let p(x) = 2x³ – ax² + 9x - 8

                 When p(x) is divided by(x – 3) the remainder is p(3).

                 Given that p(3) = 28

ð  2(3)³ – a(3)² + 9(3) – 8 = 28

         54 – 9a + 27 – 8    = 28

               73 – 9a             = 28   

                                 -9a = 28 – 73

                                     a = -45/-9

                                     a = 5

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

3. Find the value of m if x^{3 -}  6x²+ mx+ 60 leaves the remainder 2 when divided by

    (x + 2).  

Solution:

                  Let p(x) = x^{3 -}  6x²+ mx+ 60

                 When p(x) is divided by (x +2) the remainder is p(-2).

                 Given that p(-2) = 2

ð  (-2)^{3 –}  6(-2)²+ m(-2)+ 60 = 2

  -8 – 24 – 2m + 60          = 2

    28 – 2m                        = 2

                                  -2m = 2 – 28

                                     m = -26/-2

                         

m = 13

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

 4. If(x – 1) divides mx3 -  2x² +  25x - 26 without remainder find the value of m.

Solution:

                  Let p(x) = mx³ -  2x² +  25x - 26

                 When p(x) is divided by (x -1) the remainder is p(1).

                 Given that p(1) = 0

ð  m(1)³ -  2(1)² +  25(1) – 26 = 0

           m – 2 + 25 – 26       = 0

                            m – 3      = 0

                                          

m = 3

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

5. If the polynomials x³ +  3x²-  m and 2x³-  mx + 9 leave the same remainder when

    they are divided by(x – 2), find the value of m. Also find the remainder.

Solution:

                          Let p(x) = x³ +  3x²-  m

When p(x) is divided by (x -2) the remainder is p(2).

                        Let  q(x) =  2x³-  mx + 9

When q(x) is divided by (x -2) the remainder is q(2).

        Given that p(2) = q(2)

ð  (2)³ +  3(2)²-  m = 2(2)³-  m(2) + 9

        8 + 12 – m = 16 – 2m +9

               20 – m = 25 – 2m

               2m – m = 25 – 20

m = 5

Now,

p(2) = (2)³ +  3(2)²-  m

                              = 8 + 12 – 5

         Remainder  = 15
---------------------------------------------------------------------------------------------------------