Question

The polynomials ax ³ +3x³and 2x³-5x+a when divided by (x-4)leaves the remainder k1 and k2 respectivley dind the value of a if k1 + k2=0

Answer 1

Given :

The first polynomial = a x³ + 3 x² - 3

The remainder of first polynomial = $K_1$

The second polynomial =  2 x³ - 5 x + a

The remainder of first polynomial = $K_2$

Both the polynomials are divided by ( x - 4 )

The sum $K_1$ + $K_2$  = 0

To Find :

The value of a

Solution :

∵ Both the polynomials are divided by ( x - 4 )

So, It must satisfy the polynomials ,

    i.e   x = 4

So, Put the value of x in both polynomials

Or, a x³ + 3 x² - 3 = a × (4 )³ + 3 × (4 )² - 3

                        = 64 a + 48 - 3

                        = 64 a + 45

And

   2 x³ - 5 x + a = 2 × (4 )³ - 5 × 4 + a

                        = 128 - 20 + a

                        = 108 + a

∵ The sum of remainder of both polynomials = 0

i.e   $K_1$ + $K_2$  = 0

So, ( 64 a + 45 ) = ( 108 + a )

Or,  ( 64 a - a ) = ( - 45 + 108 )

Or,  63 a = 63

∴          a = $\dfrac{63}{63}$

i.e        a = 1

Hence, The value of a when sum of remainder zero is 1  Answer