when 16x^3 - 32x^2 + cx - 8 is divided by ( 2x-1) the remiander is 1.Find the value of c

Question

it is another maths of remainder problem.

how could i find  c????

Answer 1

or:

f( X ) = 16 X^3 - 32 x^2 + cX - 8

 

 f ( 1/2 ) = 1

 

   16 ( 1/2) ^3 - 32 ( 1/2) ^2 + C ( 1/2) - 8 = 1

 

      16/8 - 32/ 4 + C/2 - 8 = 1

 

       2 - 4 + C/2 - 8= 1

 

         C/ 2 = 1 + 14

 

         C = 2( 15 ) =30

Answer 2

Hi Rodela;

16x^3 - 32x^2 + cx - 8

(2x-1)(...

FIRST...(2x)(8x²)=16x³

(2x-1)(8x²...

SECOND...(-1)(8x²)=-8x² and (2x)(-12x)=-24x²...-8x²-24x²=-32x²

(2x-1)(8x²-12x...

THIRD...(-1)(-12x)=12x and (2x)(0.5c-6)=cx-12x....12x+cx-12x=cx

(2x-1)[8x²-12x+(0.5c-6)]

(2x-1)(8x²-12x+0.5c-6)

Let's do the multiplication...

(2x)(8x²-12x+0.5c-6)=16x³-24x²+cx-12x

(-1)(8x²-12x+0.5c-6)=.......-8x²+12x-0.5c+6

(2x-1)(8x²-12x+0.5c-6)=16x³-32x²+cx-0.5c+6

 

The remainder is 1.

The constant is -8 which is equal to -9+1.

-0.5c+6=-9

-0.5c+6=-9

-0.5c=-15

c=30

 

The equation is now...

16x³ - 32x² + 30x - 9+1

16x³-32x²+30x-8

Answer 3

            8x^2 - 12x + (c-12)/2 

2x-1 | 16x^3 - 32x^2 + cx - 8

          16x^3 -   8x^2

                    - 24x^2   

                    - 24x^2 + 12x

                                  (c-12)x-8 

                                  (c-12)x-(c-12)/2

                                              (c-12)/2-8 = R

 

(c-12)/2-8 = R = 1

 

(c-12)/2 = 9

 

c-12 = 18

 

c = 30

=== OR ===

p(x) = (2x-1)Q(x) + R

 

p(x) = (x-1/2)(2Q(x)) + R

p(1/2) = ((1/2)-1/2)(2Q(x)) + R

p(1/2) = (0)(2Q(x)) + R

R = p(1/2)

1/2 | 16 -32   c         -8
                8 -12     (c-12)/2    
        16 -24 c-12 | (c-12)/2 - 8 = R

 

Note that 2Q(x) = 16x^2 - 24x + c-12.

(c-12)/2 - 8 = 1

c = 30