What is the remainder when 4x4 − 10x3 + 14x2 + 7x − 19 is divided by 2x − 1

Using long division, we have

       
2x-1 | 4x^4-10x³+14x²+7x-19

Next, we divide the first term of 2x-1 by the first term of 4x^4-10x³+14x²+7x-19 to get 4x^4/2x=2x³ (since 4/2=2 and to divide exponents you subtract the denominator from the numerator). Next, we put our 2x³ on the top and multiply (2x-1) by that. With the result of (2x-1) and 2x³, we multiply that by -1 and add it to 4x^4-10x³+14x²+7x-19, looking like

        2x³
       
2x-1 | 4x^4-10x³+14x²+7x-19

       - (4x^4-2x³)
         
                  -8x³+14x²+7x-19. Repeating the process, we get

        2x³-4x²+5x+6
       
2x-1 | 4x^4-10x³+14x²+7x-19

       - (4x^4-2x³)
         
                  -8x³+14x²+7x-19. 
                  -(-8x³+4x²)
       
                         10x²+7x-19
                        -(10x²-5x)
                       
                                 12x-19
                               -(12x-6)
                             
                                       -13 as our remainder

The probability would be a 1 out of 6, if it is wrong, then i take all blame for it! ^_^