Question 458026
When ax^3-2x^2+3x-5 is divided by 2x-1, the remainder is -8
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      _________________________
2x - 1)ax³ - 2x² - 2x² + 3x - 5

We won't divide that out, but suppose we did.
We would get some quotient Q(x) and a remainder of -8

      <u>          Q(x)            </u>
2x - 1)ax³ - 2x² - 2x² + 3x - 5
         ..... 
           ......
                        _______
                             -8

Then use 

               DIVIDEND = QUOTIENT*DIVISOR + REMAINDER

     ax³ - 2x² + 3x - 5 = Q(x)*(2x - 1) + (-8)

We want to make the factor (2x - 1) equal to zero so that
term will be eliminated. 

2x - 1 = 0
    2x = 1
     x = ½
 
Now substitute x = ½  in

     ax³ - 2x² + 3x - 5 = Q(x)*(2x - 1) + (-8)

     a(½)³ - 2(½)² + 3(½) - 5 = Q(½)*(2*½ - 1) + (-8)

     a(&#8539;) - 2(¼) + {{{3/2}}} - 5 = Q(½)*(1 - 1) - 8

    {{{a/8}}} - ½ + {{{3/2}}} - 5 = Q(½)*(0) - 8

    {{{a/8}}} - ½ + {{{3/2}}} - 5 = 0 - 8

    {{{a/8}}} - ½ + {{{3/2}}} - 5 = -8

Clear of fractions by multiplying through by 8

         a - 4 + 12 - 40 = -64
                  a - 32 = -64
                       a = -32    


Checking:    

      <u>        -16x² - 9x - 3</u> 
2x - 1)-32x³ -  2x² + 3x - 5
       <u>-32x³ + 16x²</u>
              -18x² + 3x
              <u>-18x² + 9x</u>
                     -6x - 5
                     <u>-6x + 3</u>
                          -8    
 
It leaves the remainder -8, so we are correct!
Edwin</pre>