Question 1186024
<pre>

Here's another way to do it, by dividing directly with long division:

       <u>             ax<sup>2</sup>+       bx+    (-8+a)</u>
x<sup>2</sup>+0x-1)ax<sup>4</sup>+bx<sup>3</sup>-     8x<sup>2</sup>+      6x-         6
        <u>ax<sup>4</sup>+0x<sup>3</sup>-     ax<sup>2</sup></u>
            bx<sup>3</sup>+(-8+a)x<sup>2</sup>+      6x
            <u>bx<sup>3</sup>+     0x<sup>2</sup>-      bx</u>
                (-8+a)x<sup>2</sup>+  (6+b)x-         6
                <u>(-8+a)x<sup>2</sup>+      0x-    (-8+a)</u>
                           (6+b)x+(-6+(-8+a))

For this remainder to be 2x+1,

6+b = 2        -6+(-8+a) = 1
  b = -4          -6-8+a = 1
                   -14+a = 1
                       a = 15
                       
Edwin</pre>