Question 978038
Find the value of k for which a-3b is a factor of a⁴-7a²b²+kb⁴ . 
Hence for this value of k, factorise a⁴-7a²b²+kb⁴ completely.
<pre><font size=4>
Divide by long division, inserting zero terms for missing terms +0a³b and
+0ab³:

    <u>     a³ +3a²b +2ab²+ 6b³</u>   
a-3b)a&#8308;+0a³b-7a²b²+0ab³+ kb&#8308;
     <u>a&#8308;-3a³b</u>
        3a³b-7a²b²
        <u>3a³b-9a²b²</u>
             2a²b²+0ab³
             <u>2a²b²-6ab³</u>
                   6ab³+ kb&#8308;
                   <u>6ab³-18b&#8308;</u>
                         kb&#8308;+18b&#8308;  <-- remainder

The remainder must = 0 so that a-3b 
will be a factor of a&#8308;-7a²b²+kb&#8308;.

kb&#8308;+18b&#8308; = 0
(k+18)b&#8308; = 0
k+18=0; b&#8308;=0
 k=-18;  b=0  

There is a trivial case for b=0 and k is any number, 
which we ignore.

So k = -18

a&#8308;-7a²b²-18b&#8308;

(a²+2b²)(a²-9b²)

(a²+2b²)(a-3b)(a+3b)

Edwin</pre>