Question 601655
<pre>
10x² - 7xy - 5y²

The factorization, if there is one, has to be of the form

(&#8591;x &#8591; &#8591;y)(&#8591;x &#8591; &#8591;y)

Since there is just one factorization of the number 5 in the 
last term as 5·1, we can put this:

(&#8591;x &#8591; 5y)(&#8591;x &#8591; 1y)

The 10 can be factored as 10·1, 5·2, 2·5 and 1·10, so we try all these
using tho O and I of "FOIL"

(10x &#8591; 5y)(1x &#8591; 1y) <--- outers = 10xy,  Inners = 5xy, difference = 5xy
(5x &#8591; 5y)(2x &#8591; 1y)  <--- outers = 5xy, Inners = 10xy, difference = 5xy
(2x &#8591; 5y)(5x &#8591; 1y)  <--- outers = 2xy, inners = 25xy, difference = 23xy 
(1x &#8591; 5y)(10x &#8591; 1y) <--- outers = 1xy, inners = 50xy, difference = 49xy

Since the last sign of 10x² - 7xy - 5y² is -, we look through
those to find one which two have a difference of the middle term,
7xy, ignoring the sign.  

There are none, so the trinomial 10x² - 7xy - 5y² is PRIME, i.e., it 
cannot be factored using integer coefficients.

Edwin</pre>