Question 723479
This is the first I've ever heard of the "diamond method". I'll try to explain how it will factor the numerator and denominator.<br>
{{{2x^2+11x+5}}}
<ol><li>a*c = 10 so a 10 goes in the top box</li><li>b is 11 so an 11 goes in the bottom box</li><li>The factors of the 10 in the top box that add up to the 11 in the bottom box are 10 and 1. So a 10 and a 1 go in the left and right boxes.</li><li>Next we make fractions in the left and right boxes. The numbers that are there, 10 and 1, go into the denominators of these fractions. The numerators will be a*x. Since the a is 2 in this problem, the numerators will be "2x". So the left and right boxes now have "2x/10" and "2x/1" in them.</li><li>Reduce but don't eliminate the fractions in the left and right boxes. The "2x/10" fraction will reduce to "x/5". Normally we would simplify "2x/1" to just "2x" but we do not want to eliminate the fraction so we will leave it as "2x/1".</li><li>The factors are formed from the fractions in the left and right boxes. The factors will be of the form: (numerator + denominator). So "x/5" becomes the factor (x+5) and the "2x/1" becomes the factor (2x+1)</li></ol>So {{{2x^2+11x+5 = (x+5)(2x+1)}}}<br>
Repeating this with {{{3x^2-5x+2}}}:<ol><li>Top box: 6</li><li>Bottom box: -5</li><li>Left and right box: -2 and -3 (Don't forget the negative factors of positive numbers!)</li><li>Left and right boxes: 3x/-2 and 3x/-3</li><li>Left and right boxes: 3x/-2 and x/-1</li><li>Factors: (3x + (-2)) and (x + (-1))</li></ol> So {{{3x^2-5x+2 = (3x+(-2))(x+(-1))}}} or (3x-2)(x-1)<br>
So  {{{(2x^2+11x+5)/(3x^2-5x+2) = ((x+5)(2x+1))/((3x-2)(x-1))}}}. Since there are no common factors between the numerator and denominator, this fraction will not reduce.