<p>For instance, how would you go about factorizing this:</p>
<p>3(x^3/2) - 9(x^1/2) + 6(x^-1/2)</p>
<p>Thanks. :)</p>
<p>I would factor out a 3 (it’s the greatest common factor of 3, 9, and 6). That leaves you with:</p>
<p>3(x^3/2 - 3x^1/2 + 2x^-1/2)</p>
<p>How far do you need to take this problem?</p>
<p>I don’t think you can take this problem any further. If you make x^1/2 = y, you get:</p>
<p>3(y^3 - 3y - 2/y)</p>
<p>^That clearly isn’t anything special (e.g. difference of cubes).</p>
<p>If he factored out the x^-1/2, couldn’t he have a factorable polynomial with the solutions x=1 and x=2?</p>
<p>Nevermind. That would only be if it was set equal to 0. x=0 would be a solution, as well.</p>
<p>^^ Tyler is on the right track.</p>
<p>Rewrite as: 3x^1/2 (3x-9+6/x)=3x^1/2 [(x^2-3x+2)/x]=<a href=“x-1”>3x^-1/2</a>(x-2)</p>
<p>Validate with x=1 and you get 0. Then validate with x=2, and you get 0.</p>
<p>(Note that for x=0 the result is “undefined”. Don’t assume that “factoring” always means that the result must be a polynomial.)</p>
<p>Of course, I forgot that if x was 0, 3x^-1/2 would cause a domain error, as it is actually 3/sqrt(0), or 3/0.</p>