<p>What is the largest possible value of the following expression?
(x+2)(3-x)(2+x)^2(2x-6)(2x+4)</p>
<p>The answer is 0 but I would like to know how to go about such a question.</p>
<p>(2+x) is the only item squared.</p>
<p>easiest way to solve this is to graph it and find the maximums</p>
<p>…or you could multiply it all out and solve for the maximums</p>
<p>Sneaky question, but no graphing or foiling needed…</p>
<p>If (x+2)(3-x)(2+x)^2(2x-6)(2x+4),</p>
<p>Notice that you can factor a 2 out of two of the terms and a minus sign out of (3-x) and get:</p>
<p>-4(x+2)(x-3)(2+x)^2(x-3)(x+2) </p>
<p>or -4(x+2)^4(x-3)^2</p>
<p>Which will never be positive because of the even exponents and the minus sign out front. </p>
<p>So the biggest it can be is zero.</p>
<p>Thanks pckeller but I don’t understand why and how you factorized a minus sign out of (3-x)</p>
<p>So that after I factored a 2 out of (2x-6) I would have two “(x-3)”'s that I could pair up as
(x-3)^2, which can never be negative.</p>
<p>Thanks pckeller. I understand it now.</p>