<PRE><B>Question 924006</B>
&lt;pre&gt;
a&lt;sup&gt;2&lt;/sup&gt;(a+1)-2a(a+1)+a+1

Put parentheses around the last two terms

a&lt;sup&gt;2&lt;/sup&gt;(a+1)-2a(a+1)+(a+1)

Put a 1 in front of the last parentheses:

a&lt;sup&gt;2&lt;/sup&gt;(a+1)-2a(a+1)+1(a+1)

Now you see that each term contains a &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;

a&lt;sup&gt;2&lt;/sup&gt;&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;-2a&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;+1&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;

And you can factor out the red &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;.

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;(a&lt;sup&gt;2&lt;/sup&gt;-2a+1)

Next, the black part factors:

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;(a-1)(a-1)

or

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(a+1)&lt;/b&gt;&lt;/font&gt;(a-1)&lt;sup&gt;2&lt;/sup&gt;

or dropping the red coloring:

(a+1)(a-1)&lt;sup&gt;2&lt;/sup&gt;

becomes



Edwin&lt;/pre&gt;</PRE>