<PRE><B>Question 1031275</B>

How do you factor 5x^2+21x+4=0? The AC method has been slightly puzzling me, and I cannot quite figure it out.
&lt;pre&gt;ac = + 5 * + 4 = + 20
Looking for a PAIR of factors that multiply to give: ac, or + 20, and SUM to &quot;b&quot;, or + 21 (value of b, since the general form of a quadratic is: {{{ab^2 + bx + c = 0}}}).
These 2 factors are + 20 and + 1
{{{5x^2 + 21x + 4 = 0}}}
{{{5x^2 + 20x + x + 4 = 0}}} ------- Replacing middle term: + 21x with + 20x and + x
&lt;b&gt;&lt;u&gt;5x&lt;sup&gt;2&lt;/sup&gt; + 20x&lt;/b&gt;&lt;/u&gt; &lt;b&gt;&lt;u&gt;+ x + 4&lt;/b&gt;&lt;/u&gt; -------- Separating polynomials in order to factor each
5x(x + 4) + 1(x + 4)
{{{highlight_green((5x + 1)(x + 4) = 0)}}}
Continue from here, setting each binomial equal to 0, and then solving for x</PRE>