<PRE><B>Question 703839</B>


Looking at the expression {{{2x^2+5x+3}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{5}}}, and the last term is {{{3}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{3}}} to get {{{(2)(3)=6}}}.



Now the question is: what two whole numbers multiply to {{{6}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{5}}}?



To find these two numbers, we need to list &lt;font size=4&gt;&lt;b&gt;all&lt;/b&gt;&lt;/font&gt; of the factors of {{{6}}} (the previous product).



Factors of {{{6}}}:

1,2,3,6

-1,-2,-3,-6



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{6}}}.

1*6 = 6
2*3 = 6
(-1)*(-6) = 6
(-2)*(-3) = 6


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{5}}}:



&lt;table border=&quot;1&quot;&gt;&lt;th&gt;First Number&lt;/th&gt;&lt;th&gt;Second Number&lt;/th&gt;&lt;th&gt;Sum&lt;/th&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+6=7&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2+3=5&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-6)=-7&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-3)=-5&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{2}}} and {{{3}}} add to {{{5}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{3}}} both multiply to {{{6}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{5}}}



Now replace the middle term {{{5x}}} with {{{2x+3x}}}. Remember, {{{2}}} and {{{3}}} add to {{{5}}}. So this shows us that {{{2x+3x=5x}}}.



{{{2x^2+highlight(2x+3x)+3}}} Replace the second term {{{5x}}} with {{{2x+3x}}}.



{{{(2x^2+2x)+(3x+3)}}} Group the terms into two pairs.



{{{2x(x+1)+(3x+3)}}} Factor out the GCF {{{2x}}} from the first group.



{{{2x(x+1)+3(x+1)}}} Factor out {{{3}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(2x+3)(x+1)}}} Combine like terms. Or factor out the common term {{{x+1}}}



===============================================================



Answer:



So {{{2x^2+5x+3}}} factors to {{{(2x+3)(x+1)}}}.



In other words, {{{2x^2+5x+3=(2x+3)(x+1)}}}.



Note: you can check the answer by expanding {{{(2x+3)(x+1)}}} to get {{{2x^2+5x+3}}} or by graphing the original expression and the answer (the two graphs should be identical).
</PRE>