<PRE><B>Question 423054</B>
I&#39;m assuming you mean factor into a product of binomials.





Looking at the expression {{{3x^2+10x+3}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{10}}}, and the last term is {{{3}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{3}}} to get {{{(3)(3)=9}}}.



Now the question is: what two whole numbers multiply to {{{9}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{10}}}?



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 {{{9}}} (the previous product).



Factors of {{{9}}}:

1,3,9

-1,-3,-9



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{9}}}.

1*9 = 9
3*3 = 9
(-1)*(-9) = 9
(-3)*(-3) = 9


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{10}}}:



&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=red&gt;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;9&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;1+9=10&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+3=6&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;-9&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-9)=-10&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;-3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+(-3)=-6&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{1}}} and {{{9}}} add to {{{10}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{9}}} both multiply to {{{9}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{10}}}



Now replace the middle term {{{10x}}} with {{{x+9x}}}. Remember, {{{1}}} and {{{9}}} add to {{{10}}}. So this shows us that {{{x+9x=10x}}}.



{{{3x^2+highlight(x+9x)+3}}} Replace the second term {{{10x}}} with {{{x+9x}}}.



{{{(3x^2+x)+(9x+3)}}} Group the terms into two pairs.



{{{x(3x+1)+(9x+3)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(3x+1)+3(3x+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.



{{{(x+3)(3x+1)}}} Combine like terms. Or factor out the common term {{{3x+1}}}



===============================================================



Answer:



So {{{3x^2+10x+3}}} factors to {{{(x+3)(3x+1)}}}.



In other words, {{{3x^2+10x+3=(x+3)(3x+1)}}}.



Note: you can check the answer by expanding {{{(x+3)(3x+1)}}} to get {{{3x^2+10x+3}}} or by graphing the original expression and the answer (the two graphs should be identical).



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Jim</PRE>