<PRE><B>Question 618477</B>


Looking at the expression {{{3y^2+13y+14}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{13}}}, and the last term is {{{14}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{14}}} to get {{{(3)(14)=42}}}.



Now the question is: what two whole numbers multiply to {{{42}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{13}}}?



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 {{{42}}} (the previous product).



Factors of {{{42}}}:

1,2,3,6,7,14,21,42

-1,-2,-3,-6,-7,-14,-21,-42



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{42}}}.

1*42 = 42
2*21 = 42
3*14 = 42
6*7 = 42
(-1)*(-42) = 42
(-2)*(-21) = 42
(-3)*(-14) = 42
(-6)*(-7) = 42


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{13}}}:



&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;42&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+42=43&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;21&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+21=23&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;14&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+14=17&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;6+7=13&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;-42&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-42)=-43&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;-21&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-21)=-23&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;-14&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+(-14)=-17&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;-7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-6+(-7)=-13&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{6}}} and {{{7}}} add to {{{13}}} (the middle coefficient).



So the two numbers {{{6}}} and {{{7}}} both multiply to {{{42}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{13}}}



Now replace the middle term {{{13y}}} with {{{6y+7y}}}. Remember, {{{6}}} and {{{7}}} add to {{{13}}}. So this shows us that {{{6y+7y=13y}}}.



{{{3y^2+highlight(6y+7y)+14}}} Replace the second term {{{13y}}} with {{{6y+7y}}}.



{{{(3y^2+6y)+(7y+14)}}} Group the terms into two pairs.



{{{3y(y+2)+(7y+14)}}} Factor out the GCF {{{3y}}} from the first group.



{{{3y(y+2)+7(y+2)}}} Factor out {{{7}}} 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.



{{{(3y+7)(y+2)}}} Combine like terms. Or factor out the common term {{{y+2}}}



===============================================================



Answer:



So {{{3y^2+13y+14}}} factors to {{{(3y+7)(y+2)}}}.



In other words, {{{3y^2+13y+14=(3y+7)(y+2)}}}.



Note: you can check the answer by expanding {{{(3y+7)(y+2)}}} to get {{{3y^2+13y+14}}} or by graphing the original expression and the answer (the two graphs should be identical).
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