<PRE><B>Question 283487</B>


{{{3y^2+24y+45}}} Start with the given expression.



{{{3(y^2+8y+15)}}} Factor out the GCF {{{3}}}.



Now let&#39;s try to factor the inner expression {{{y^2+8y+15}}}



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Looking at the expression {{{y^2+8y+15}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{8}}}, and the last term is {{{15}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{15}}} to get {{{(1)(15)=15}}}.



Now the question is: what two whole numbers multiply to {{{15}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{8}}}?



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 {{{15}}} (the previous product).



Factors of {{{15}}}:

1,3,5,15

-1,-3,-5,-15



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{15}}}.

1*15 = 15
3*5 = 15
(-1)*(-15) = 15
(-3)*(-5) = 15


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{8}}}:



&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;15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+15=16&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;3+5=8&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;-15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-15)=-16&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;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+(-5)=-8&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{3}}} and {{{5}}} add to {{{8}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{5}}} both multiply to {{{15}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{8}}}



Now replace the middle term {{{8y}}} with {{{3y+5y}}}. Remember, {{{3}}} and {{{5}}} add to {{{8}}}. So this shows us that {{{3y+5y=8y}}}.



{{{y^2+highlight(3y+5y)+15}}} Replace the second term {{{8y}}} with {{{3y+5y}}}.



{{{(y^2+3y)+(5y+15)}}} Group the terms into two pairs.



{{{y(y+3)+(5y+15)}}} Factor out the GCF {{{y}}} from the first group.



{{{y(y+3)+5(y+3)}}} Factor out {{{5}}} 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.



{{{(y+5)(y+3)}}} Combine like terms. Or factor out the common term {{{y+3}}}



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So {{{3(y^2+8y+15)}}} then factors further to {{{3(y+5)(y+3)}}}



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Answer:



So {{{3y^2+24y+45}}} completely factors to {{{3(y+5)(y+3)}}}.



In other words, {{{3y^2+24y+45=3(y+5)(y+3)}}}.



Note: you can check the answer by expanding {{{3(y+5)(y+3)}}} to get {{{3y^2+24y+45}}} or by graphing the original expression and the answer (the two graphs should be identical).
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