<PRE><B>Question 599416</B>


Looking at the expression {{{15y^2-y-2}}}, we can see that the first coefficient is {{{15}}}, the second coefficient is {{{-1}}}, and the last term is {{{-2}}}.



Now multiply the first coefficient {{{15}}} by the last term {{{-2}}} to get {{{(15)(-2)=-30}}}.



Now the question is: what two whole numbers multiply to {{{-30}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-1}}}?



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 {{{-30}}} (the previous product).



Factors of {{{-30}}}:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-30}}}.

1*(-30) = -30
2*(-15) = -30
3*(-10) = -30
5*(-6) = -30
(-1)*(30) = -30
(-2)*(15) = -30
(-3)*(10) = -30
(-5)*(6) = -30


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-1}}}:



&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;-30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-30)=-29&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;-15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-15)=-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;3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+(-10)=-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;5&lt;/font&gt;&lt;/td&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;5+(-6)=-1&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;30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+30=29&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;15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+15=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;-3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+10=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;-5&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;-5+6=1&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{5}}} and {{{-6}}} add to {{{-1}}} (the middle coefficient).



So the two numbers {{{5}}} and {{{-6}}} both multiply to {{{-30}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-1}}}



Now replace the middle term {{{-1y}}} with {{{5y-6y}}}. Remember, {{{5}}} and {{{-6}}} add to {{{-1}}}. So this shows us that {{{5y-6y=-1y}}}.



{{{15y^2+highlight(5y-6y)-2}}} Replace the second term {{{-1y}}} with {{{5y-6y}}}.



{{{(15y^2+5y)+(-6y-2)}}} Group the terms into two pairs.



{{{5y(3y+1)+(-6y-2)}}} Factor out the GCF {{{5y}}} from the first group.



{{{5y(3y+1)-2(3y+1)}}} Factor out {{{2}}} 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.



{{{(5y-2)(3y+1)}}} Combine like terms. Or factor out the common term {{{3y+1}}}



===============================================================



Answer:



So {{{15y^2-y-2}}} factors to {{{(5y-2)(3y+1)}}}.



In other words, {{{15y^2-y-2=(5y-2)(3y+1)}}}.



Note: you can check the answer by expanding {{{(5y-2)(3y+1)}}} to get {{{15y^2-y-2}}} or by graphing the original expression and the answer (the two graphs should be identical).
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