<PRE><B>Question 694628</B>


Looking at the expression {{{2t^2-t-3}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{-1}}}, 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 {{{-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 {{{-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 {{{-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;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-6)=-5&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)=-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;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+6=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;-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=1&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 {{{-1}}} (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 {{{-1}}}



Now replace the middle term {{{-1t}}} with {{{2t-3t}}}. Remember, {{{2}}} and {{{-3}}} add to {{{-1}}}. So this shows us that {{{2t-3t=-1t}}}.



{{{2t^2+highlight(2t-3t)-3}}} Replace the second term {{{-1t}}} with {{{2t-3t}}}.



{{{(2t^2+2t)+(-3t-3)}}} Group the terms into two pairs.



{{{2t(t+1)+(-3t-3)}}} Factor out the GCF {{{2t}}} from the first group.



{{{2t(t+1)-3(t+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.



{{{(2t-3)(t+1)}}} Combine like terms. Or factor out the common term {{{t+1}}}



===============================================================



Answer:



So {{{2t^2-t-3}}} factors to {{{(2t-3)(t+1)}}}.



In other words, {{{2t^2-t-3=(2t-3)(t+1)}}}.



Note: you can check the answer by expanding {{{(2t-3)(t+1)}}} to get {{{2t^2-t-3}}} or by graphing the original expression and the answer (the two graphs should be identical).
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