<PRE><B>Question 293740</B>


Looking at the expression {{{t^2+3t-10}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{3}}}, and the last term is {{{-10}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{-10}}} to get {{{(1)(-10)=-10}}}.



Now the question is: what two whole numbers multiply to {{{-10}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{3}}}?



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 {{{-10}}} (the previous product).



Factors of {{{-10}}}:

1,2,5,10

-1,-2,-5,-10



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-10}}}.

1*(-10) = -10
2*(-5) = -10
(-1)*(10) = -10
(-2)*(5) = -10


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{3}}}:



&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-10)=-9&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;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-5)=-3&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+10=9&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;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-2+5=3&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-2}}} and {{{5}}} add to {{{3}}} (the middle coefficient).



So the two numbers {{{-2}}} and {{{5}}} both multiply to {{{-10}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{3}}}



Now replace the middle term {{{3t}}} with {{{-2t+5t}}}. Remember, {{{-2}}} and {{{5}}} add to {{{3}}}. So this shows us that {{{-2t+5t=3t}}}.



{{{t^2+highlight(-2t+5t)-10}}} Replace the second term {{{3t}}} with {{{-2t+5t}}}.



{{{(t^2-2t)+(5t-10)}}} Group the terms into two pairs.



{{{t(t-2)+(5t-10)}}} Factor out the GCF {{{t}}} from the first group.



{{{t(t-2)+5(t-2)}}} 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.



{{{(t+5)(t-2)}}} Combine like terms. Or factor out the common term {{{t-2}}}



===============================================================



Answer:



So {{{t^2+3t-10}}} factors to {{{(t+5)(t-2)}}}.



In other words, {{{t^2+3t-10=(t+5)(t-2)}}}.



Note: you can check the answer by expanding {{{(t+5)(t-2)}}} to get {{{t^2+3t-10}}} or by graphing the original expression and the answer (the two graphs should be identical).
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