<PRE><B>Question 309886</B>
I&#39;m assuming that you want to factor this.



{{{2t^2+4t-30}}} Start with the given expression.



{{{2(t^2+2t-15)}}} Factor out the GCF {{{2}}}.



Now let&#39;s try to factor the inner expression {{{t^2+2t-15}}}



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Looking at the expression {{{t^2+2t-15}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{2}}}, 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 {{{2}}}?



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 {{{2}}}:



&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)=-14&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)=-2&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=14&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=2&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 {{{2}}} (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 {{{2}}}



Now replace the middle term {{{2t}}} with {{{-3t+5t}}}. Remember, {{{-3}}} and {{{5}}} add to {{{2}}}. So this shows us that {{{-3t+5t=2t}}}.



{{{t^2+highlight(-3t+5t)-15}}} Replace the second term {{{2t}}} with {{{-3t+5t}}}.



{{{(t^2-3t)+(5t-15)}}} Group the terms into two pairs.



{{{t(t-3)+(5t-15)}}} Factor out the GCF {{{t}}} from the first group.



{{{t(t-3)+5(t-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.



{{{(t+5)(t-3)}}} Combine like terms. Or factor out the common term {{{t-3}}}



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So {{{2(t^2+2t-15)}}} then factors further to {{{2(t+5)(t-3)}}}



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Answer:



So {{{2t^2+4t-30}}} completely factors to {{{2(t+5)(t-3)}}}.



In other words, {{{2t^2+4t-30=2(t+5)(t-3)}}}.



Note: you can check the answer by expanding {{{2(t+5)(t-3)}}} to get {{{2t^2+4t-30}}} or by graphing the original expression and the answer (the two graphs should be identical).
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