<PRE><B>Question 318015</B>


{{{3c^2+6c-105}}} Start with the given expression.



{{{3(c^2+2c-35)}}} Factor out the GCF {{{3}}}.



Now let&#39;s try to factor the inner expression {{{c^2+2c-35}}}



---------------------------------------------------------------



Looking at the expression {{{c^2+2c-35}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{2}}}, and the last term is {{{-35}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{-35}}} to get {{{(1)(-35)=-35}}}.



Now the question is: what two whole numbers multiply to {{{-35}}} (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 {{{-35}}} (the previous product).



Factors of {{{-35}}}:

1,5,7,35

-1,-5,-7,-35



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-35}}}.

1*(-35) = -35
5*(-7) = -35
(-1)*(35) = -35
(-5)*(7) = -35


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;-35&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-35)=-34&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;-7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;5+(-7)=-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;35&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+35=34&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;7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-5+7=2&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-5}}} and {{{7}}} add to {{{2}}} (the middle coefficient).



So the two numbers {{{-5}}} and {{{7}}} both multiply to {{{-35}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{2}}}



Now replace the middle term {{{2c}}} with {{{-5c+7c}}}. Remember, {{{-5}}} and {{{7}}} add to {{{2}}}. So this shows us that {{{-5c+7c=2c}}}.



{{{c^2+highlight(-5c+7c)-35}}} Replace the second term {{{2c}}} with {{{-5c+7c}}}.



{{{(c^2-5c)+(7c-35)}}} Group the terms into two pairs.



{{{c(c-5)+(7c-35)}}} Factor out the GCF {{{c}}} from the first group.



{{{c(c-5)+7(c-5)}}} Factor out {{{7}}} 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.



{{{(c+7)(c-5)}}} Combine like terms. Or factor out the common term {{{c-5}}}



--------------------------------------------------



So {{{3(c^2+2c-35)}}} then factors further to {{{3(c+7)(c-5)}}}



===============================================================



Answer:



So {{{3c^2+6c-105}}} completely factors to {{{3(c+7)(c-5)}}}.



In other words, {{{3c^2+6c-105=3(c+7)(c-5)}}}.



Note: you can check the answer by expanding {{{3(c+7)(c-5)}}} to get {{{3c^2+6c-105}}} or by graphing the original expression and the answer (the two graphs should be identical).
</PRE>