<PRE><B>Question 195817</B>


Looking at the expression {{{k^2-6k+5}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-6}}}, and the last term is {{{5}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{5}}} to get {{{(1)(5)=5}}}.



Now the question is: what two whole numbers multiply to {{{5}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-6}}}?



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 {{{5}}} (the previous product).



Factors of {{{5}}}:

1,5

-1,-5



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{5}}}.

1*5
(-1)*(-5)


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-6}}}:



&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;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+5=6&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-1&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;-1+(-5)=-6&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-1}}} and {{{-5}}} add to {{{-6}}} (the middle coefficient).



So the two numbers {{{-1}}} and {{{-5}}} both multiply to {{{5}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-6}}}



Now replace the middle term {{{-6k}}} with {{{-k-5k}}}. Remember, {{{-1}}} and {{{-5}}} add to {{{-6}}}. So this shows us that {{{-k-5k=-6k}}}.



{{{k^2+highlight(-k-5k)+5}}} Replace the second term {{{-6k}}} with {{{-k-5k}}}.



{{{(k^2-k)+(-5k+5)}}} Group the terms into two pairs.



{{{k(k-1)+(-5k+5)}}} Factor out the GCF {{{k}}} from the first group.



{{{k(k-1)-5(k-1)}}} 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.



{{{(k-5)(k-1)}}} Combine like terms. Or factor out the common term {{{k-1}}}


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Answer:



So {{{k^2-6k+5}}} factors to {{{(k-5)(k-1)}}}.



Note: you can check the answer by FOILing {{{(k-5)(k-1)}}} to get {{{k^2-6k+5}}} or by graphing the original expression and the answer (the two graphs should be identical).
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