<PRE><B>Question 551194</B>
I&#39;ll do the first one to get you started




Looking at the expression {{{22n^2+n-5}}}, we can see that the first coefficient is {{{22}}}, the second coefficient is {{{1}}}, and the last term is {{{-5}}}.



Now multiply the first coefficient {{{22}}} by the last term {{{-5}}} to get {{{(22)(-5)=-110}}}.



Now the question is: what two whole numbers multiply to {{{-110}}} (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 {{{-110}}} (the previous product).



Factors of {{{-110}}}:

1,2,5,10,11,22,55,110

-1,-2,-5,-10,-11,-22,-55,-110



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-110}}}.

1*(-110) = -110
2*(-55) = -110
5*(-22) = -110
10*(-11) = -110
(-1)*(110) = -110
(-2)*(55) = -110
(-5)*(22) = -110
(-10)*(11) = -110


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;-110&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-110)=-109&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;-55&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-55)=-53&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;-22&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;5+(-22)=-17&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;-11&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;10+(-11)=-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;110&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+110=109&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;55&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+55=53&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;22&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-5+22=17&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;11&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-10+11=1&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-10}}} and {{{11}}} add to {{{1}}} (the middle coefficient).



So the two numbers {{{-10}}} and {{{11}}} both multiply to {{{-110}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{1}}}



Now replace the middle term {{{1n}}} with {{{-10n+11n}}}. Remember, {{{-10}}} and {{{11}}} add to {{{1}}}. So this shows us that {{{-10n+11n=1n}}}.



{{{22n^2+highlight(-10n+11n)-5}}} Replace the second term {{{1n}}} with {{{-10n+11n}}}.



{{{(22n^2-10n)+(11n-5)}}} Group the terms into two pairs.



{{{2n(11n-5)+(11n-5)}}} Factor out the GCF {{{2n}}} from the first group.



{{{2n(11n-5)+1(11n-5)}}} Factor out {{{1}}} 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.



{{{(2n+1)(11n-5)}}} Combine like terms. Or factor out the common term {{{11n-5}}}



===============================================================



Answer:



So {{{22n^2+n-5}}} factors to {{{(2n+1)(11n-5)}}}.



In other words, {{{22n^2+n-5=(2n+1)(11n-5)}}}.



Note: you can check the answer by expanding {{{(2n+1)(11n-5)}}} to get {{{22n^2+n-5}}} or by graphing the original expression and the answer (the two graphs should be identical).
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