<PRE><B>Question 707276</B>
# 1




{{{3x^3y^2-3x^2y^2+3xy^2}}} Start with the given expression.



{{{3xy^2(x^2-x+1)}}} Factor out the GCF {{{3xy^2}}}.



Now let&#39;s try to factor the inner expression {{{x^2-x+1}}}



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Looking at the expression {{{x^2-x+1}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-1}}}, and the last term is {{{1}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{1}}} to get {{{(1)(1)=1}}}.



Now the question is: what two whole numbers multiply to {{{1}}} (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 {{{1}}} (the previous product).



Factors of {{{1}}}:

1

-1



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{1}}}.

1*1 = 1
(-1)*(-1) = 1


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;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+1=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;-1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-1)=-2&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that there are no pairs of numbers which add to {{{-1}}}. So {{{x^2-x+1}}} cannot be factored.



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&lt;a name=&quot;ans&quot;&gt;


Answer:



So {{{3x^3y^2-3x^2y^2+3xy^2}}} simply factors to {{{3xy^2(x^2-x+1)}}}



In other words, {{{3x^3y^2-3x^2y^2+3xy^2=3xy^2(x^2-x+1)}}}.


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# 2




Looking at the expression {{{35x^2-2x-1}}}, we can see that the first coefficient is {{{35}}}, the second coefficient is {{{-2}}}, and the last term is {{{-1}}}.



Now multiply the first coefficient {{{35}}} by the last term {{{-1}}} to get {{{(35)(-1)=-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=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;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;/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 {{{-2x}}} with {{{5x-7x}}}. Remember, {{{5}}} and {{{-7}}} add to {{{-2}}}. So this shows us that {{{5x-7x=-2x}}}.



{{{35x^2+highlight(5x-7x)-1}}} Replace the second term {{{-2x}}} with {{{5x-7x}}}.



{{{(35x^2+5x)+(-7x-1)}}} Group the terms into two pairs.



{{{5x(7x+1)+(-7x-1)}}} Factor out the GCF {{{5x}}} from the first group.



{{{5x(7x+1)-1(7x+1)}}} 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.



{{{(5x-1)(7x+1)}}} Combine like terms. Or factor out the common term {{{7x+1}}}



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Answer:



So {{{35x^2-2x-1}}} factors to {{{(5x-1)(7x+1)}}}.



In other words, {{{35x^2-2x-1=(5x-1)(7x+1)}}}.



Note: you can check the answer by expanding {{{(5x-1)(7x+1)}}} to get {{{35x^2-2x-1}}} or by graphing the original expression and the answer (the two graphs should be identical).

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