<PRE><B>Question 203897</B>
{{{3x^2 = 2 - x}}} Start with the given equation



{{{3x^2 +x-2=0}}} Add &quot;x&quot; to both sides. Subtract 2 from both sides.



Now let&#39;s factor {{{3x^2+x-2}}}



Looking at the expression {{{3x^2+x-2}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{1}}}, and the last term is {{{-2}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{-2}}} to get {{{(3)(-2)=-6}}}.



Now the question is: what two whole numbers multiply to {{{-6}}} (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 {{{-6}}} (the previous product).



Factors of {{{-6}}}:

1,2,3,6

-1,-2,-3,-6



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-6}}}.

1*(-6)
2*(-3)
(-1)*(6)
(-2)*(3)


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;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-6)=-5&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;-3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-3)=-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;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+6=5&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-2&lt;/font&gt;&lt;/td&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;-2+3=1&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-2}}} and {{{3}}} add to {{{1}}} (the middle coefficient).



So the two numbers {{{-2}}} and {{{3}}} both multiply to {{{-6}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{1}}}



Now replace the middle term {{{1x}}} with {{{-2x+3x}}}. Remember, {{{-2}}} and {{{3}}} add to {{{1}}}. So this shows us that {{{-2x+3x=1x}}}.



{{{3x^2+highlight(-2x+3x)-2}}} Replace the second term {{{1x}}} with {{{-2x+3x}}}.



{{{(3x^2-2x)+(3x-2)}}} Group the terms into two pairs.



{{{x(3x-2)+(3x-2)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(3x-2)+1(3x-2)}}} 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.



{{{(x+1)(3x-2)}}} Combine like terms. Or factor out the common term {{{3x-2}}}



So {{{3x^2+x-2}}} factors to {{{(x+1)(3x-2)}}}.



Note: the order of the factors does not matter.
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