<PRE><B>Question 180578</B>
It looks like you have a good understanding of what you are doing. It seems that the only thing that is hanging you up is the factoring. 



So let&#39;s factor {{{-3x^2+5x+2}}}



{{{-3x^2+5x+2}}} Start with the given expression.



{{{-(3x^2-5x-2)}}} Factor out a negative 1 (to make the leading coefficient positive; this isn&#39;t required, but it helps)


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Now let&#39;s factor the inner expression {{{3x^2-5x-2}}}



Looking at the expression {{{3x^2-5x-2}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{-5}}}, 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 {{{-5}}}?



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 {{{-5}}}:



&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=red&gt;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&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=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;/table&gt;



From the table, we can see that the two numbers {{{1}}} and {{{-6}}} add to {{{-5}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{-6}}} both multiply to {{{-6}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-5}}}



Now replace the middle term {{{-5x}}} with {{{x-6x}}}. Remember, {{{1}}} and {{{-6}}} add to {{{-5}}}. So this shows us that {{{x-6x=-5x}}}.



{{{3x^2+highlight(x-6x)-2}}} Replace the second term {{{-5x}}} with {{{x-6x}}}.



{{{(3x^2+x)+(-6x-2)}}} Group the terms into two pairs.



{{{x(3x+1)+(-6x-2)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(3x+1)-2(3x+1)}}} Factor out {{{2}}} 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-2)(3x+1)}}} Combine like terms. Or factor out the common term {{{3x+1}}}



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So this means that {{{-(3x^2-5x-2)}}} factors further down to {{{-(x-2)(3x+1)}}}



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Answer:



So {{{-3x^2+5x+2}}} completely factors to {{{-(x-2)(3x+1)}}}.



Now from here, simply set each factor equal to zero and solve for &quot;x&quot;</PRE>