<PRE><B>Question 214983</B>
First, let&#39;s factor {{{2x^2-3x-5}}}



Looking at the expression {{{2x^2-3x-5}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{-3}}}, and the last term is {{{-5}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{-5}}} to get {{{(2)(-5)=-10}}}.



Now the question is: what two whole numbers multiply to {{{-10}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-3}}}?



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 {{{-10}}} (the previous product).



Factors of {{{-10}}}:

1,2,5,10

-1,-2,-5,-10



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-10}}}.

1*(-10) = -10
2*(-5) = -10
(-1)*(10) = -10
(-2)*(5) = -10


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-3}}}:



&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-10)=-9&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;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2+(-5)=-3&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+10=9&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;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+5=3&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{2}}} and {{{-5}}} add to {{{-3}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{-5}}} both multiply to {{{-10}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-3}}}



Now replace the middle term {{{-3x}}} with {{{2x-5x}}}. Remember, {{{2}}} and {{{-5}}} add to {{{-3}}}. So this shows us that {{{2x-5x=-3x}}}.



{{{2x^2+highlight(2x-5x)-5}}} Replace the second term {{{-3x}}} with {{{2x-5x}}}.



{{{(2x^2+2x)+(-5x-5)}}} Group the terms into two pairs.



{{{2x(x+1)+(-5x-5)}}} Factor out the GCF {{{2x}}} from the first group.



{{{2x(x+1)-5(x+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.



{{{(2x-5)(x+1)}}} Combine like terms. Or factor out the common term {{{x+1}}}



So {{{2x^2-3x-5}}} factors to {{{(2x-5)(x+1)}}}.



In other words, {{{2x^2-3x-5=(2x-5)(x+1)}}}.



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Now let&#39;s use this factorization to solve {{{2x^2-3x-5=0}}}



{{{2x^2-3x-5=0}}} Start with the given equation



{{{(2x-5)(x+1)=0}}} Factor the left side (using the factorization from above)




Now set each factor equal to zero:

{{{2x-5=0}}} or  {{{x+1=0}}} 



{{{x=5/2}}} or  {{{x=-1}}}    Now solve for x in each case



So the solutions are {{{x=5/2}}} or {{{x=-1}}}</PRE>