<PRE><B>Question 640407</B>


Looking at the expression {{{2x^2+5x-12}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{5}}}, and the last term is {{{-12}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{-12}}} to get {{{(2)(-12)=-24}}}.



Now the question is: what two whole numbers multiply to {{{-24}}} (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 {{{-24}}} (the previous product).



Factors of {{{-24}}}:

1,2,3,4,6,8,12,24

-1,-2,-3,-4,-6,-8,-12,-24



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-24}}}.

1*(-24) = -24
2*(-12) = -24
3*(-8) = -24
4*(-6) = -24
(-1)*(24) = -24
(-2)*(12) = -24
(-3)*(8) = -24
(-4)*(6) = -24


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=black&gt;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-24&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-24)=-23&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;-12&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-12)=-10&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;-8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+(-8)=-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;4&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;4+(-6)=-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;24&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+24=23&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;12&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+12=10&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-3+8=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;-4&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;-4+6=2&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-3}}} and {{{8}}} add to {{{5}}} (the middle coefficient).



So the two numbers {{{-3}}} and {{{8}}} both multiply to {{{-24}}} &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 {{{-3x+8x}}}. Remember, {{{-3}}} and {{{8}}} add to {{{5}}}. So this shows us that {{{-3x+8x=5x}}}.



{{{2x^2+highlight(-3x+8x)-12}}} Replace the second term {{{5x}}} with {{{-3x+8x}}}.



{{{(2x^2-3x)+(8x-12)}}} Group the terms into two pairs.



{{{x(2x-3)+(8x-12)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x-3)+4(2x-3)}}} Factor out {{{4}}} 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+4)(2x-3)}}} Combine like terms. Or factor out the common term {{{2x-3}}}



===============================================================



Answer:



So {{{2x^2+5x-12}}} factors to {{{(x+4)(2x-3)}}}.



In other words, {{{2x^2+5x-12=(x+4)(2x-3)}}}.



Note: you can check the answer by expanding {{{(x+4)(2x-3)}}} to get {{{2x^2+5x-12}}} or by graphing the original expression and the answer (the two graphs should be identical).
</PRE>