<PRE><B>Question 586045</B>
I&#39;ll do the first one to get you stared.


a)


Looking at the expression {{{10x^2-x-3}}}, we can see that the first coefficient is {{{10}}}, the second coefficient is {{{-1}}}, and the last term is {{{-3}}}.



Now multiply the first coefficient {{{10}}} by the last term {{{-3}}} to get {{{(10)(-3)=-30}}}.



Now the question is: what two whole numbers multiply to {{{-30}}} (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 {{{-30}}} (the previous product).



Factors of {{{-30}}}:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-30}}}.

1*(-30) = -30
2*(-15) = -30
3*(-10) = -30
5*(-6) = -30
(-1)*(30) = -30
(-2)*(15) = -30
(-3)*(10) = -30
(-5)*(6) = -30


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;-30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-30)=-29&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;-15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-15)=-13&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+(-10)=-7&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;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;5+(-6)=-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;30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+30=29&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;15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+15=13&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+10=7&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;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-5+6=1&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{5}}} and {{{-6}}} add to {{{-1}}} (the middle coefficient).



So the two numbers {{{5}}} and {{{-6}}} both multiply to {{{-30}}} &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 {{{5x-6x}}}. Remember, {{{5}}} and {{{-6}}} add to {{{-1}}}. So this shows us that {{{5x-6x=-1x}}}.



{{{10x^2+highlight(5x-6x)-3}}} Replace the second term {{{-1x}}} with {{{5x-6x}}}.



{{{(10x^2+5x)+(-6x-3)}}} Group the terms into two pairs.



{{{5x(2x+1)+(-6x-3)}}} Factor out the GCF {{{5x}}} from the first group.



{{{5x(2x+1)-3(2x+1)}}} Factor out {{{3}}} 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-3)(2x+1)}}} Combine like terms. Or factor out the common term {{{2x+1}}}



===============================================================



Answer:



So {{{10x^2-x-3}}} factors to {{{(5x-3)(2x+1)}}}.



In other words, {{{10x^2-x-3=(5x-3)(2x+1)}}}.



Note: you can check the answer by expanding {{{(5x-3)(2x+1)}}} to get {{{10x^2-x-3}}} or by graphing the original expression and the answer (the two graphs should be identical).


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Jim
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