<PRE><B>Question 752945</B>


Looking at the expression {{{5d^2+6d-8}}}, we can see that the first coefficient is {{{5}}}, the second coefficient is {{{6}}}, and the last term is {{{-8}}}.



Now multiply the first coefficient {{{5}}} by the last term {{{-8}}} to get {{{(5)(-8)=-40}}}.



Now the question is: what two whole numbers multiply to {{{-40}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{6}}}?



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 {{{-40}}} (the previous product).



Factors of {{{-40}}}:

1,2,4,5,8,10,20,40

-1,-2,-4,-5,-8,-10,-20,-40



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-40}}}.

1*(-40) = -40
2*(-20) = -40
4*(-10) = -40
5*(-8) = -40
(-1)*(40) = -40
(-2)*(20) = -40
(-4)*(10) = -40
(-5)*(8) = -40


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{6}}}:



&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;-40&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-40)=-39&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;-20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-20)=-18&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;4+(-10)=-6&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;-8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;5+(-8)=-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;40&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+40=39&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;20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+20=18&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-4&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-4+10=6&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;8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-5+8=3&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-4}}} and {{{10}}} add to {{{6}}} (the middle coefficient).



So the two numbers {{{-4}}} and {{{10}}} both multiply to {{{-40}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{6}}}



Now replace the middle term {{{6d}}} with {{{-4d+10d}}}. Remember, {{{-4}}} and {{{10}}} add to {{{6}}}. So this shows us that {{{-4d+10d=6d}}}.



{{{5d^2+highlight(-4d+10d)-8}}} Replace the second term {{{6d}}} with {{{-4d+10d}}}.



{{{(5d^2-4d)+(10d-8)}}} Group the terms into two pairs.



{{{d(5d-4)+(10d-8)}}} Factor out the GCF {{{d}}} from the first group.



{{{d(5d-4)+2(5d-4)}}} 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.



{{{(d+2)(5d-4)}}} Combine like terms. Or factor out the common term {{{5d-4}}}



===============================================================



Answer:



So {{{5d^2+6d-8}}} factors to {{{(d+2)(5d-4)}}}.



In other words, {{{5d^2+6d-8=(d+2)(5d-4)}}}.



Note: you can check the answer by expanding {{{(d+2)(5d-4)}}} to get {{{5d^2+6d-8}}} or by graphing the original expression and the answer (the two graphs should be identical).
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