<PRE><B>Question 251546</B>


Looking at the expression {{{8x^2-22x+5}}}, we can see that the first coefficient is {{{8}}}, the second coefficient is {{{-22}}}, and the last term is {{{5}}}.



Now multiply the first coefficient {{{8}}} by the last term {{{5}}} to get {{{(8)(5)=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 {{{-22}}}?



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 {{{-22}}}:



&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=41&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=22&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=14&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=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;-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)=-41&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;-20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-2+(-20)=-22&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)=-14&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)=-13&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-2}}} and {{{-20}}} add to {{{-22}}} (the middle coefficient).



So the two numbers {{{-2}}} and {{{-20}}} both multiply to {{{40}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-22}}}



Now replace the middle term {{{-22x}}} with {{{-2x-20x}}}. Remember, {{{-2}}} and {{{-20}}} add to {{{-22}}}. So this shows us that {{{-2x-20x=-22x}}}.



{{{8x^2+highlight(-2x-20x)+5}}} Replace the second term {{{-22x}}} with {{{-2x-20x}}}.



{{{(8x^2-2x)+(-20x+5)}}} Group the terms into two pairs.



{{{2x(4x-1)+(-20x+5)}}} Factor out the GCF {{{2x}}} from the first group.



{{{2x(4x-1)-5(4x-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)(4x-1)}}} Combine like terms. Or factor out the common term {{{4x-1}}}



===============================================================



Answer:



So {{{8x^2-22x+5}}} factors to {{{(2x-5)(4x-1)}}}.



In other words, {{{8x^2-22x+5=(2x-5)(4x-1)}}}.



Note: you can check the answer by expanding {{{(2x-5)(4x-1)}}} to get {{{8x^2-22x+5}}} or by graphing the original expression and the answer (the two graphs should be identical).
</PRE>