<PRE><B>Question 201235</B>
# 1




Looking at the expression {{{8x^2+6x-5}}}, we can see that the first coefficient is {{{8}}}, the second coefficient is {{{6}}}, 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 {{{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)
2*(-20)
4*(-10)
5*(-8)
(-1)*(40)
(-2)*(20)
(-4)*(10)
(-5)*(8)


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 {{{6x}}} with {{{-4x+10x}}}. Remember, {{{-4}}} and {{{10}}} add to {{{6}}}. So this shows us that {{{-4x+10x=6x}}}.



{{{8x^2+highlight(-4x+10x)-5}}} Replace the second term {{{6x}}} with {{{-4x+10x}}}.



{{{(8x^2-4x)+(10x-5)}}} Group the terms into two pairs.



{{{4x(2x-1)+(10x-5)}}} Factor out the GCF {{{4x}}} from the first group.



{{{4x(2x-1)+5(2x-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.



{{{(4x+5)(2x-1)}}} Combine like terms. Or factor out the common term {{{2x-1}}}


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Answer:



So {{{8x^2+6x-5}}} factors to {{{(4x+5)(2x-1)}}}.



Note: you can check the answer by FOILing {{{(4x+5)(2x-1)}}} to get {{{8x^2+6x-5}}} or by graphing the original expression and the answer (the two graphs should be identical).



&lt;hr&gt;


# 2





Looking at the expression {{{x^2+17x+16}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{17}}}, and the last term is {{{16}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{16}}} to get {{{(1)(16)=16}}}.



Now the question is: what two whole numbers multiply to {{{16}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{17}}}?



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 {{{16}}} (the previous product).



Factors of {{{16}}}:

1,2,4,8,16

-1,-2,-4,-8,-16



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{16}}}.

1*16
2*8
4*4
(-1)*(-16)
(-2)*(-8)
(-4)*(-4)


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{17}}}:



&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=red&gt;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;16&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;1+16=17&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;8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+8=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;4&lt;/font&gt;&lt;/td&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;4+4=8&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;-16&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-16)=-17&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;-8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-8)=-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;-4&lt;/font&gt;&lt;/td&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;-4+(-4)=-8&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{1}}} and {{{16}}} add to {{{17}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{16}}} both multiply to {{{16}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{17}}}



Now replace the middle term {{{17x}}} with {{{x+16x}}}. Remember, {{{1}}} and {{{16}}} add to {{{17}}}. So this shows us that {{{x+16x=17x}}}.



{{{x^2+highlight(x+16x)+16}}} Replace the second term {{{17x}}} with {{{x+16x}}}.



{{{(x^2+x)+(16x+16)}}} Group the terms into two pairs.



{{{x(x+1)+(16x+16)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+1)+16(x+1)}}} Factor out {{{16}}} 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+16)(x+1)}}} Combine like terms. Or factor out the common term {{{x+1}}}


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Answer:



So {{{x^2+17x+16}}} factors to {{{(x+16)(x+1)}}}.



Note: you can check the answer by FOILing {{{(x+16)(x+1)}}} to get {{{x^2+17x+16}}} or by graphing the original expression and the answer (the two graphs should be identical).
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