<PRE><B>Question 887725</B>


Looking at the expression {{{x^2+10x+16}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{10}}}, 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 {{{10}}}?



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 = 16
2*8 = 16
4*4 = 16
(-1)*(-16) = 16
(-2)*(-8) = 16
(-4)*(-4) = 16


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{10}}}:



&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;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=red&gt;2&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;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 {{{2}}} and {{{8}}} add to {{{10}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{8}}} both multiply to {{{16}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{10}}}



Now replace the middle term {{{10x}}} with {{{2x+8x}}}. Remember, {{{2}}} and {{{8}}} add to {{{10}}}. So this shows us that {{{2x+8x=10x}}}.



{{{x^2+highlight(2x+8x)+16}}} Replace the second term {{{10x}}} with {{{2x+8x}}}.



{{{(x^2+2x)+(8x+16)}}} Group the terms into two pairs.



{{{x(x+2)+(8x+16)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+2)+8(x+2)}}} Factor out {{{8}}} 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+8)(x+2)}}} Combine like terms. Or factor out the common term {{{x+2}}}



===============================================================



Answer:



So {{{x^2+10x+16}}} factors to {{{(x+8)(x+2)}}}.



In other words, {{{x^2+10x+16=(x+8)(x+2)}}}.



Note: you can check the answer by expanding {{{(x+8)(x+2)}}} to get {{{x^2+10x+16}}} or by graphing the original expression and the answer (the two graphs should be identical).
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