<PRE><B>Question 316714</B>


Looking at the expression {{{4x^2+4x+1}}}, we can see that the first coefficient is {{{4}}}, the second coefficient is {{{4}}}, and the last term is {{{1}}}.



Now multiply the first coefficient {{{4}}} by the last term {{{1}}} to get {{{(4)(1)=4}}}.



Now the question is: what two whole numbers multiply to {{{4}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{4}}}?



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 {{{4}}} (the previous product).



Factors of {{{4}}}:

1,2,4

-1,-2,-4



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{4}}}.

1*4 = 4
2*2 = 4
(-1)*(-4) = 4
(-2)*(-2) = 4


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{4}}}:



&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;4&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+4=5&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;2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2+2=4&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;-4&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-4)=-5&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;-2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-2)=-4&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{2}}} and {{{2}}} add to {{{4}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{2}}} both multiply to {{{4}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{4}}}



Now replace the middle term {{{4x}}} with {{{2x+2x}}}. Remember, {{{2}}} and {{{2}}} add to {{{4}}}. So this shows us that {{{2x+2x=4x}}}.



{{{4x^2+highlight(2x+2x)+1}}} Replace the second term {{{4x}}} with {{{2x+2x}}}.



{{{(4x^2+2x)+(2x+1)}}} Group the terms into two pairs.



{{{2x(2x+1)+(2x+1)}}} Factor out the GCF {{{2x}}} from the first group.



{{{2x(2x+1)+1(2x+1)}}} Factor out {{{1}}} 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+1)(2x+1)}}} Combine like terms. Or factor out the common term {{{2x+1}}}



{{{(2x+1)^2}}} Condense the terms.



===============================================================



Answer:



So {{{4x^2+4x+1}}} factors to {{{(2x+1)^2}}}.



In other words, {{{4x^2+4x+1=(2x+1)^2}}}.



Note: you can check the answer by expanding {{{(2x+1)^2}}} to get {{{4x^2+4x+1}}} or by graphing the original expression and the answer (the two graphs should be identical).
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