Question 221042
We can solve this by factoring.


{{{4x^2+8x+4}}} Start with the given expression.



{{{4(x^2+2x+1)}}} Factor out the GCF {{{4}}}.



Now let's try to factor the inner expression {{{x^2+2x+1}}}



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Looking at the expression {{{x^2+2x+1}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{2}}}, and the last term is {{{1}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{1}}} to get {{{(1)(1)=1}}}.



Now the question is: what two whole numbers multiply to {{{1}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{2}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{1}}} (the previous product).



Factors of {{{1}}}:

1

-1



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{1}}}.

1*1 = 1
(-1)*(-1) = 1


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{2}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=red>1</font></td><td  align="center"><font color=red>1</font></td><td  align="center"><font color=red>1+1=2</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-1+(-1)=-2</font></td></tr></table>



From the table, we can see that the two numbers {{{1}}} and {{{1}}} add to {{{2}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{1}}} both multiply to {{{1}}} <font size=4><b>and</b></font> add to {{{2}}}



Now replace the middle term {{{2x}}} with {{{x+x}}}. Remember, {{{1}}} and {{{1}}} add to {{{2}}}. So this shows us that {{{x+x=2x}}}.



{{{x^2+highlight(x+x)+1}}} Replace the second term {{{2x}}} with {{{x+x}}}.



{{{(x^2+x)+(x+1)}}} Group the terms into two pairs.



{{{x(x+1)+(x+1)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+1)+1(x+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.



{{{(x+1)(x+1)}}} Combine like terms. Or factor out the common term {{{x+1}}}



{{{(x+1)^2}}} Condense the terms.



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So {{{4(x^2+2x+1)}}} then factors further to {{{4(x+1)^2}}}




So {{{4x^2+8x+4}}} completely factors to {{{4(x+1)^2}}}.



In other words, {{{4x^2+8x+4=4(x+1)^2}}}.



So {{{4x^2+8x+4=0}}} is equivalent to {{{4(x+1)^2=0}}}



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Now let's solve {{{4(x+1)^2=0}}}



{{{4(x+1)^2=0}}} Start with the given equation.



{{{(x+1)^2=(0)/(4)}}} Divide both sides by {{{4}}}.



{{{(x+1)^2=0}}} Reduce.



{{{x+1=""+-sqrt(0)}}} Take the square root of both sides.



{{{x+1=sqrt(0)}}} or {{{x+1=-sqrt(0)}}} Break up the "plus/minus" to form two equations.



{{{x+1=0}}} or {{{x+1=-0}}}  Take the square root of {{{0}}} to get {{{0}}}.



Since -0 is really 0, this means we have one equation: {{{x+1=0}}}



{{{x=0-1}}} Subtract {{{1}}} from both sides.



{{{x=-1}}} Combine like terms.



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Answer:



So the only solution is {{{x=-1}}}.