Question 251756


{{{4y^2+16y+16}}} Start with the given expression.



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



Now let's try to factor the inner expression {{{y^2+4y+4}}}



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Looking at the expression {{{y^2+4y+4}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{4}}}, and the last term is {{{4}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{4}}} to get {{{(1)(4)=4}}}.



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



To find these two numbers, we need to list <font size=4><b>all</b></font> 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's add up each pair of factors to see if one pair adds to the middle coefficient {{{4}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>1+4=5</font></td></tr><tr><td  align="center"><font color=red>2</font></td><td  align="center"><font color=red>2</font></td><td  align="center"><font color=red>2+2=4</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-1+(-4)=-5</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-2+(-2)=-4</font></td></tr></table>



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}}} <font size=4><b>and</b></font> add to {{{4}}}



Now replace the middle term {{{4y}}} with {{{2y+2y}}}. Remember, {{{2}}} and {{{2}}} add to {{{4}}}. So this shows us that {{{2y+2y=4y}}}.



{{{y^2+highlight(2y+2y)+4}}} Replace the second term {{{4y}}} with {{{2y+2y}}}.



{{{(y^2+2y)+(2y+4)}}} Group the terms into two pairs.



{{{y(y+2)+(2y+4)}}} Factor out the GCF {{{y}}} from the first group.



{{{y(y+2)+2(y+2)}}} Factor out {{{2}}} 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.



{{{(y+2)(y+2)}}} Combine like terms. Or factor out the common term {{{y+2}}}



{{{(y+2)^2}}} Condense the terms.



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



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Answer:



So {{{4y^2+16y+16}}} completely factors to {{{4(y+2)^2}}}.



In other words, {{{4y^2+16y+16=4(y+2)^2}}}.



Note: you can check the answer by expanding {{{4(y+2)^2}}} to get {{{4y^2+16y+16}}} or by graphing the original expression and the answer (the two graphs should be identical).