Question 144025


{{{5x^3+20x^2+20x}}} Start with the given expression



{{{5x(x^2+4x+4)}}} Factor out the GCF {{{5x}}}



Now let's focus on the inner expression {{{x^2+4x+4}}}





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Looking at {{{1x^2+4x+4}}} we can see that the first term is {{{1x^2}}} and the last term is {{{4}}} where the coefficients are 1 and 4 respectively.


Now multiply the first coefficient 1 and the last coefficient 4 to get 4. Now what two numbers multiply to 4 and add to the  middle coefficient 4? Let's list all of the factors of 4:




Factors of 4:

1,2


-1,-2 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 4

1*4

2*2

(-1)*(-4)

(-2)*(-2)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to 4? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 4


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



From this list we can see that 2 and 2 add up to 4 and multiply to 4



Now looking at the expression {{{1x^2+4x+4}}}, replace {{{4x}}} with {{{2x+2x}}} (notice {{{2x+2x}}} adds up to {{{4x}}}. So it is equivalent to {{{4x}}})


{{{1x^2+highlight(2x+2x)+4}}}



Now let's factor {{{1x^2+2x+2x+4}}} by grouping:



{{{(1x^2+2x)+(2x+4)}}} Group like terms



{{{x(x+2)+2(x+2)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{2}}} out of the second group



{{{(x+2)(x+2)}}} Since we have a common term of {{{x+2}}}, we can combine like terms


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



So this also means that {{{1x^2+4x+4}}} factors to {{{(x+2)(x+2)}}} (since {{{1x^2+4x+4}}} is equivalent to {{{1x^2+2x+2x+4}}})



note:  {{{(x+2)(x+2)}}} is equivalent to  {{{(x+2)^2}}} since the term {{{x+2}}} occurs twice. So {{{1x^2+4x+4}}} also factors to {{{(x+2)^2}}}




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So our expression goes from {{{5x(x^2+4x+4)}}} and factors further to {{{5x(x+2)^2}}}



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Answer:


So {{{5x^3+20x^2+20x}}} factors to {{{5x(x+2)^2}}}