Question 201040


{{{x^3y+4x^2y^2-21xy^3}}} Start with the given expression



{{{xy(x^2+4xy-21y^2)}}} Factor out the GCF {{{xy}}}



Now let's focus on the inner expression {{{x^2+4xy-21y^2}}}





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Looking at {{{1x^2+4xy-21y^2}}} we can see that the first term is {{{1x^2}}} and the last term is {{{-21y^2}}} where the coefficients are 1 and -21 respectively.


Now multiply the first coefficient 1 and the last coefficient -21 to get -21. Now what two numbers multiply to -21 and add to the  middle coefficient 4? Let's list all of the factors of -21:




Factors of -21:

1,3,7,21


-1,-3,-7,-21 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -21

(1)*(-21)

(3)*(-7)

(-1)*(21)

(-3)*(7)


note: remember, the product of a negative and a positive number is a negative 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">-21</td><td>1+(-21)=-20</td></tr><tr><td align="center">3</td><td align="center">-7</td><td>3+(-7)=-4</td></tr><tr><td align="center">-1</td><td align="center">21</td><td>-1+21=20</td></tr><tr><td align="center">-3</td><td align="center">7</td><td>-3+7=4</td></tr></table>



From this list we can see that -3 and 7 add up to 4 and multiply to -21



Now looking at the expression {{{x^2+4xy-21y^2}}}, replace {{{4xy}}} with {{{-3xy+7xy}}} (notice {{{-3xy+7xy}}} adds up to {{{4xy}}}. So it is equivalent to {{{4xy}}})


{{{x^2+highlight(-3xy+7xy)-21y^2}}}



Now let's factor {{{x^2-3xy+7xy-21y^2}}} by grouping:



{{{(x^2-3xy)+(7xy-21y^2)}}} Group like terms



{{{x(x-3y)+7y(x-3y)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{7y}}} out of the second group



{{{(x+7y)(x-3y)}}} Since we have a common term of {{{x-3y}}}, we can combine like terms


So {{{x^2-3xy+7xy-21y^2}}} factors to {{{(x+7y)(x-3y)}}}



So this also means that {{{x^2+4xy-21y^2}}} factors to {{{(x+7y)(x-3y)}}} (since {{{x^2+4xy-21y^2}}} is equivalent to {{{x^2-3xy+7xy-21y^2}}})




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So our expression goes from {{{xy(x^2+4xy-21y^2)}}} and factors further to {{{xy(x+7y)(x-3y)}}}



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Answer:


So {{{x^3y+4x^2y^2-21xy^3}}} completely factors to {{{xy(x+7y)(x-3y)}}}