Question 191403


{{{14x^3y+21x^2y^2-35xy^3}}} Start with the given expression



{{{7xy(2x^2+3xy-5y^2)}}} Factor out the GCF {{{7xy}}}



Now let's focus on the inner expression {{{2x^2+3xy-5y^2}}}





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Looking at {{{2x^2+3xy-5y^2}}} we can see that the first term is {{{2x^2}}} and the last term is {{{-5y^2}}} where the coefficients are 2 and -5 respectively.


Now multiply the first coefficient 2 and the last coefficient -5 to get -10. Now what two numbers multiply to -10 and add to the  middle coefficient 3? Let's list all of the factors of -10:




Factors of -10:

1,2,5,10


-1,-2,-5,-10 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -10

(1)*(-10)

(2)*(-5)

(-1)*(10)

(-2)*(5)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to 3? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 3


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-10</td><td>1+(-10)=-9</td></tr><tr><td align="center">2</td><td align="center">-5</td><td>2+(-5)=-3</td></tr><tr><td align="center">-1</td><td align="center">10</td><td>-1+10=9</td></tr><tr><td align="center">-2</td><td align="center">5</td><td>-2+5=3</td></tr></table>



From this list we can see that -2 and 5 add up to 3 and multiply to -10



Now looking at the expression {{{2x^2+3xy-5y^2}}}, replace {{{3xy}}} with {{{-2xy+5xy}}} (notice {{{-2xy+5xy}}} adds up to {{{3xy}}}. So it is equivalent to {{{3xy}}})


{{{2x^2+highlight(-2xy+5xy)+-5y^2}}}



Now let's factor {{{2x^2-2xy+5xy-5y^2}}} by grouping:



{{{(2x^2-2xy)+(5xy-5y^2)}}} Group like terms



{{{2x(x-y)+5y(x-y)}}} Factor out the GCF of {{{2x}}} out of the first group. Factor out the GCF of {{{5y}}} out of the second group



{{{(2x+5y)(x-y)}}} Since we have a common term of {{{x-y}}}, we can combine like terms


So {{{2x^2-2xy+5xy-5y^2}}} factors to {{{(2x+5y)(x-y)}}}



So this also means that {{{2x^2+3xy-5y^2}}} factors to {{{(2x+5y)(x-y)}}} (since {{{2x^2+3xy-5y^2}}} is equivalent to {{{2x^2-2xy+5xy-5y^2}}})




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So our expression goes from {{{7xy(2x^2+3xy-5y^2)}}} and factors further to {{{7xy(2x+5y)(x-y)}}}



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Answer:


So {{{14x^3y+21x^2y^2-35xy^3}}} completely factors to {{{7xy(2x+5y)(x-y)}}}