Question 168806


{{{2x^2y^3+3xy^2-5xy}}} Start with the given expression



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



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





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Looking at {{{2xy^2+3y-5}}} we can see that the first term is {{{2xy^2}}} and the last term is {{{-5}}} 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 {{{2xy^2+3y-5}}}, replace {{{3y}}} with {{{-2y+5y}}} (notice {{{-2y+5y}}} adds up to {{{3y}}}. So it is equivalent to {{{3y}}})


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



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



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



{{{2y(xy-1)+5(y-1)}}} Factor out the GCF of {{{2y}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



So this also means that {{{2xy^2+3y-5}}} factors to {{{(2y+5)(xy-1)}}} (since {{{2xy^2+3y-5}}} is equivalent to {{{2xy^2-2y+5y-5}}})




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So our expression goes from {{{xy(2xy^2+3y-5)}}} and factors further to {{{xy(2y+5)(xy-1)}}}



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Answer:


So {{{2x^2y^3+3xy^2-5xy}}} factors to {{{xy(2y+5)(xy-1)}}}