Question 164438
Are you sure that the expression is not {{{x^3-3x^2y-10xy^2}}}???




{{{x^3-3x^2y-10xy^2}}} Start with the given expression



{{{x(x^2-3xy-10y^2)}}} Factor out the GCF {{{x}}}



Now let's focus on the inner expression {{{x^2-3xy-10y^2}}}





------------------------------------------------------------




Looking at {{{1x^2-3xy-10y^2}}} we can see that the first term is {{{1x^2}}} and the last term is {{{-10y^2}}} where the coefficients are 1 and -10 respectively.


Now multiply the first coefficient 1 and the last coefficient -10 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 {{{1x^2-3xy-10y^2}}}, replace {{{-3xy}}} with {{{2xy+-5xy}}} (notice {{{2xy+-5xy}}} adds up to {{{-3xy}}}. So it is equivalent to {{{-3xy}}})


{{{1x^2+highlight(2xy+-5xy)+-10y^2}}}



Now let's factor {{{1x^2+2xy-5xy-10y^2}}} by grouping:



{{{(1x^2+2xy)+(-5xy-10y^2)}}} Group like terms



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



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


So {{{1x^2+2xy-5xy-10y^2}}} factors to {{{(x-5y)(x+2y)}}}



So this also means that {{{1x^2-3xy-10y^2}}} factors to {{{(x-5y)(x+2y)}}} (since {{{1x^2-3xy-10y^2}}} is equivalent to {{{1x^2+2xy-5xy-10y^2}}})




------------------------------------------------------------





So our expression goes from {{{x(x^2-3xy-10y^2)}}} and factors further to {{{x(x-5y)(x+2y)}}}



------------------

Answer:


So {{{x^3-3x^2y-10xy^2}}} factors to {{{x(x-5y)(x+2y)}}}