Question 143549
Do you want to factor?


I'll do the first two to get you started



# 1




Looking at {{{1x^2-2xy-15y^2}}} we can see that the first term is {{{1x^2}}} and the last term is {{{-15y^2}}} where the coefficients are 1 and -15 respectively.


Now multiply the first coefficient 1 and the last coefficient -15 to get -15. Now what two numbers multiply to -15 and add to the  middle coefficient -2? Let's list all of the factors of -15:




Factors of -15:

1,3,5,15


-1,-3,-5,-15 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -15

(1)*(-15)

(3)*(-5)

(-1)*(15)

(-3)*(5)


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



Now which of these pairs add to -2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -2


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-15</td><td>1+(-15)=-14</td></tr><tr><td align="center">3</td><td align="center">-5</td><td>3+(-5)=-2</td></tr><tr><td align="center">-1</td><td align="center">15</td><td>-1+15=14</td></tr><tr><td align="center">-3</td><td align="center">5</td><td>-3+5=2</td></tr></table>



From this list we can see that 3 and -5 add up to -2 and multiply to -15



Now looking at the expression {{{1x^2-2xy-15y^2}}}, replace {{{-2xy}}} with {{{3xy+-5xy}}} (notice {{{3xy+-5xy}}} adds up to {{{-2xy}}}. So it is equivalent to {{{-2xy}}})


{{{1x^2+highlight(3xy+-5xy)+-15y^2}}}



Now let's factor {{{1x^2+3xy-5xy-15y^2}}} by grouping:



{{{(1x^2+3xy)+(-5xy-15y^2)}}} Group like terms



{{{x(x+3y)-5y(x+3y)}}} 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+3y)}}} Since we have a common term of {{{x+3y}}}, we can combine like terms


So {{{1x^2+3xy-5xy-15y^2}}} factors to {{{(x-5y)(x+3y)}}}



So this also means that {{{1x^2-2xy-15y^2}}} factors to {{{(x-5y)(x+3y)}}} (since {{{1x^2-2xy-15y^2}}} is equivalent to {{{1x^2+3xy-5xy-15y^2}}})




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     Answer:

So {{{x^2-2xy-15y^2}}} factors to {{{(x-5y)(x+3y)}}}





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# 2


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


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




Factors of 1:

1


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


These factors pair up and multiply to 1

1*1

(-1)*(-1)


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



Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1


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

None of these pairs of factors add to 1. So the expression {{{x^2+x+1}}} cannot be factored