Question 201048


Looking at {{{x^2+3xy-208y^2}}} we can see that the first term is {{{x^2}}} and the last term is {{{-208y^2}}} where the coefficients are 1 and -208 respectively.


Now multiply the first coefficient 1 and the last coefficient -208 to get -208. Now what two numbers multiply to -208 and add to the  middle coefficient 3? Let's list all of the factors of -208:




Factors of -208:

1,2,4,8,13,16,26,52,104,208


-1,-2,-4,-8,-13,-16,-26,-52,-104,-208 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -208

(1)*(-208)

(2)*(-104)

(4)*(-52)

(8)*(-26)

(13)*(-16)

(-1)*(208)

(-2)*(104)

(-4)*(52)

(-8)*(26)

(-13)*(16)


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">-208</td><td>1+(-208)=-207</td></tr><tr><td align="center">2</td><td align="center">-104</td><td>2+(-104)=-102</td></tr><tr><td align="center">4</td><td align="center">-52</td><td>4+(-52)=-48</td></tr><tr><td align="center">8</td><td align="center">-26</td><td>8+(-26)=-18</td></tr><tr><td align="center">13</td><td align="center">-16</td><td>13+(-16)=-3</td></tr><tr><td align="center">-1</td><td align="center">208</td><td>-1+208=207</td></tr><tr><td align="center">-2</td><td align="center">104</td><td>-2+104=102</td></tr><tr><td align="center">-4</td><td align="center">52</td><td>-4+52=48</td></tr><tr><td align="center">-8</td><td align="center">26</td><td>-8+26=18</td></tr><tr><td align="center">-13</td><td align="center">16</td><td>-13+16=3</td></tr></table>



From this list we can see that -13 and 16 add up to 3 and multiply to -208



Now looking at the expression {{{1x^2+3xy-208y^2}}}, replace {{{3xy}}} with {{{-13xy+16xy}}} (notice {{{-13xy+16xy}}} adds up to {{{3xy}}}. So it is equivalent to {{{3xy}}})


{{{x^2+highlight(-13xy+16xy)-208y^2}}}



Now let's factor {{{x^2-13xy+16xy-208y^2}}} by grouping:



{{{(x^2-13xy)+(16xy-208y^2)}}} Group like terms



{{{x(x-13y)+16y(x-13y)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{16y}}} out of the second group



{{{(x+16y)(x-13y)}}} Since we have a common term of {{{x-13y}}}, we can combine like terms


So {{{x^2-13xy+16xy-208y^2}}} factors to {{{(x+16y)(x-13y)}}}



So this also means that {{{x^2+3xy-208y^2}}} factors to {{{(x+16y)(x-13y)}}} (since {{{x^2+3xy-208y^2}}} is equivalent to {{{x^2-13xy+16xy-208y^2}}})




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




     Answer:

So {{{x^2+3xy-208y^2}}} factors to {{{(x+16y)(x-13y)}}}