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A number is expressed as N=(x^2-169)+(x-13)^2 1. Factorise (x^2-169)+(x-13)^2 completely 2. Explain why N is an even number for any value of x 3. Hence, given that N=(83^2-169)+70^2, without evaluating for N, find the 2 factors of N which are between 10 and 100

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So have you expanded N first by doing (x-13)(x-13)?
And then you will collect like terms for N and then you can factor it

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2x(x-13) right? But for me the real problem is with the second and last number
Where did the extra x come from?
wait sorry
There is something missing though
\[x^2-169+(x-13)^2\] \[x^2-169+(x^2-2(13)x+169)\] Is this what you did?
Gosh you are right Its late here
Ok So you have 2x(x-13) And it has the factors 2, x, x-13 Since it has the factor 2 , then that means N is divisble by 2 which means that N is ____?
Eveeen ohohhhhh, of course, dang can't believe I missed that
first two parts have been explained by @myininaya for the last part put x=83 in 2x(x+13) u get 2*83*96..this means two factors are 83 and 96
sorry the eq is 2x(x-13)..factors are 83 and 70
Well, thanks to you guys, sorry I only got one medal to give

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