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Factorise $3x^2 + 26x + 51$. Hence , find the two factors of $32651$

Workings $$3x^2 + 26x + 51= (x+3)(3x+17) $$

I don't understand how can I use the answer above to help me find factors of $32651$

choco_addicted
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user307640
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1 Answers1

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Let $x=100$,

Then $3x^2+26x+51=32651$

Hence $32651=(100+3)(3*100+17)$

It is easy to check both 317 and 103 are prime numbers.

lEm
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    I would credit the answer to @gammatester's hint. – barak manos Mar 02 '16 at 08:19
  • A question for you: prove that 2438100000001 isn't prime! Further, if you know why, you can find it's factors! This is the inverse question, one far harder to answer if I give you a general number (say 667502283127 ) – Sarvesh Ravichandran Iyer Mar 02 '16 at 10:43
  • @abdefghijklmnopqrtxyz-stoo Clever question! $2438100000001=3013^(4)10^(8)+1=3^(5)10^(10)+3^(4)10^(8)+1$, $k^5 + k^4 +1 =(k^5+k^4+k^3+k^2+k+1)-k(k^2+k+1)=(k^3-k+1)(k^2+k+1)$, $let k=300, 2438100000001=26999701*90301$ Thanks for the cool problem – lEm Mar 02 '16 at 15:29
  • Im a new user, sorry if I messed up typing – lEm Mar 02 '16 at 15:30
  • congratulations, and that's fine if you occasionally slip up while typing. try the same for the other one (note: it is a crazy polynomial). – Sarvesh Ravichandran Iyer Mar 05 '16 at 00:47