jdfwright jdfwright
  • 22-12-2014
  • Mathematics
contestada

4n^2-16 factor the quadratic

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jemimahclaire98 jemimahclaire98
  • 22-12-2014
(2n+4)(2n-4)
This way when you distribute it out, you get 4n^2-8n+8n-16.
Then the -8n and the 8n cancel out and you are left with 4n^2-16.
Another way of factoring it is to start out by taking out a common factor which is 4.
This then becomes 4(n^2-4) You can then factor that by using the method above into 4(n+2)(n-2).
The second way is just a bit more factored but they're both an answer.
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Аноним Аноним
  • 24-12-2014
[tex]4n^2-16=4(n^2-4)=4(n^2-2^2)=(*)\\\\use:a^2-b^2=(a-b)(a+b)\\\\(*)=4(n-2)(n+2)[/tex]
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