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State the missing factor 4n^2+8n-60=(?)(n+5)
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what times \(n\) is \(4n^2\)?
4n?
yes
and what times \(5\) gives \(-60\)?
-12
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ok put them together are you are done
(4n-12)
yup
at the back of my textbook the answer is 4(n-3)
not really factored completely that would be \[4(n-3)(n+5)\]
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yeah because the terms of \(4n-12\) have a common factor of 4
so you divide (4n-12) by 4 to get 4(n-3) ?
that is not "division" that is factoring, aka the distributive property \[4\times x-4\times 3=4(x-3)\]
ohh okay
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