SOLUTION: How do you factor this completely? 6rcubed+15rsquared+8r+20

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Question 723112: How do you factor this completely? 6rcubed+15rsquared+8r+20
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The polynomial P%28r%29=6r%5E3%2B15r%5E2%2B8r%2B20 would have a factor of the form %28qr%2Bp%29
if -p%2Fq is a zero of the polynomial.
Zeros of that polynomial must be negative because for all r%3E=0 ,6r%5E3%2B15r%5E2%2B8r%3E=0 and 6r%5E3%2B15r%5E2%2B8r%2B20%3E=20
Any rational zeros will be of the form -p%2Fq ,
where p is a factor of the independent term, 20 , and
q is a factor of the leading coefficient 6.
So, q could be 1, 2, 3, or 6, and p could be 1, 2, 4, 5, 10, or 20.
P%28-1%29=-6%2B15-8%2B20=21
P%28-2%29=-8%2A6%2B15%2A4-2%2A8%2B20=-48%2B60-16%2B20=16
P%28-4%29=-64%2A6%2B15%2A16-4%2A8%2B20=-286%2B90-32%2B20=-58
Since P%28-2%29%3E0 and P%28-4%29%3C0 , there must be a zero of P%28r%29 between -2 and -4. Of all the rational possibilities, only -5%2F2 and -10%2F3 are between -2 and -4.

so %282r%2B5%29 is a factor of P%28r%29 and dividing P%28r%29 by %282r%2B5%29 we get 3r%5E2%2B4, which cannot be factored further, so
6r%5E3%2B15r%5E2%2B8r%2B20=highlight%28%282r%2B5%29%283r%5E2%2B4%29%29

NOTE: If you have gone beyond real numbers and are expected to use complex numbers, you could factor %283r%5E2%2B4%29 further.