Stephanie M. answered • 05/13/15
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(k2+10k+21)(k2+9k)
k4 + 9k3 + 10k3 + 90k2 + 21k2 + 189k
k4 + 19k3 + 111k2 + 189k
(k2+16k+63)(k2+7k+12)
k4 + 7k3 + 12k2 + 16k3 + 112k2 + 192k + 63k2 + 441k + 756
k4 + 23k3 + 187k2 + 633k + 756
Regina G.
Never mind. I had to drop a cobweb or two to figure out what the 633k came from.
You're terrific.
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05/13/15
Stephanie M.
tutor
So, I've been going over my old answers, and in doing so, I've discovered that I answered this one incorrectly. I must not have noticed that you mentioned they were fractions!
What I should have suggested (which won't help you a year later, but just for correctness' sake) is:
1. Factor everything you can (numerators and denominators):
k2+10k+21 = (k+7)(k+3)
k2+9k = k(k+9)
k2+16k+63 = (k+9)(k+7)
k2+7k+12 = (k+3)(k+4)
2. Cancel out any terms that appear in both the top and bottom
That's the (k+3) terms, the (k+9) terms, and the (k+7) terms.
3. Multiply by just combining the remaining numerator and denominator terms
You're left with k in the numerator and k+4 in the denominator:
k
-------
k+4
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04/22/16
Regina G.
05/13/15