Question 217753: factor by grouping
12a^3-9a^2+4a-3
2p^3+5p^2+6p+15
Found 2 solutions by RAY100, stanbon:
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! 12a^3 -9a^2 +4a -3
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3a^2(4a -3) +1(4a-3)
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,,,,,,hard to visualize but if b=(4a-3),,,,this becomes 3a^2b +1b,,,,,or b(3a^2 +1),,,,which becomes by subst back,,,,(4a-3)(3a^2 +1)
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(4a-3)(3a^2 +1),,,,answer
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check,,,,FOIL,,,,12a^3 +4a -9a^2 -3,,,,,ok
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2p^3 +5p^2 +6p +15
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p^2(2p +5),,,,,+ 3(2p+5)
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(p^2 +3) (2p+5),,,,,,answer
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check,,,,FOIL,,,,,2p^3 +5p^2 +6p +15,,,,,ok
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after factoring by grouping,,,,,check to see if either of the factors can be factored further,,,,,in our problems,,,,no
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Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 12a^3-9a^2+4a-3
3a^2(4a-3) + (4a-3)
= (4a-3)(3a^2+1)
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2p^3+5p^2+6p+15
Rearrange:
2p^3 + 6p + 5p^2+15
2p(p^2+3) + 4(p^2+3)
(p^2+3)(2p+4)
2(p+2)(p^2+3)
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Cheers,
Stan H.
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