Simplify. Please help!?
3u^2-u-4/u^2 divided by 3u^2+12u+4/u^4-3u^3
Thanks!!!
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(3u^2 – u – 4)/u^2 TIMES the reciprocal of “blah”….
= (3u – 4)(u +1) / u^2 TIMES (u^4 – 3u^3) / (3u^2 + 12u + 4)
= (3u – 4)(u+1) / u^2 TIMES u^3 (u – 3) / (3u^2 + 12u + 4)
= (3u – 4)(u+1) x u(u-3) / (3u^2 + 12u + 4)
= (u)(u+1)(u-3)(3u – 4) / (3u^2 + 12u + 4)
ummm… (3u^2 + 12u + 4) is not factorable because (3u+1)(u+4) = (3u^2 + 13u + 4)…. and (3u+4)(u+1) = (3u^2 + 7u + 4)…. and (3u+2)(u+2) = (3u^2 + 8u + 4)…. and we have (3u^2 + 12u + 4)
… so this can’t be reduced anymore…
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