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1. Simplify (b2c +7bc – 3) – (12b2c – 2bc + 5). –11bc2 + 9bc – 8 –11b2c + 9bc – 8 13b2c + 5bc + 2 –11b2c + 5bc + 2 2. Simplify 3fg(f2g + 5f – 4). 3f2g2 +15fg – 12 3f3g3 + 15f2g – 12 3f3g2 + 15f2g – 12 3f3g2 + 15f2g – 12fg
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question(1) –11b2c + 9bc – 8 question(2) 3f3g2 + 15f2g – 12fg
Thank You so much :D
we see here in question 1 I add the same terms together (b2c with b2c) and (bc with bc) but, I cannot add (b2c with bc) question2 I multiplied 3fg with the every component ex. 3fg(g) = 3fgg= 3fg2 3fg(4fg)= (3)(4)(f)(f)(g)(g)=12f2g2 best of luck with your future studies
Thank you:D
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