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Still stuck on #20

20. Don't know what to do after this: [(16a^4/12a^3)]-[(40a^2/12a^3)]+[(24a/12a^3)]

RainbowPanda Sep 11, 2018

#2**+3 **

18. \(4(2-3w)(w^2-2w+10)\), we take a good look at the last two terms, excluding the 4: (2-3w) and (w^2-2w+10)

We can expand or distribute the first two terms(2 and -3w) with the rest of the terms. So, we yield: \(2w^2+2\left(-2w\right)+2\cdot \:10+\left(-3w\right)w^2+\left(-3w\right)\left(-2w\right)+\left(-3w\right)\cdot \:10\) . We simplify this, to get: \(-3w^3+8w^2-34w+20\).

Now, finally, we multiply this terms by 4: \(\:4\left(-3w^3+8w^2-34w+20\right)= \boxed{-12w^3+32w^2-136w+80}\), and that's our answer!

I have to go now, bye!

Oops, didn't see your reply, sorry.

tertre Sep 11, 2018

#4**+3 **

Wait, I can try 20!

20. \(\frac{16a^4-40a^2+24a}{12a^3}\), we can factor out a common term 8a from the numerator, so we get: \(8a\left(2a^3-5a+3\right)\).

Now, we are left with \(\frac{8a\left(2a^3-5a+3\right)}{12a^3}\) . We can factor a common term 4, and cancel \(a\) , a common factor.

Cancel common factor 4: \(\frac{2a\left(2a^3-5a+3\right)}{3a^3}\).

Cancel common factor \(a\): \(\frac{2\left(2a^3-5a+3\right)}{3a^2}\).

So, \(\boxed{\frac{2\left(2a^3-5a+3\right)}{3a^2}} \)

is our answer!

tertre Sep 11, 2018