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# Still stuck on these >.<

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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)]

Sep 11, 2018
edited by RainbowPanda  Sep 11, 2018

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Nvm I got 18 but still need help with 20

Sep 11, 2018
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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!

Sep 11, 2018
edited by tertre  Sep 11, 2018
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Oh darn >.< I just needed 20. It's alright

RainbowPanda  Sep 11, 2018
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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, $$\boxed{\frac{2\left(2a^3-5a+3\right)}{3a^2}}$$

Sep 11, 2018
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Notice: In the cancellation of a common factor a, we canceled 2a and 3a^3, yielding us with: 2 and 3a^2.

tertre  Sep 11, 2018
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Thank you so much! ^-^ have an amazing day!

RainbowPanda  Sep 11, 2018