## Exploding Dots

### 6.9 Solutions

As promised, here are my solutions to the question posed.

1.

a) $$\left(2x^4+3x^3+5x^2+4x+1\right)\div \left(2x+1\right)=x^3+x^2+2x+1$$

b) $$\left(x^4+3x^3+6x^2+5x+3\right)\div \left(x^2+x+1\right)=x^2+2x+3$$

And if $$x$$ happens to be 10, we’ve just computed $$23541\div 21 =1121$$ and $$13653 \div 111 = 123$$.

2. We can do it. The answer is $$x^2+2x+1$$.

3.

a) For $$x=10$$ it says $$14641 \div 11 = 1331$$.

b) For $$x=2$$ it says $$81 \div 3 = 27$$.

c) For $$x=3$$ it says $$256 \div 4 = 64$$.

For $$x=4$$ it says $$625 \div 5 = 125$$.

For $$x=5$$ it says $$1296 \div 6 = 216$$.

For $$x=6$$ it says $$2401 \div 7 = 343$$.

For $$x=7$$ it says $$4096 \div 8 = 512$$.

For $$x=8$$ it says $$6561 \div 9 = 729$$.

For $$x=9$$ it says $$10000 \div 10 = 1000$$.

For $$x=11$$ it says $$20736 \div 12 = 1728$$.

d) For $$x=0$$ it says $$1 \div 1 = 1$$.

e) For $$x=-1$$ it says $$0 \div 0 = 0$$. Hmm! That’s fishy! (Can you have a $$1 \leftarrow 0$$ machine?)

4. $$\dfrac{x^3-3x^2+3x-1}{x-1}=x^2-x+1$$

5. $$\dfrac{4x^3-14x^2+14x-3}{2x-3}=2x^2-4x+1$$

6. $$\dfrac{4x^5-2x^4+7x^3-4x^2+6x-1}{x^2-x+1}=4x^3+2x^2+5x-1$$

7. $$\dfrac{x^{10}-1}{x^2-1}=x^8+x^6+x^4+x^2+1$$

8. We know that $$\left(2x^2+7x+6\right) \div \left(x+2\right) = 2x+3$$ so I bet $$\left(2x^2+7x+7\right) \div \left(x+2\right)$$ turns out to be $$2x+3+\dfrac{1}{x+2}$$. Does it?

9. $$\dfrac{x^4}{x^2-3}=x^2+3+\dfrac{9}{x^2-3}$$

10. $$5x^2-2x+21+\dfrac{-14x^2+82x-14}{x^3-4x+1}$$

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