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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