## anonymous one year ago Long division for x^4 - 2x^3 + 3x^2 - 4x + 6 by x^2 + 2x -1?

1. anonymous

$x ^{2} + 2x - 1 \div x ^{4} - 2x ^{3} + 3x ^{2} - 4x + 6$

2. triciaal

switch

3. anonymous

$x ^{4} - 2x^3 + 3x^2 - 4x + 6 \div x^2 + 2x - 1$

4. triciaal

yes

5. calculusxy

$\ \large \frac{ x^4 - 2x^3 + 3x^2 - 4x + 6 }{ x^2 +2x - 1 }$ when u r dividing, you are basically subtracting the exponents with the same root for example $\large \frac{ x^6 }{ x^2 } = x^4$

6. calculusxy

and if we have $\large 2x$ then the exponent is 1 $\large 2x^1$

7. anonymous

Would x^4 end up divided (subtracting, I guess) by x^2, turning into x^2?

8. calculusxy

yes

9. anonymous

|dw:1441163299335:dw|

10. anonymous

|dw:1441164156039:dw|

11. anonymous

You also need a place holder for the x² term|dw:1441164237673:dw|

12. anonymous

|dw:1441164097336:dw|

13. anonymous

I have no idea where or why I put -1/4x.

14. anonymous