## anonymous one year ago What is y^4-y^3=0 How do you factor this out?

1. Nnesha

what's the common factor ? well its soo easy to find common factor of variables variable with the smallest degree would be the common factor

2. anonymous

How would you do it though?

3. DanJS

you are really multiplying by y^2 / y^2 this will pull a y^2 term out of each of those

4. Nnesha

we should take out the common factor so first ) what's a common factor ???

5. DanJS

left with y^2*(each divided by y^2)

6. anonymous

@DanJS my book did it by y(y^3-1) =y(y-1)(y^2+y+1) I dont understand how they do it

7. Nnesha

well ^^take out y from y^4 -y^3 $\huge\rm y(y^3-1)$ y^3-1 is same as (y^3 -1^3)

8. Nnesha

then you can apply the rule $\huge\rm x^3-y^3=(x-y)(x^2+xy+y^2)$

9. DanJS

is the original problem y^4 - y^3

10. DanJS

= y^3 * (y-1) = 0

11. DanJS

y is 0 or 1

12. anonymous

Yes

13. Nnesha

i guess it's y(y^3-1) ??

14. anonymous

Shouldn't it factor out as y^3(y-1)?

15. anonymous

Doesn't y(y^3-1) = y^4-y

16. Nnesha

ye....... and then set it equal to zero if you want to solve to for y but factor is y^3(y-1) that looks correct 2 me :/

17. mathstudent55

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18. mathstudent55

$$y^4 - y^3 = 0$$ $$y^3(y - 1) = 0$$ $$y^3 = 0$$ or $$y - 1 = 0$$ Now solve both equations for y.

19. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Anzhbar my book did it by y(y^3-1) =y(y-1)(y^2+y+1) I dont understand how they do it $$\color{blue}{\text{End of Quote}}$$ i guess you're looking at the wrong question .....:/

20. mathstudent55

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