## anonymous one year ago Factor x^3-5x^2-25x+125

1. anonymous

5x(x^2-x-5+25)

2. welshfella

- no - thats gives 5x^3

3. welshfella

x+ 5 or x-5 might be a factor find f(5) and see if it = 0

4. anonymous

So (x+5)(x-5)(x-5)

5. anonymous

What type of roots of those?

6. Michele_Laino

we can factor out x^2 between the first two terms, and we can factor out -25x between the third and the fourth term, so we get this: $\Large {x^3} - 5{x^2} - 25x + 125 = {x^2}\left( {x - 5} \right) - 25x\left( {x - 5} \right)$

7. welshfella

OOOH - did you guess that?? (x-5) is a factor because f(5) = 0

8. anonymous

No. I just did (x^2-25)(x-5)

9. welshfella

well you are correct

10. anonymous

What type of roots are those?

11. anonymous

And thanks youu

12. Michele_Laino

oops... we can factor out 25 between the third and the fourth terms, not 25 x: $\Large {x^3} - 5{x^2} - 25x + 125 = {x^2}\left( {x - 5} \right) - 25\left( {x - 5} \right)$

13. welshfella

right

14. welshfella

- you beat me too it!!

15. anonymous

So would the roots be -5,-5,5 or -5,5,5?

16. welshfella

the roots are 5 and -5 root 5 is a duplicate

17. anonymous

Ok but since the polynomial is 3 there has to be 3 roots?

18. welshfella

- correction - that was rubbish the roots are 5 , 5 and -5

19. anonymous

How?

20. welshfella

yes -5 and 5 duplicity 2

21. welshfella

factors are (x + 5)(x - 5)(x - 5) the fist gives x = -5 and the other 2 give x = 5

22. anonymous

Ohhh<3

23. anonymous

Thanks! :)

24. welshfella

no 3 roots - because there are 2 5's

25. welshfella

the graph will look a bit like |dw:1436970554356:dw|

26. welshfella

- just touching the point (5,0)

27. anonymous

:) Thank you!!

28. welshfella

yw