## yociyoci Group Title How to factor: x^5-6x^4+40x^2-13x-70 Please show me the steps! one year ago one year ago

1. ArkGoLucky Group Title

There is a theorem thing that states that plus or minus the factors of the constant(70 in this case) divided by plus or minus the constant in front of the term with the largest power are all the possibilities of the factors of the polynomial. So just start crunching the numbers.

2. ArkGoLucky Group Title

try $\pm1,2,5,7,10,35,70$then use synthetic division or polynomial long division to divide out the factors

3. yociyoci Group Title

can you explain synthetic divsion please?

4. ArkGoLucky Group Title

let finish solving the problem and then I will explain

5. yociyoci Group Title

Then, can you please show the steps for doing the polynomial long division to divide out the factors

6. ArkGoLucky Group Title

there is only one integer factor: (x-2). |dw:1357004129003:dw|

7. ArkGoLucky Group Title

the other factor is (x-2)

8. yociyoci Group Title

What are the numbers of the thrid and forth row?

9. yociyoci Group Title

oh, I am beginning to see how synethetic divsion works

10. yociyoci Group Title

Please correct me if I am going to wrong way: x^5-6x^4+40x^2-13x-70 =(x-2)(x^4-4x^3-8x^2+24x+35) The next step would be to divide (x^4-4x^3-8x^2+24x+35) by 5 using synethetic divsion?

11. LogicalApple Group Title

The remaining factors are not rational, so the rational roots theorem would not assist us any further.

12. yociyoci Group Title

oh, no wonder it didn't work anymore

13. ArkGoLucky Group Title

so that's how you factor it

14. yociyoci Group Title

(x-2)(x^4-4x^3-8x^2+24x+35) This is the simplest form? Can't (x^4-4x^3-8x^2+24x+35) be factor too?

15. LogicalApple Group Title

Since the remaining roots are complex, you couldn't factor out a single term, you would have to somehow factor out a quadratic. I do not know any techniques that will do this.

16. LogicalApple Group Title

Not complex, irrational. But they exist as conjugates so you would have to factor out a quadratic instead of a single term.

17. yociyoci Group Title

ok, i will continue to try

18. yociyoci Group Title

I got it (x^4-4x^3-8x^2+24x+35)=(x^2-2x-7)(x^2-2x-5)

19. yociyoci Group Title

so when (x^5-6x^4+40x^2-13x-70) is factored, it will be (x-2)(x^2-2x-7)(x^2-2x-5) Is it correct?

20. LogicalApple Group Title

Yes. That's it !

21. mathmate Group Title

@yociyoci just curious, how did you factorize it?

22. yociyoci Group Title

Thank you so much everyone for helping me with factoring! I learnt so much from you guys!! Thanks!!

23. yociyoci Group Title

I plug in variables and did some algebra and trial and error

24. LogicalApple Group Title

o_o

25. yociyoci Group Title

ya, it was long...

26. mathmate Group Title

You can also use some facts as we usually do in quadratics. Assume (x^2+ax+b)(x^2+cx+d) then (a+c)=-4 ac+b+d=-8 bc+ad=24 bd=35 Hopefully this will help you eliminate cases quickly without having to multiply out the whole expression.

27. yociyoci Group Title

ya, I did something similar

28. LogicalApple Group Title

That's a really good technique.

29. mathmate Group Title

Great! Thank you for the enjoyable session!

30. yociyoci Group Title

Thank you very much, guys!! :)