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

1. ArkGoLucky

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

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

3. yociyoci

can you explain synthetic divsion please?

4. ArkGoLucky

let finish solving the problem and then I will explain

5. yociyoci

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

6. ArkGoLucky

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

7. ArkGoLucky

the other factor is (x-2)

8. yociyoci

What are the numbers of the thrid and forth row?

9. yociyoci

oh, I am beginning to see how synethetic divsion works

10. yociyoci

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

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

12. yociyoci

oh, no wonder it didn't work anymore

13. ArkGoLucky

so that's how you factor it

14. yociyoci

(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

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

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

17. yociyoci

ok, i will continue to try

18. yociyoci

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

19. yociyoci

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

Yes. That's it !

21. mathmate

@yociyoci just curious, how did you factorize it?

22. yociyoci

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

23. yociyoci

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

24. LogicalApple

o_o

25. yociyoci

ya, it was long...

26. mathmate

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

ya, I did something similar

28. LogicalApple

That's a really good technique.

29. mathmate

Great! Thank you for the enjoyable session!

30. yociyoci

Thank you very much, guys!! :)

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