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anonymous
 4 years ago
How to factor: x^56x^4+40x^213x70
Please show me the steps!
anonymous
 4 years ago
How to factor: x^56x^4+40x^213x70 Please show me the steps!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0try \[\pm1,2,5,7,10,35,70\]then use synthetic division or polynomial long division to divide out the factors

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you explain synthetic divsion please?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let finish solving the problem and then I will explain

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then, can you please show the steps for doing the polynomial long division to divide out the factors

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is only one integer factor: (x2). dw:1357004129003:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the other factor is (x2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What are the numbers of the thrid and forth row?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, I am beginning to see how synethetic divsion works

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Please correct me if I am going to wrong way: x^56x^4+40x^213x70 =(x2)(x^44x^38x^2+24x+35) The next step would be to divide (x^44x^38x^2+24x+35) by 5 using synethetic divsion?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The remaining factors are not rational, so the rational roots theorem would not assist us any further.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, no wonder it didn't work anymore

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so that's how you factor it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(x2)(x^44x^38x^2+24x+35) This is the simplest form? Can't (x^44x^38x^2+24x+35) be factor too?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not complex, irrational. But they exist as conjugates so you would have to factor out a quadratic instead of a single term.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, i will continue to try

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got it (x^44x^38x^2+24x+35)=(x^22x7)(x^22x5)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so when (x^56x^4+40x^213x70) is factored, it will be (x2)(x^22x7)(x^22x5) Is it correct?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1@yociyoci just curious, how did you factorize it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you so much everyone for helping me with factoring! I learnt so much from you guys!! Thanks!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I plug in variables and did some algebra and trial and error

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1You 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya, I did something similar

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's a really good technique.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1Great! Thank you for the enjoyable session!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you very much, guys!! :)
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