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ArkGoLuckyBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

yociyociBest ResponseYou've already chosen the best response.0
can you explain synthetic divsion please?
 one year ago

ArkGoLuckyBest ResponseYou've already chosen the best response.0
let finish solving the problem and then I will explain
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
Then, can you please show the steps for doing the polynomial long division to divide out the factors
 one year ago

ArkGoLuckyBest ResponseYou've already chosen the best response.0
there is only one integer factor: (x2). dw:1357004129003:dw
 one year ago

ArkGoLuckyBest ResponseYou've already chosen the best response.0
the other factor is (x2)
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
What are the numbers of the thrid and forth row?
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
oh, I am beginning to see how synethetic divsion works
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
Please 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?
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
The remaining factors are not rational, so the rational roots theorem would not assist us any further.
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
oh, no wonder it didn't work anymore
 one year ago

ArkGoLuckyBest ResponseYou've already chosen the best response.0
so that's how you factor it
 one year ago

yociyociBest 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?
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
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.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
Not complex, irrational. But they exist as conjugates so you would have to factor out a quadratic instead of a single term.
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
ok, i will continue to try
 one year ago

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

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

mathmateBest ResponseYou've already chosen the best response.1
@yociyoci just curious, how did you factorize it?
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
Thank you so much everyone for helping me with factoring! I learnt so much from you guys!! Thanks!!
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
I plug in variables and did some algebra and trial and error
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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.
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
ya, I did something similar
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
That's a really good technique.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Great! Thank you for the enjoyable session!
 one year ago

yociyociBest ResponseYou've already chosen the best response.0
Thank you very much, guys!! :)
 one year ago
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