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factor the polynomial in the picture

Anonymous Mon 28 Mar 19:43:54 2022 No.14344763 View Reply Original Report

Quoted By: >>14346252 >>14348715

the less factors in your answer the better.

the polynomial is considered factored if there is only 1 term (read this if you don't remember what a term is https://www.nelson.com/school/elementary/mathK8/quebec/0176237879/documents/NM8SB_8A.pdf )

good luck.

Anonymous

Anonymous Mon 28 Mar 2022 23:44:14 No.14345302 Report

Quoted By: >>14345454

(x-1)^4

Anonymous

Anonymous Tue 29 Mar 2022 00:49:48 No.14345454 Report

Quoted By: >>14345483 >>14345506 >>14345519

>>14345302
correct, how did you arrive at that answer?

Anonymous

Anonymous Tue 29 Mar 2022 01:13:11 No.14345483 Report

Quoted By:

>>14345454
Ur mom told me

Anonymous

Anonymous Tue 29 Mar 2022 01:33:35 No.14345506 Report

Quoted By: >>14345517 >>14346249

>>14345454
There is probably a more elegant way but the first step is to expand everything out and simplify.
Since all coefficients are integers, you can apply the rational root theorem to find a root.
Since the leading coefficient and constant term are both 1, a solution has no other choice but to be 1 as well.
After that you do polynomial long division and repeat the process until you get to a quadratic equation from where on it's trivial to solve.

Anonymous

Anonymous Tue 29 Mar 2022 01:39:45 No.14345517 Report

Quoted By: >>14345522

>>14345506
In hindsight applying the rational root theorem once is enough, since no other solution but 1 works out. but since the degree of the polynomial is 4 and it's leading coefficient is 1, one can immediately deduce that the polynomial must be equal to (x-1)^4

Anonymous

Anonymous Tue 29 Mar 2022 01:41:56 No.14345519 Report

Quoted By:

>>14345454
1, -4, 6, -4, 1
luckily you can notice these are just binomial coefficients

Anonymous

Anonymous Tue 29 Mar 2022 01:45:08 No.14345522 Report

Quoted By: >>14346197 >>14346249

>>14345517
Something seems fishy, there are lots of other polynomials with the features you listed but they aren’t equal to (x-1)^4

Anonymous

Anonymous Tue 29 Mar 2022 10:32:31 No.14346197 Report

Quoted By:

>>14345522
I hope he didn't just use wolfram alpha, that would be cringe

Anonymous

Anonymous Tue 29 Mar 2022 11:00:04 No.14346249 Report

Quoted By:

>>14345522
The theorem he used only works if the roots are rational. It is not really possible to know whether irrational roots are involved (most other polynomials with these properties have irrational roots). However if they are rational, then trying +1 or -1 by plugging that into the equation should verify if +-1 is a root. This is probably what was done here
If you assume that all other roots are rational as well, then the argument holds (if -1 is also not a root) and it has to be (x-1)^4.
Probably just took an educated guess that worked out, or as suggested in >>14345506 you can do polynomial long division to get rid of the root you just found and have a new polynomial of degree 3, with which you do the same thing: plugging in 1 and seeing whether it is still a root. (if you really do have rational roots only, then they would be "carried" over to the polynomial of lesser degrees because no other choice but 1 worked beforehand)

Anonymous

Anonymous Tue 29 Mar 2022 11:03:28 No.14346252 Report

Quoted By: >>14346319

>>14344763
>algebra2

Anonymous

Anonymous Tue 29 Mar 2022 11:41:14 No.14346319 Report

Quoted By:

>>14346252
basic maths actually (middle school level)

Anonymous

Anonymous Tue 29 Mar 2022 11:55:40 No.14346338 Report

Quoted By: >>14346355

>factored if there is only 1 term
So if it does happen to be (x-1)^4, then it isn't factored, because then there are two terms?

Anonymous

Anonymous Tue 29 Mar 2022 12:05:17 No.14346351 Report

Quoted By: >>14346376 >>14346378 >>14346422 >>14347819 >>14348715

ok since someone already gave the correct answer and more than 12 huors have passed, I will give the official solution

(1 - 2x - x^2)(1 - 2x + 3x^2) + 4x^2 = (- x^2 - 2x + 1)( 3x^2 - 2x + 1) + 4x^2 =

= ( (- 2x^2 + x^2) - 2x + 1)( (2x^2 + x^2) - 2x + 1) + 4x^2 =

= ( - 2x^2 + (x^2 - 2x + 1) )( 2x^2 + (x^2 - 2x + 1) ) + 4x^2 =

= [ - 2x^2 + (x - 1)^2 ] * [ 2x^2 + (x - 1)^2 ] = [ (x - 1)^2 - 2x^2 ] * [ (x - 1)^2 + 2x^2 ] =

= ((x - 1)^2 )^2 - (2x^2)^2 + 4x^2 = (x-1)^4

If you got the right answer congratulations, that's the type of factoring you need to do to solve national military insitutions exams in my country

Anonymous

Anonymous Tue 29 Mar 2022 12:08:27 No.14346355 Report

Quoted By: >>14346364

>>14346338
>because then there are two terms
(x-1)^4 is only one term, that term has a subexpression (inside the parentheses) which indeed has 2 terms
https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables#cc-6th-evaluating-expressions
https://www.khanacademy.org/math/algebra-home/alg-intro-to-algebra/alg-manipulating-expressions/a/combining-like-terms-reminder

Anonymous

Anonymous Tue 29 Mar 2022 12:16:31 No.14346364 Report

Quoted By: >>14346422

>>14346355
Not according to the link supplied by OP:
(x-1)^4 is neither "a signed number, a variable, or a constant multiplied by a variable or variables".

Anonymous

Anonymous Tue 29 Mar 2022 12:28:13 No.14346376 Report

Quoted By: >>14346422

>>14346351
>that's the type of factoring you need to do to solve national military insitutions exams in my country
Based. Have you ever committed a warcrime?

Anonymous

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Anonymous Tue 29 Mar 2022 12:29:50 No.14346378 Report

Quoted By: >>14346422

>>14346351
i dont get the brackets

Anonymous

Anonymous Tue 29 Mar 2022 13:04:14 No.14346422 Report

Quoted By:

>>14346351
>= [ - 2x^2 + (x - 1)^2 ] * [ 2x^2 + (x - 1)^2 ] = [ (x - 1)^2 - 2x^2 ] * [ (x - 1)^2 + 2x^2 ] =
made a mistake (forgot the + 4x^4 ) on thse lines, it's supposed to be

= [ - 2x^2 + (x - 1)^2 ] * [ 2x^2 + (x - 1)^2 ] + 4x^4 = [ (x - 1)^2 - 2x^2 ] * [ (x - 1)^2 + 2x^2 ] + 4x^4 =

>>14346364
>Not according to the link supplied by OP:
the link I gave doesn't mention sub-expressions at all, notice that are no parentheses in any examples, my bad

though from
>If a term has more than one variable, the degree is equal to the sum of the exponents of all its variables. The degree of 8x^3 * y^2 * z is 6.
You could infer 8x^3 * y^2 * z = 2(5x^3 - x^3) * y^2 * z which is still just one term

>>14346376
>Based. Have you ever committed a warcrime?
I ould never even dream of entering a military academy, besides, the most prestigious one has an age limit of 25 I believe

>>14346378
>i dont get the brackets
it's just to look more understandable, you can use only parentheses and get the same result following the right order of operations

Anonymous

Anonymous Wed 30 Mar 2022 00:15:13 No.14347819 Report

Quoted By: >>14347826

>>14346351
>((x - 1)^2 )^2 - (2x^2)^2 + 4x^2 = (x-1)^4
I'm probably a brainlet but why is this equality obvious? As in, why stop showing the steps here? Should I recognize the form of the right-hand side?

Anonymous

Anonymous Wed 30 Mar 2022 00:18:11 No.14347826 Report

Quoted By: >>14348715

>>14347819
wait is there some kind of error?
((x-1)^2)^2 = (x-1)^4,
but we have an additional
-4x^4 + 4x^2

Anonymous

Anonymous Wed 30 Mar 2022 10:15:27 No.14348715 Report

Quoted By:

>>14347826
>wait is there some kind of error?
>((x-1)^2)^2 = (x-1)^4,
>but we have an additional
>-4x^4 + 4x^2
yeah turns out I wrote it incorrectly since the beginning
>>14346351
>+ 4x^2
it's not supposed to be 4x^2, if you look at the original picture you see it's 4x^4
>>14344763

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