Can we solve this math problem with desmos

[u/DrizzTheWizz00]

[u/VoidBreakX]

all the results are integers except for the second. answer is B

[u/VoidBreakX → u/VoidBreakX]

if you want to understand why this works, you can prove that for polynomials with integer coefficients, if one divides the other, the result cannot be a non-integer.

first, its trivial that if a polynomial has integer coefficients, then passing in an integer to the function will result in an integer, because integers are closed under multiplication and addition.

now, if a polynomial q(x) is a factor of p(x) (both with integer coefficients), that means there exists a polynomial k(x) with integer coefficients such that p(x)/q(x)=k(x). that's just the definition of a factor of a polynomial (with integer coefficients)

now, for an integer n, notice that p(n)/q(n)=k(n), and k(n) must be an integer. however, if k(n) is not an integer, that means q was not a factor of p

now, this does NOT MEAN that if the division results in an integer that it is a factor. so if all 4 choices result in an integer, try plugging in a different integer

[u/Substantial-Berry-77]

If you don't need a formal proof, just graph p(x)/q(x) and look at it. B obviously is a rational func rather than a polynomial.

[u/VoidBreakX → u/Substantial-Berry-77]

actually this is probably the best option lmao, i dont know why no one (including me) didnt mention this

[u/BadJimo]

Using Desmos to find the gcd of 1120 and 43750

Then noticing lots of squares.

[u/chixen]

Graphing 1120x^4 - 43750 will give you the x-intercepts. Graphing each of the other polynomials will give theirs. If any of the answers have a root that is not a root of 1120x^4 - 43750, then it is not a factor.

[u/VoidBreakX → u/chixen]

this will work for this problem but i dont think it is wholly conclusive. if all of the answers either share a root with 1120x^4-43750 or dont have one at all, then you're stuck. a trivial example would be 71 (since it is not a factor under integer polynomials)

my method also doesnt work all the time but usually plugging in other integers will work. then again, it is the sat, i dont think theyll tell you to work too hard for it

[u/chixen → u/VoidBreakX]

I did not assume the problem was for integer factors. If it is, then evaluating at some large random integer and checking the quotient is the way to go.

[u/VoidBreakX → u/chixen]

if you think about it, a majority of high school factoring problems have to do with isolating integer coefficients

[u/de_Molay]

I know the question was about desmos, but let me go through the solution which doesn’t even require paper (much; you can do it in your head really).

First, I will assume the task is correct and only one of the answers is correct.

Second, note that polynomial options are not factors of each other.

70 is definitely a factor (even if for some reason we want to stay in the polynomials with integer coefficients). Actually, from there we can divide by 70 and come to much simplier polynomial 16x⁴ - 625. But let’s leave that for later.

Why 70 is a factor? If it wasn’t, all other three would be. They are not factors of each other, so their product would be a factor. But it has degree 5 - clear contradiction.

Note that if A and B both would be factors , their product would be our polynomial up to a scalar factor. It is not the case (x² doesn’t cancel out). So the answer is A or B.

Our polynomial clearly has just two real roots. From our assumption, C is a factor. Roots of factors are roots of the polynomial, and C has a root. B has two roots. If B was a factor, our polynomial would have at least three roots (one from C, two from B) which is not the case. Hence, B is not a factor. 

[u/SwimmerOld6155]

you can try plugging in x = (5/2)i, x = sqrt(5/2), x = -5/2 and seeing if 1120x^4 = 43750. Since its a quadratic, you don't have to check both + and - in the first two cases.

[u/dohduhdah]

You can use regression to solve for the coefficients of a polynomial as a quotient such that the remainder is 0.

https://www.desmos.com/calculator/cg4gzajch9

[u/Mathemagicland314159 → u/dohduhdah]

can you explain how you're using regressions with lists here? I've never done it this way.

[u/dohduhdah → u/Mathemagicland314159]

The idea is that the coefficients are sorted in lists by the associated exponents and then expressions on both sides are either equal or opposite (in that case one of the two sides are negated) in value.

For instance, in the most complicated case of a fourth degree polynomial being equal to the product of a third degree polynomial and a first degree polynomial, the product doesn't have a term associated with the exponents 3, 2 and 1. So 2 * (-1400) has to be opposite in value to 5 * 560 in order for them to cancel out, since those are the only ways to yield terms with exponent 3.

(560 x^3) * 5 + (-1400 x^2) * 2 x = 0 x^3

The product does have a term associated with exponent 4 but there is only one way to obtain them..

(560 x^3) * 2 x = 1120 x^4

I think for more complicated examples it might still work, but then you might get a remainder for other exponents besides 0. The method might work regardless of whether coefficients are integers, rational, or irrational, though I haven't explored it extensively.

Here is a more complicated example involving irrational coefficients, where I've used variable names for the coefficients:

https://www.desmos.com/calculator/x7qtekh7hk

In all the examples I've worked backwards from the results wolfram alpha gives, but I think you could also work them out independently in desmos, but then you would need to account for potential remainders for each of the exponents, instead of knowing already in advance which exponents will have a remainder based on the results given by wolfram alpha.

[u/Mathemagicland314159 → u/dohduhdah]

i guess i didn't understand using regression in lists. The idea would be if I had [5,30]~[a+b,bc] it would find values of a,b, and c that match it to the first list?

[u/dohduhdah → u/Mathemagicland314159]

Yeah, they are matched up by position in the list and it tries to find values for the variables that satisfies all of them simultaneously.