LOL!!!!!!

Another possibility is to simply subtract 81 from both sides. This gives

x^4 - 81 = 0     factor

(x^2 + 9) (x^2 - 9) = 0

If we set both factors = to 0, the first has no real solution. By factoring again, the second gives us

(x + 3) (x - 3) = 0

And the solutions are -3 and 3

Nice job on that second one Melody !!!

There may be some factoring trick for the first, but I would have probably taken Melody's approach and used the Factor Theorem, realizing that there can be no positive real roots because all the linear terms in the original question would be > 0. And multiplying these together and adding a "+15" to them could never give us "0."

The expansion of the original question is x^4 + 16X^3 + 86x^2 +176x + 120, and the  factorization of this  is, indeed, (x+2) (x+6) (x^2 + 8x + 10)

Using "p/q,"  the first root ... (-2)... could be found pretty easily...and after some synthetic division, the resulting polynomial would be  ...   x^3 + 14x^2 + 58x + 60......and the next rational roots that would come to mind to "test" would be -3, -4, -5 and -6. And, as Melody found, -6 is the correct one.

Then, we would use synthetic division again to reduce the polynomial to x^2 + 8x + 10.... and this polynomial has no rational zeroes. (But it does have two "real" roots!!)

Again...good job, Melody !!!   Gin and tonics really DO help one's math skills.....(it's an "inside" joke !!!)

I propose a "toast".........

1225 = 25 X 49 = 5² X 7²

We have confirmed it ......the answer is   2 !!!!!!!

We have

2log25 - 3log5 + log20

2log(5*5) - 3log 5 + log (4*5)

2log 5 + 2log 5 - 3log 5 + log4 + log 5      combine like terms

5log5 - 3log 5 + log4 =

2log5 + log 4 =

log5² + log 4 =

log 25 + log 4 =

log (25*4) =

log 100 =

2

This question doesn't make sense...we have more people "ticking" than are in the survey !!!!

Maybe we could figure this out with  "clock" arithmetic ?????

We're just looking at numbers such as

4242 + 1

42424242 + 1

4242 - 1

424242 - 1

etc.

In other words, numbers where a plus (or minus) 1 is being added to a string of digits with a "42" pattern repeated at some length. I just tested a few random ones in WA to see if these sort of numbers were prime or not. Some are and some aren't. Thus...there doesn't appear to be any "smallest" factor - in general - that divides any number of this sort.

Maybe there is some algorithm that will tell us when such numbers are prime, but I'm not aware of any........

I think the questioner wants to know, if we add or subtract 1 from a string of digits having a pattern of 424242 etc., do such numbers  have some  prime factorization pattern???  Using WolframAlpha and choosing some at random, it seems as if some are prime numbers and some aren't. Thus, there is no "smallest" factor that I can detect. Perhaps there is some algorithm to tell when these may be prime and when they aren't???  I don't know.......

Does that help??

Just for the heck of it, let's plug the values we found for the triangle's sides back into the original polynomial to see what the solution might be:

(3b -2c)x^2 + 2bx + (4c - 3b) = 0

[3(4) -2(5)] x^2 + 2(4)x  + [4(5) - 3(4)] = 0

2x^2 + 8x + 8 = 0

x^2 + 4x + 4 = 0

(x + 2)  (x + 2) = 0

x + 2 = 0

x = -2

Ah ha !!!....just as quinn noted in his original question.....our polynomial indeed has only one (repeated) root. This seems to further confirm that our answer was correct .......