Math Is Fun Forum

If you divide 2x³+4qx²-3q²x-2 by x-q you will get 3q²-2 and from the information in the question you will have:
3q²-2=10 ⇒ 3q²=12 ⇒ q²=4 ⇒ q=±2
That’s it.

The answer for part 1 can be accomplished as follows: The condition that f(x) has the zeros -3,4 and 8 entails that:
f(x)=(x+3)ᵃ×(x-4)ᵇ×(x-8)ᶜ
And since the zeros of g(x) are -5,-3, 2, 4 and 8 we also have:
g(x)=(x+5)ᵈ×(x+3)ᵉ×(x-2)ᶠ×(x-4)ᶢ×(x-8)ʰ
and hence:
g(x)/f(x)=[(x+5)ᵈ×(x+3)ᵉ×(x-2)ᶠ×(x-4)ᶢ×(x-8)ʰ]/[x+3)ᵃ×(x-4)ᵇ×(x-8)ᶜ]
so since the result of the division is (x+5)ᵈ×(x-2)ᶠ×(x+3)ᵝ×(x-4)ᵞ×(x-8)ᵟ, where β=e-a, γ=g-b and δ=h-c. Therefore, if β, γ and δ  are nonnegative integers then g(x) is divisible by f(x). Otherwise it is not.  The answer of part 2 is very clear from the answer of part 1. For a polynomial g(x) to be divisible by a polynomial f(x) at least all the roots (zeros) of f(x) should be also roots (zeros) of g(x) and their repetetion in g(x) should be greater than or equal to their repetetion in f(x). This is the complete proof in any case.

Hi Zetafunc, if there are duplicate roots of a polymolial you should counts them as distinct roots. For example, if you say that a give polynomial has “exactly” the roots -3, 4 and 8 that means there is no duplicate roots. If there are a power for a given factor of the polynomial then there are repeted zeros and you should say something like “the polynomial f(x) has exactly the roots 4, 8 and the two duplicate roots -3”.  I hope this clears your confusion.

Alternative solution: the manager may tell them, in his announcement, that when they come to work the next day after the day off, he will make a lot in which he will mix x pieces of papers numbered from 1 to x (where x is the number of employees who do not attend) and let each one of the employees who did not come to work draws one piece and the one who draws the piece with a certain number (to be indicated before drawing) will be charged with the day off. Now since any one of them has the same chance (5%) to be the charged person, if all of them did not come to work and more than that if some of them attend, then (assuming they are rational) most of them will come to work the next day.

Hi, I guess he said in his announcement that "the first employee who accept his advice will be charged with the day off" so since no one of them would like to be charged with the leave day, no one will accept the advice first and all of them will come to work the next day hence the manager achieves his objective. (This is a kind of "foxy" advice)...

Sorry but it does not imply that 2=3!! To show you why this is the case, consider the following two systems:

2x+3y=2
4x-y=3

And

2x+3y=3
4x-y=4

Both of these systems have solutions and they are different. However, this does not imply that 2=3 and 3=4 because 2 and 3 equals 2x+3y and 3 and 4 equals 4x-y respectively. On the other hand, the one who said that the solution of a system of equations are the values which make each equation in the system hold is the definition of the solutions of a system of equations: "the solutions of a system of equations are the values which satisfy each equation simultaneously". Finally who did say that the solutions of a system of equations, if they exist, have to satisfy one and only one system. This claim has to be poven, in fact, the provided system is a counter-example which disprove that claim.

Note: If you would like to exclude the deduced solutions form being a solution of the provided system, you have to modify the definition of the solutions of a system of equations to include the uniqueness to that system which are not included in the actual definition. Also in that case, we would have a solution to the system according to the previous and unmodified definition. "It's a matter of definition sir".

If someone present a solution to a problem, you just have to ask him for a proof of his results. if he gave you a correct argument, but you did not accept it, then you have to provide plausible reasons to justify your situation. And I think you did not do that. "Any one can reject any angument even if it is correct without reason". Sorry but your situation is weak.

I hate to disagree with you, but in my opinion it is correct!! Let's solve the problem in different way: suppose we have an educated guess that the solutions are x=0 and y=∞ so to check that our supposed solutions are correct, we just have to see if they satisfy the system of equations or not. Let's do that:

Therefore the porposed solutions are correct.

It is much easier than what you have thought of. Since all the greyhounds have constant speeds therefore the difference between the first and the last greyhounds will change at a rate 3/4/30. Now the first possible formation of the square (if it could form) will be when the difference between the first and the last is d₃=1/4 (after the first lap since the difference between the last and the starting point is less then a quarter of a circle) hence: 0.25=3/4/30×t₁-1 ⇒ t₁=1.25/(3/4/30) ⇒ t₁=50 s
But in this case the difference between the first and the second greyhounds will be: d₁=1/4/30×50=5/12 lap which is not a multiple of a quarter of a circle. So let's move to the next chance, this will occur when the difference between the first and the last is 2/4 circle, and the time t₂ to accomplish this will be: 2/4=3/4/30×t₂-1 ⇒ t₂=1.5/(3/4/30) ⇒ t₂=60 s. In this time the differences between the first and the second and the third respectively will be: d₂=1/4/30×60=0.5 lap and d₃= 2/4/30×60=1 lap
Since all the differences between the three greyhounds and the first one are multiples of a quarter circle therefore 60 s is the next time, after the first, in which they form a square again (90 s from the start).

Sorry, but it is not "totally wrong"  the first parts are correct in both arguments:

And:

So the corners have the same number of ways in both arguments, also you have devised the same technique.

By saying "by symmery". So the mistake just in "counting" which is a trivial one.

20) If we let v1,v2 and v3 be the speeds of the car, bus and van respectively, and if t is the time when the car and van meet then we have:
680-[v1(2+t)+v3t]=0
680-[70(2+t)+65t]=0 ⇒ 680-140-70t-65t=0 ⇒ 540-135t=0 ⇒ t=540/135=4 hours
Now since the bus left town M at 1:30 pm that is 0.5 hours before the van then:
d1=4.5V2=405 miles ⇒ d2=d-d1⇒ d2=680-405=275 miles
Where d,d1 and d2 are the distance from towns M and N, the distance the bus traveled form town M when the car meet the van and the required distance respectively.

21) let us suppose that A1,A2,A3 and A4 is the areas of the square ODCE, the trangles OBA,BDC and ACE respectively. Where the points E and D have the coordinates (5,0) and (0,6) respectively. Then we have:
Area(ABC)=A1-(A2+A3+A4)=5×6×4_(3×2×3×2/2+3×2×5×2/2+6×2×2×2/2)=120-(18+30+24)=48 cm^2

It is easy. Suppose that D is the domain of y and R is its range, since y is one to one then ∀ m ∋ D there is one and only one image in R and since y is outo ∀ n ∋ R there is a distinct inverse image "x" in D hence:

(a∘y)∘yˉ¹=(b∘y)∘yˉ¹ ⇒ a∘(y∘yˉ¹)=b∘(y∘yˉ¹) ⇒ a∘x= b∘x ⇒ a=b ( since x is a variable)

Note: the "ay" and "by" is a composition "∘" operation not multiplication "×" since if it was so, it could be solved by division as ordinary numbers.