Tricky math problem if you are bored.?

Hello,

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DISCLAIMER:
I am not trying to get my homework done. This quiz is provided as it is, and I do have the answer. So it is a challenge for anyone. Best answer will be awarded in approximatively two days. Have fun gals and guys!

= = = = = = = =
Let π‘Ž and 𝑏 be two real values.
Let 𝑃 be a polynomial such that:
Β Β Β {   𝑃(π‘Ž) = 0
Β Β Β {   𝑃(𝑏) = 0

A. What is the value of 𝑃'(π‘Ž)+𝑃'(𝑏) ?
B. What is the minimal degree π‘šπ‘–π‘› of 𝑃?
C. If 𝑃 has a degree of π‘šπ‘–π‘›+1, what can you conclude?

Regards,
Dragon.Jade :-)

2016-07-29T10:22:12Z

Mmm... After consideration. Question A is faulty and is hereby cancelled. It will not count toward Best Answer.
Questions B and C stand though. And they seem so trivial that no one wants to answer them correctly...?

2016-07-30T07:20:50Z

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
That was the trick in the question.

There's no doubt the Zero polynomial defined by:
Β Β Β βˆ€π‘₯βˆˆβ„, Β Zero(π‘₯) = 0
fits the given definition of 𝑃.

2016-07-30T07:21:47Z

Now what is the degree of this polynomial?
It is indeed a convention that some think it should be undefined, 0, -1 or -∞.

If π‘šπ‘–π‘› were undefined or π‘šπ‘–π‘›=-∞, then π‘šπ‘–π‘›+1 would have no meaning. Which means question C would be meaningless.
So those conventions need to be discarded in order to answer question C.

2016-07-30T07:25:44Z

If π‘šπ‘–π‘›=-1, then π‘šπ‘–π‘›+1=0 and such 𝑃 would be a nonzero constant polynomial then 𝑃(π‘Ž)β‰ 0 and 𝑃(𝑏)β‰ 0 which is contrary to the problem.
The conclusion would then be there is no 𝑃 with degree π‘šπ‘–π‘›+1=0.

If π‘šπ‘–π‘›=0, then π‘šπ‘–π‘›+1=1 and such 𝑃 would be a linear polynomial which can only have an unique zero.
The conclusion would then be that π‘Ž=𝑏.

β–Ί Thanks for participating. The BA is given to Barry G for being the only one seeing that the Zero polynomial did fit the definition of 𝑃.

Barry G2016-07-29T10:31:31Z

Favorite Answer

B. The minimum degree is zero :
P(x) = x^0 - 1
which is zero for all real values of x.

C. If P has degree 1 then
P(x) = Ax+B with A <>0.
P(a) = Aa+B = 0 --> B = -Aa
P(b) = Ab+B = 0 --> Ab-Aa = A(b-a) = 0
so b=a.

00

Jeff Aaron2016-07-29T09:29:03Z

A. Insufficient information given. There are infinity possible answers.

B. Assuming a and b are different, then P must have at least degree 2. But if a and b can be the same, then P can have degree 1.

C. There are min+1 roots (but not necessarily distinct roots; some or all may be equal), but some of those may be non-real roots.

10

husoski2016-07-29T09:36:57Z

Seems underspecified to me.
p(x) = x^2 - x and q(x) = x^3 - 2x^2 + x are zero at a=0 and b=1, and nowhere else.

p'(x) = 2x - 1 --> p'(0) = -1, p'(1) = 1, p'(0) + p'(1) = -1 + 1 = 0
q'(x) = 3x^2 - 2x + 1 --> q'(0) = 1, p'(1) = 2, p'(0) + p'(1) = -1 + 1 = 3

So, question A has no definite answer. The others two have pretty obvious answers.

10

Sqdancefan2016-07-29T10:54:11Z

B. Assuming a and b are distinct, P must be of degree 2 or greater. The minimum degree is 2.

C. A degree 3 polynomial with 2 real roots also has a third real root. Conclusion: there must be one more real root, though it isn't necessarily distinct from both a and b.

00

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