Lesson Math circle level problems on finding polynomials with prescribed roots

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Math circle level problems on finding polynomials with prescribed roots

Problem 1

Find the polynomial with roots alpha

and

, if

= 6,

= 5, and

=21.

Solution

We are given 

     +  +  = 5,            (1)
  
     +  +  = 21.        (2)


It implies

     =  - ( +  + ) =  -  = 4.


Hence,

     = 2.     (3)


Since  ,    and    satisfy  equations 

     +  +  = 5,

     = 2,

    .. = 6,

then by the Vieta's theorem,  ,    and   are the roots of the cubic equation

     = 0.


Answer.  The polynomial with roots  ,    and   satisfying given conditions is   = 0.

Problem 2

Let the roots of the equation x^3 -2x^2 -3x-7=0 be r, s, and t.
Find an equation whose roots are r^2, s^2 and t^2.

Solution

The given equation is

     = 0        (1)


Equation (1) has the roots  r,  s  and  t.  Therefore, due to to Vieta's theorem

    r + s + t = 2,  r*s + r*t + s*t = -3,  r*s*t = 7.      (2)


Now, an equation with the roots  ,    and    is

     = 0.                     (3)


By the Vieta's theorem (or by applying FOIL directly), the coefficients of the left side polynomial are

      at  x^2;                          (4)

      at x;   and                 (5)

      as the constant term.                (6)


So, my task now is to express the coefficient (4), (5) and (6)  via  the coefficients (2) of the equation (1).


Regarding   ,  it is easy:

     =  -  =  -  = 4 + 6 = 10.


So, the coefficient at x^2 of the polynomial (3)  is   = -10.


Regarding  ,  it is easy, too :

     =  =  = 49.


So, the constant term of the polynomial (3)  is   = -49.


Regarding  ,  it is slightly more long way :

     = -3  of (2)  implies (squaring both sides)

    9 =  = 

         =  +  = substituting the known values from (4) = 

         =  + 2*7*2,

which implies

     = 9 - 28 = -19.


So, the coefficient at x of the polynomial (3)  is   = -11.


Thus we know all three coefficients of the polynomial (3)

     = -10  at  x^2;          

     = -19 at x;   and 

     = -49 as the constant term.   


Answer.  The polynomial equation under the question is   = 0.

Problem 3

The roots of the polynomial equation 2x%5E3+-+8x%5E2+%2B+3x+%2B+5

= 0 are

and

.
Find the polynomial equation with roots alpha%5E2

.

Solution

The given equation

     = 0        (1)

is equivalent to

     = 0     (2)  (all the coefficients of (1) are divided by 2)


Equation (2) has the same roots  ,    and    as equation (1).  Therefore, 

     = ,                       (3)

and, according to Vieta's theorem

     = 4,   = 1.5,   = -2.5.      (4)


Now, an equation with the roots  ,    and    is

     = 0.                     (5)


By the Vieta's theorem (or by applying FOIL directly), the coefficients of the left side polynomial are

      at  x^2;                          (6)

      at x;   and                 (7)

      as the constant term.                (8)


So, my task now is to express the coefficient (6), (7) and (8)  via  the coefficients (4) of the equation (2).


Regarding   ,  it is easy:

     =  =  = 16-3 = 13.


So, the coefficient at x^2 of the polynomial (5)  is   = -13.


Regarding  ,  it is easy, too :

     =  =  = 6.25.


So, the constant term of the polynomial (5)  is   = -6.25.


Regarding  , it is slightly more long way :

     = 1.5  of (4)  implies (squaring both sides)

    2.25 =  = 

         =  +  = substituting the known values from (4) = 

         =  + 2*(-2.5)*4,

which implies

     = 2.25 + 20 = 22.25.


Thus we know all three coefficients of the polynomial (5)

     = -13  at  x^2;          

     = 22.25 at x;   and 

     = -6.25 as the constant term.   


Answer.  The polynomial equation under the question is   = 0.

Problem 4

Use this identify

tan4Q = %284tanQ-4%28tanQ%29%5E3%29%2F%281-6%28tanQ%29%5E2%2B%28tanQ%29%5E4%29

to find the polynomial of least degree that has zeroes %28tan%28pi%2F24%29%29%5E2

.

Solution

So, according to the problem formulation, I should accept the given identity "as is", and based on it,
to construct the polynomial.

Since this problem is designed/intendent for advanced students, I will only show the major idea and the major steps
without going in details.

Let me start noticing that of four numbers %28tan%28pi%2F24%29%29%5E2

,
the first is equal to the third, while the second is equal to the fourth.

So, I will construct the polynomial which has two zeroes as the first and the second numbers of the four numbers listed:
then the third and the fourth numbers will be the roots of this polynomial automatically . . .

Take Q = .  Then  4Q =   and  tan(4Q) =  = .


According to the given identity, I have then


     =   or

     = .    (2)


Next, in (2), I will replace (tan(Q))^2  by x everywhere (at each appearance).

I will get then


     = .    (3)


Now, for   there is the formula (the known expression)  

          =   (see the link  https://brainly.in/question/9263846 )

and/or   

          =   (see the link  https://mathworld.wolfram.com/TrigonometryAnglesPi24.html)


Replacing    in  formula  (3)  by any of these constant expressions (they are equal (!)),

I get the polynomial of the degree 2


     = ,           (4)


whose two roots are    and  .


Probably, it is more correctly in English to say "a polynomial P(x)", because every product of such a polynomial by a constant
satisfies the problem's conditions, too.