Question

Find all rational polynomials p(x) = x3 + ax2 + bx + c such that a, b, c are

Answer 1

Answer:

• p(x) = x³ +0x² +0x +0
• p(x) = x³ +x² -x -1
• p(x) = x³ -x² -2x

Step-by-step explanation:

If the roots of p(x) are a, b, c, then it factors as ...

  p(x) = (x -a)(x -b)(x -c)

and the expanded form is ...

  p(x) = x^3 -(a+b+c)x^2 +(ac +b(a+c))x -abc

The requirement that the coefficients match the roots means we have three equations in three unknowns:

  a = -(a+b+c)

  b = ac +b(a+c)

  c = -abc

The attachment shows the solutions to these equations. The solutions are found by making the substitutions ...

• b = x
• c = y
• a = -(x+y)/2

This leaves us with two equations in two unknowns that can be graphed on a Cartesian plane. The graph of the second equation is a circle:

  x(1 -a -y) -ay = 0   ⇒   (x +1)^2 +y^2 = 1

The graph of the first equation is the union of the line y=0 and two hyperbolas. The points of intersection between this graph and the circle are ...

  (x, y) = (-2, 0), (-1, -1), (0, 0)   ⇒   (a, b, c) = (1, -2, 0), (1, -1, -1), (0, 0, 0)

The three polynomials that correspond to these values are shown above and in the attachment.

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Additional comment

There is an irrational fourth solution to the requirement that the roots match the coefficients.