EASY QUESTION HURRY!!!! When solving a system of linear equations algebraically, Meredith obtained 0 = 0. How can Meredith inter
2 answers:
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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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<em>Additional comment</em>
There is an irrational fourth solution to the requirement that the roots match the coefficients.
