Question 1131984: Find an equation for f(x), the polynomial of smallest degree with real coefficients such that f(x) breaks through the x-axis at −4, breaks through the x-axis at 5, has complex roots of −3−i and 5+5i and passes through the point (0,−54).
Found 4 solutions by Edwin McCravy, MathLover1, ikleyn, solver91311:
Answer by Edwin McCravy(20066)   (Show Source):
Answer by MathLover1(20850)  (Show Source):
Answer by ikleyn(52926)  (Show Source):
You can put this solution on YOUR website! .
The "solution" by @MathLover1 is not exactly correct, since she mistakenly assumed that -54 is the root.
It is not so. y = -54 is "y-intercept"; it is not "x-intercept".
She may want to edit her post . . .
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Corrolaries to the Fundamental Theorem of Algebra tell us the following things:
1. A polynomial function with complex coefficients of degree  has exactly complex zeros.
2. There is always an even number of complex zeros with non-zero imaginary parts, and these zeros always appear as conjugate pairs. That is if  is a zero of a polynomial function with complex coefficients, then  is also a zero of that polynomial.
Both of these assertions apply to polynomials with strictly real coefficients because real numbers are the subset of the complex numbers where the imaginary part is zero.
We also know that if  is a zero of a polynomial function, then  is a factor of the polynomial. Further, any such polynomial has an additional non-zero constant factor  . For example, the general third-degree polynomial function \ =\ a_1x^3\ +\ a_2x^2\ +\ a_3x\ +\ a_4) has the following factors: (x\ -\ z_2)(x\ -\ z_3)) where 
With the given information, we can identify six roots of the desired polynomial, namely 
Hence, the polynomial of the smallest degree that has these zeros is a polynomial of degree six.
Next, we can identify the seven factors of our desired polynomial:
,\ (x\,-\,5),\ (x\,+\,3\,-\,i),\ (x\,+\,3\,+\,i)\ (x\,-\,5\,-\,5i),\ (x\,-\,5\,+\,5i))
Multiplying the factors in pairs:
(x\ -\ 5)\ =\ x^2\ -\ x\ +\ 20)
(x\,+\,3\,+\,i)\ =\ x^2 +\ 3x\ +\ xi\ +\ 3x\ +\ 9\ +\ 3i\ -\ xi\ -\ 3i\ -\ (-1)\ =\ x^2\ +\ 6x\ +\ 10)
(x\,-\,5\,+\,5i)\ =\ x^2\ -\ 5x\ +\ 5ix\ -\ 5x\ +\ 25\ -\ 25i\ -\ 5ix\ +\ 25i\ -\ (-1)25\ =\ x^2\ -\ 10x\ +\ 50))
So far we have \ =\ \alpha(x^2\ -\ x\ -\ 20)(x^2\ +\ 6x\ +\ 10)(x^2\ -\ 10x\ +\ 50))
But we are given that the point ) is on the graph. Since any point on the graph is of the form \)) we can see that \ =\ -54) , hence:
\ =\ \alpha\((0)^2\ -\ (0)\ -\ 20\)\((0)^2\ +\ 6(0)\ +\ 10\)\((0)^2\ -\ 10(0)\ +\ 50\)\ =\ \alpha(-20)(10)(50)\ =\ -54)
Solving for :
And now we have:
\ =\ 0.0054(x^2\ -\ x\ -\ 20)(x^2\ +\ 6x\ +\ 10)(x^2\ -\ 10x\ +\ 50))
The last step is multiplying the three quadratic trinomials and distributing the lead coefficient to derive the final form of the sixth-degree polynomial function in standard form. I leave this as an exercise for the student.
John
My calculator said it, I believe it, that settles it
|
|
|
