Question 1027752: Please help me with these problems, I have never understood this and I am really trying to learn this. Thanks
Sketch a graph of each of the following. If not possible, write 'not possible' and explain why it is not possible.
a. A polynomial function, degree 3, with zeros at -2,2,and 4.
b. A polynomial function, degree 3, with zeros at -2,0,2 and 4.
c. A polynomial function, degree 3, with only two real zeros.
d. A polynomial function, degree 3 with only one real zero.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The Fundamental Theorem of Algebra: every non-zero, single-variable, degree  polynomial with complex coefficients has, counted with multiplicity, exactly roots.
The Complex Conjugate Root Theorem: if P is a polynomial in one variable with real coefficients, and a + bi is a root of P with a and b real numbers, then its complex conjugate a − bi is also a root of P.
A corollary to the FTA is that if  is a zero of a single-variable polynomial with real coefficients, then ) is a factor of the polynomial.
Assuming real coefficients, then a representation of a third-degree polynomial with zeros -2, 2, and 4 would be:
\ =\ a(x\ +\ 2)(x\ -\ 2)(x\ -\ 4))
Which has  -intercepts of ) , ) , and ) , a  -intercept that is the additive inverse of the product of the zeros and the common coefficient  , in this case  , and opposite behavior as decreases without bound from the behavior as increases without bound.
Example for  (magenta curve) and  (blue curve).
b) is clearly impossible because a polynomial function with four zeros, must by the FTA, be a 4th degree polynomial.
c) is impossible IF you are restricted to real coefficients. The Complex Conjugate Theorem demands that complex roots of polynomial functions with real coefficients come in pairs, and the FTA demands exactly 3 roots. Therefore, any 3rd degree polynomial function with real coeficients must have either 3 real roots or 1 real root and a conjugate pair of complex roots. On the other hand, in the highly unlikely event that your instructor is having you consider complex polynomials, then c) is possible but well beyond my capability to illustrate it.
d) Try  ,  , and 
I'll leave verifying the expansion as an exercise for you, but the polynomial comes out to be:
\ =\ x^3\ +\ x\ +\ 10)
John
My calculator said it, I believe it, that settles it
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