1. The degree three polynomial f(x) with real coefficients and leading coefficient 1, has -3 and + 4i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients.

3. If a piece of real estate purchased for $50,000 in 1998 appreciates at the rate of 5% per year, then its value t years after the purchase will be f (t) = 50,000(1.05t ) . According to this model, by how much will the value of this piece of property increase between the years 2007 and 2008?

4. Solve loga (7x +1) = loga (4x +16).

5. Find the domain of f (x) = 7 + 3x + 21 , and express it using interval notation.

6. If points A, B, and C lie on a coordinate line and points A and B have coordinates 15 and 7 respectively, then which of the possible coordinates for point C satisfy(ies) d(A, C) < d(B, C)?

Solution Summary

Polynomials with real and complex solutions are investigated. The solution is detailed and well presented.

... F(x)=x^3-8x^2+29x-52 Use the complex zeros to write f ... see the attached file for the complete solution) 1. Form a polynomial f(x) with real coefficients having ...

...Polynomial. By:- Thokchom Sarojkumar Sinha. Let be a field of real numbers. Prove that is a field isomorphic to the field of complex numbers. Solution:- Let be ...

... answer with the help of one of the identities of factorization of polynomials. ... Is the result real, complex, or both? ... Neat and step-wise solutions are provided. ...

... P into linear and irreducible quadratic factors with real coefficients. ... 12 = (x-3)(x^2+4). The polynomial (x^2 ... and x^2+16 is irreducible over reals, because its ...

... Any polynomial equation has n complex roots, in general, in ... axis is set up where the real number is ... The complex number system is defined and supplemented with ...

... All variables represent positive real numbers. 14. Solution. ... So, there are two complex solutions: x=4+i and x=4-i. ... Evaluate each polynomial as indicated. ...

... (If an answer does not exist, enter DNE.) real zero x ... so, x=5 is a root of this polynomial. ... x^2-4x+5 = 0 D=16 - 4*1*5 = -4, we get two complex solutions: x = (4 ...

... problems: first using intermediate value theorem to show the existance of zeros, then finding zeros (both real and complex) for specific polynomials. ...

... some fraction of their time solving polynomial equations. ... are of interest, whether they are real or complex... The techniques in complex analysis are just one more ...

... all values of x which do not cause our polynomial denominator to ... x2 + 1), where x is considered to be a real number (imaginary or complex numbers not ...