#  Evaluate a polynomial  Direct Substitution  Synthetic Substitution  Polynomial Division  Long Division  Synthetic Division  Remainder Theorem 

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 Evaluate a polynomial  Direct Substitution  Synthetic Substitution  Polynomial Division  Long Division  Synthetic Division  Remainder Theorem  Factor Theorem

 If a polynomial is divided by x-k, then the remainder r = f(k).

 Polynomial f(x) has a factor x – k, if and only if f(k) = 0.  In other words…  If r is 0, then x-k is a factor of f(x)  If x-k is a factor, then k is a zero (root) of f(x).

 3 is a zero of f(x).  f(3) = 0  3 is an x ‑ intercept of the graph of f(x).  ( x ‑ 3) is a factor of f(x).  f(x) divided by (x ‑ 3) has a remainder of 0.  3 is a root of f(x)

 The profit P (in millions of dollars) for a T-shirt manufacturer can be modeled by P = -x 3 + 4x 2 + x where x is the # of t-shirts produced (in millions). Currently the company produces 4 million t-shirts and makes a profit of 4 million. What lesser number of t-shirts could the company produce and still make the same profit?

 If f(x) = a n x n + …+ a 1 x + a o has integer coefficients, then every rational zero of f(x) has the following form:

 Every polynomial of degree n where n>0, has at least one zero, where a zero may be a complex number (a + bi).  Corollary:  Every polynomial of degree n where n>0, has exactly n zeros, including multiplicities.

Complex Conjugates Theorem  If f is a polynomial function with real coefficients, and a + bi is an imaginary zero of f, then a – bi is also a zero. Irrational Conjugates Theorem  Suppose f is a polynomial function with rational coefficients and a and b are rational numbers such that √b is irrational. If a +√b is a zero, then a –√b is also a rational zero of f.

 If f(x) is a polynomial if degree n where n>0, then the equation f(x) = 0 has exactly n solutions provided each solution repeated twice is counted as two solutions, each solution repeated three time is counted as three solutions…

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