Presentation on theme: " Evaluate a polynomial Direct Substitution Synthetic Substitution Polynomial Division Long Division Synthetic Division Remainder Theorem "— Presentation transcript:
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…