Z EROS OF P OLYNOMIAL F UNCTIONS

The Fundamental Theorem of Algebra The f(x) is a polynomial of degree n, where n > 0, then f has at least one zero in the complex number system. That is, in the complex number system, every nth-degree polynomial function has precisely n zeros. Linear Factorization Theorem One obtains the linear factorization theorem through the use of the fundamental theorem of Algebra and the equivalence of zeros and factors. If f(x) is a polynomial of degree n, where n > 0, then f has precisely n linear factors. When are complete numbers

Z EROS OF P OLYNOMIALS a. The first-degree polynomial f(x)=x-2 has exactly one zero: x=2. b. Counting multiplicity, the second-degree polynomial function Has exactly two zeros: x=3 and x=3. (This is considered a repeated zero) c. The third-degree polynomial function Has exactly three zeros: x=0, x=2i, and x=-2i.

R ATIONAL Z ERO T EST The Rational Zero Test relates the possible rational zeros of a polynomial (having integer coefficients) to the leading coefficient and to the constant terms of the polynomial. If the polynomial has integer coefficients, every rational zero of f has the form. Where p and q have no common factors other than 1, and P = a factor of the constant term Q = a factor of the leading coefficient

R ATIONAL Z ERO T EST WITH L EADING C OEFFICIENT OF 1 Because the leading coefficient is 1, the possible rational zeros are the factors of the constant term Possible rational zeros: By applying synthetic division successfully, you can determine that x=-1 and x=2 are the only rational zeros So f(x) factors as Because the factor produces no real zeros, you can conclude that x=-1 and x=2 are the only real zeros of f. - 0 is remainder, so x = -1 is a zero- 0 is remainder, so x = 2 is a zero

C ONJUGATE P AIRS Complex Zeros Occur in Conjugate Pairs Let f(x) be a polynomial function that has real coefficients. If a+bi, where b 0, is a zero of the function, the conjugate a-bi is also a zero of the function. This result is true only if the polynomial function has real coefficients.

F INDING THE Z ERO OF A P OLYNOMIAL F UNCTION The possible rational zeros are. Synthetic division produces the following: So, you have You can factor as, and by factoring as You obtain Which gives the following five zeros of f. x=1, x=1, x=-2, x=2i, and x=-2i The real zeros are the only ones that appear as x-intercepts. X=1 is a repeated zero - 1 is a zero - -2 is a zero

D ESCARTES ’ R ULE OF S IGNS Let be a polynomial with real coefficients and. The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by any even integer. The number of negative real zeros of f is either equal to the number of variations in sign of f(-x) or less than that number by an even integer. A variation in sign means that two consecutive coefficients have opposite signs. When using the Descartes’ Rule of Signs, a zero of multiplicity k should be counted as k zeros. For instance, the polynomial has two variations in sign, and so has either two positive of no positive real zeros. Because You can see that the two positive real zeros are x=1 and x=1

U SING D ESCARTES ’ R ULE OF S IGNS Describe the possible real zeros of The original polynomial has three variations in sign. The polynomial has no variation in sign. So, from Descartes’ Rule of Signs, the polynomial has either three positive real zeros or one positive real zero, and no negative real zeros. The function only has one real zero(it is a positive number, near x=1). + to - - to +

U PPER AND L OWER B OUND R ULES Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x-c, using synthetic division. Is c > 0 and each number in the last row is either positive of zero, c is an upper bound for the real zeros. If c < 0 and the number in the last row are alternately positive and negative (zero entries count as positive or negative), c is lower bound for the real zeros of f

F INDING THE Z EROS OF A P OLYNOMIAL F UNCTION The possible real Zeros are as follows The original polynomial f(x) has three variations in sign. The polynomial has no variations in sign As a result of these two findings, you can apply Descartes’ rule of Signs to conclude that there are three positive real zeros or one positive real zero, and no negative zeros. Trying x = 1 produces the following So x = 1 is not a zero, but because the last row has all positive entries, you know that x = 1 is an upper bound for the real zeros. So, you can restrict the search to zeros between 0 and 1. by trial and error, you can determine that is a zero. So, Because has no real zeros, it follows that is the only real zero. Factors of 2 Factors of 6