Zeros of Polynomials 2.5

Properties of Polynomial Equations A polynomial of degree n, has n roots (counting multiple roots separately) If a + bi is a root, then a – bi is also a root. Complex roots occur in conjugate pairs!

Descartes’ Rule of Signs

Find Numbers of Positive and Negative Zeros State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x) = –x6 + 4x3 – 2x2 – x – 1. Since p(x) has degree 6, it has 6 zeros. However, some of them may be imaginary. Use Descartes’ Rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x). p(x) = –x6 + 4x3 – 2x2 – x – 1 yes – to + yes + to – no – to – no – to – 2 or 0 positive real zeros

Find Numbers of Positive and Negative Zeros Since there are two sign changes, there are 2 or 0 positive real zeros. Find p(–x) and count the number of sign changes for its coefficients. p(–x) = –(–x)6 + 4(–x)3 – 2(–x)2 – (–x) – 1 no – to – no – to – yes – to + yes + to – Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations.

Find Numbers of Positive and Negative Zeros Answer:

Find all of the zeros of f(x) = x3 – x2 + 2x + 4. Since f(x) has degree of 3, the function has three zeros. To determine the possible number and type of real zeros, examine the number of sign changes in f(x) and f(–x). f(x) = x3 – x2 + 2x + 4 2 or 0 positive real zeros yes yes no f(–x) = –x3 – x2 – 2x + 4 1 negative real zero no no yes

Rational Zero Theorem

Identify Possible Zeros A. List all of the possible rational zeros of f(x) = 3x4 – x3 + 4. Answer:

B. List all of the possible rational zeros of f(x) = x3 + 3x + 24 B. List all of the possible rational zeros of f(x) = x3 + 3x + 24. <YU TRY> A. B. C. D.

Use the Factor Theorem Show that x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the remaining factors of the polynomial. The binomial x – 3 is a factor of the polynomial if 3 is a zero of the related polynomial function. Use the factor theorem and synthetic division. 1 4 –15 –18 3 21 18 1 7 6 0

x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial. Use the Factor Theorem Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored. x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial. Answer: So, x3 + 4x2 – 15x – 18 = (x – 3)(x + 6)(x + 1).

Example Repeated zero at x = 2

Upper and Lower Bounds Suppose f(x) is divided by x – c using synthetic division If c > 0 and each number in the last row is either positive or zero, then c is an upper bound for real zeros If c < 0 and each number in the last row are alternatively positive or negative (zero counts as both), then c is a lower bound

The degree is 4, so there are 4 roots! Possible roots are ±1, ±13 The degree is 4, so there are 4 roots! Use Synthetic Division to find the roots Use Quadratic Formula to find remaining solutions Multiplicity of 2

f(x) = x3 – x2 + 2x + 4. There are 3 zeros, we know there are 2 or 0 positive and 1 negative Possible roots are ±1, ±2, ±4 -4, -2, -1, 1, 2, 4 We need either 2 or 0 positive, so 4 cannot be a zero Use Synthetic Division to find the roots -1 works, so it is our 1 negative zero. The other two have to be imaginary. Use Quadratic Formula to find remaining solutions

There are 4 zeros, We know there are 1 positive and 3 or 1 negative Possible roots are ±1, ±3, ±5, ±15 -15, -5, -3 -1, 1, 3, 5, 15 3 is a solution. This is our 1 positive zero All positive so 5 is an upper bound Now try negatives -1 is our 1 negative. Use the quadratic to find the remaining 2 zeros Alternates between positive and negative so -3 is a lower bound

Linear Factorization Theorem Find a fourth degree polynomial with real coefficients that has 2, -2 and i as zeros and such that f(3) = -150 Foil Foil Substitute f(3) = -150 Solve for a Substitute back in equation

Linear Factorization Theorem An nth – degree polynomial can be expressed as the product of a nonzero constant and n linear factors