Roots & Zeros of Polynomials How the roots, solutions, zeros, x-intercepts and factors of a polynomial function are related.
A number a is a zero or root of a function y = f (x) if and only if f (a) = 0. If y = f(x) is a polynomial function and a is a number then the following statements are equivalent. 1. x = a is a zero of f. 2. x = a is a solution of the polynomial equation f(x) = 0. 3. (x a) is a factor of the polynomial f(x). 4. (a, 0) is an x-intercept of the graph of y = f(x). 2
Example: Find all the real zeros of f (x) = x 4 x 3 2x 2. Factor completely: f (x) = x 4 x 3 2x 2 = x 2 (x + 1)(x 2). The real zeros are x = 1, x = 0, and x = 2. These correspond to the x-intercepts. 2 2 y ( 1, 0) (0, 0) (2, 0) x 3 f(x) = x 4 x 3 2x 2
Solving a Polynomial Equation Rearrange the terms to have zero on one side: x 2 + 2x = 15 x 2 + 2x 15 = 0 Factor: (x + 5)(x 3) = 0 Set each factor equal to zero and solve: (x + 5) = 0 and (x 3) = 0 x = 5 x = 3 The only way that x 2 +2x - 15 can = 0 is if x = -5 or x = 3
Here is the graph of our polynomial function: The x-intercepts for our graph are the points... y = x 2 + 2x 15 (-5, 0) and (3, 0)
Factors, Roots, Zeros, x-intercepts For our Polynomial Function: y = x 2 + 2x 15 The Factors are: (x + 5) & (x - 3) The Zeros/Roots are: x = -5 and 3 The x-intercepts are at: (-5, 0) and (3, 0)
Finding the Roots/Zeros of Polynomials The Fundamental Theorem of Algebra Descartes Rule of Signs The Complex Conjugate Theorem
Fundamental Thm. Of Algebra Every Polynomial Equation with a degree higher than zero has at least one root in the set of Complex Numbers. COROLLARY: A Polynomial Equation of the form P(x) = 0 of degree n with complex coefficients has exactly n roots in the set of Complex Numbers.
Real/Imaginary Roots If a polynomial has n complex roots will its graph have n x-intercepts? In this example, the degree n = 3, and if we factor the polynomial, the roots are x = -2, 0, 2. We can also see from the graph that there are 3 x-intercepts. y = x 3 4x
Real/Imaginary Roots Just because a polynomial has n complex roots doesn t mean that they are all Real! In this example, however, the degree is still n = 3, but there is only one Real x-intercept or root at x = -1, the other 2 roots must have imaginary components. y = x 3 2x 2 + x + 4
Descartes Rule of Signs Arrange the terms of the polynomial P(x) in descending degree: The number of times the coefficients of the terms of P(x) change sign = the number of Positive Real Roots (or less by any even number) The number of times the coefficients of the terms of P(-x) change sign = the number of Negative Real Roots (or less by any even number) In the examples that follow, use Descartes Rule of Signs to predict the number of + and - Real Roots!
EXAMPLE: Using Descartes Rule of Signs Determine the possible number of positive and negative real zeros of f (x) = x 3 + 2x 2 + 5x + 4. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f( x). We obtain this equation by replacing x with x in the given function. f(x) = x 3 + 2x 2 + 5x + 4 This is the given polynomial function. Replace x with x. f( x) = ( x) 3 + 2( x) 2 + 5( x) + 4 = x 3 + 2x 2 5x + 4
EXAMPLE: Using Descartes Rule of Signs Now count the sign changes. f( x) = x 3 + 2x 2 5x + 4 1 2 3 There are three variations in sign. # of negative real zeros of f is either equal to 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 2 = 1 negative real zero.
Find Roots/Zeros of a Polynomial If we cannot factor the polynomial, but know one of the roots, we can use the synthetic substitution. The resulting polynomial (depressed polynomial)has a lower degree and might be easier to factor. f (x) = x 3 5x 2 2x +10 one root is x = 5 We can solve the depressed polynomial x² - 2 to get the other 2 roots: x²-2=0, x²=2, x=± 2 5 1-5 -2 10 5 0-10 1 0-2 0 Don t forget your remainder should be zero So, the zeros to the problem are: x = 2, 2, 3
Complex Conjugates Theorem Roots/Zeros that are not Real are Complex with an Imaginary component. Complex roots with Imaginary components always exist in Conjugate Pairs. If a + bi (b 0) is a zero of a polynomial function with real coefficients, then its conjugate, a - bi, is also a zero of the function.
Find Roots/Zeros of a Polynomial If the known root is imaginary, we can use the Complex Conjugates Theorem. Ex: Find all the roots of f (x) = x 3 5x 2 7x + 51 If one root is 4 - i. Because of the Complex Conjugate Theorem, we know that another root must be 4 + i. If one root is 4 - i, then one factor is [x - (4 - i)], and Another root is 4 + i, & another factor is [x - (4 + i)].
[ x ( 4 i) ]x ( 4 + i) [ ]= x 2 x( 4 + i) x 4 + i ( ) + ( 4 i) ( 4 + i) = x 2 4x xi 4x + xi +16 i 2 = x 2 8x +16 ( 1) = x 2 8x +17 If the product of the two non-real factors is x 2 8x +17 then the third factor (that gives us the neg. real root) is the quotient of P(x) divided by: x +3 x 2 8x +17 x 3 5x 2 7x + 51 ) The third root is x = -3 x 3 5x 2 7x + 51 0
Finding Roots/Zeros of Polynomials We use the Fundamental Thm. Of Algebra, Descartes Rule of Signs and the Complex Conjugate Thm. to predict the nature of the roots of a polynomial. We use skills such as factoring, polynomial division and the quadratic formula to find the zeros/roots of polynomials. In future lessons you will learn other rules and theorems to predict the values of roots so you can solve higher degree polynomials!