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2.5 Zeros of Polynomial Functions Roots & Zeros of Polynomials How the roots, solutions, zeros, x-intercepts and factors of a polynomial function are related. Polynomials A Polynomial Expression can be a monomial or a sum of monomials. The Polynomial Expressions that we are discussing today are in terms of one variable. In a Polynomial Equation, two polynomials are set equal to each other. Factoring Polynomials Terms are Factors of a Polynomial if, when they are multiplied, they equal that polynomial: x 2 x 15 ( x 3)( x 5) 2 (x - 3) and (x + 5) are Factors of the polynomial x 2 x 15 2 Since Factors are a Product... …and the only way a product can equal zero is if one or more of the factors are zero… …then the only way the polynomial can equal zero is if one or more of the factors are zero. Solving a Polynomial Equation Rearrange the terms to have zero on one side: x 2 x 15 x 2 x 15 0 2 Factor: 2 ( x 5)( x 3) 0 Set each factor equal to zero and solve: ( x 5) 0 and x 5 ( x 3) 0 x 3 The only way that x2 +2x - 15 can = 0 is if x = -5 or x = 3 Solutions/Roots a Polynomial Setting the Factors of a Polynomial Expression equal to zero gives the Solutions to the Equation when the polynomial expression equals zero. Another name for the Solutions of a Polynomial is the Roots of a Polynomial ! Zeros of a Polynomial Function A Polynomial Function is usually written in function notation or in terms of x and y. 2 f (x) x 2x 15 or 2 y x 2x 15 The Zeros of a Polynomial Function are the solutions to the equation you get when you set the polynomial equal to zero. Zeros of a Polynomial Function The Zeros of a Polynomial Function ARE the Solutions to the Polynomial Equation when the polynomial equals zero. Graph of a Polynomial Function Here is the graph of our polynomial function: 2 y x 2x 15 The Zeros of the Polynomial are the values of x when the polynomial equals zero. In other words, the Zeros are the x-values where y equals zero. x-Intercepts of a Polynomial The points where y = 0 are called the x-intercepts of the graph. The x-intercepts for our graph are the points... (-5, 0) and y x 2 2x 15 (3, 0) x-Intercepts of a Polynomial When the Factors of a Polynomial Expression are set equal to zero, we get the Solutions or Roots of the Polynomial Equation. The Solutions/Roots of the Polynomial Equation are the x-coordinates for the x-Intercepts of the Polynomial Graph! Factors, Roots, Zeros For our Polynomial Function: 2 y x 2x 15 The Factors are: (x + 5) & (x - 3) The Roots/Solutions are: x = -5 and 3 The Zeros are at: (-5, 0) and (3, 0) Roots & Zeros of Polynomials II Finding the Roots/Zeros of Polynomials: • The Fundamental Theorem of Algebra • Descartes’ Rule of Signs • The Complex Conjugate Theorem Fundamental Theorem 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? 3 y x 4x 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. 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! Find Roots/Zeros of a Polynomial We can find the Roots or Zeros of a polynomial by setting the polynomial equal to 0 and factoring. Some are easier to factor than others! 3 f (x) x 4x 2 x(x 4) x(x 2)(x 2) The roots are: 0, -2, 2 Find Roots/Zeros of a Polynomial If we cannot factor the polynomial, but know one of the roots, we can divide that factor into the polynomial. The resulting polynomial has a lower degree and might be easier to factor or solve with the quadratic formula. f (x) x 5x 2x 10 3 2 one root is x 5 (x - 5) is a factor We can solve the resulting polynomial to get the other 2 roots: x 2, 2 2 x 2 3 2 x 5 x 5x 2x 10 x 3 5x 2 2x 10 2x 10 0 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, 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 If one root is 4 - i. f (x) x 3 5x 2 7x 51 Because of the Complex Conjugate Theorem, we know that another root must be 4 + i. Can the third root also be imaginary? Consider… Descartes: # of Pos. Real Roots = 2 or 0 Descartes: # of Neg. Real Roots = 1 Example (con’t) 3 2 Ex: Find all the roots of f (x) x 5x 7x 51 If one root is 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)]. Multiply these factors: x 4 i x 4 i x 2 x 4 i x 4 i 4 i 4 i x 2 4 x xi 4 x xi 16 i 2 x 2 8 x 16 (1) x 2 8 x 17 Example (con’t) 3 2 Ex: Find all the roots of f (x) x 5x 7x 51 If one root is 4 - i. 2 If the product of the two non-real factors is x 8x 17 then the third factor (that gives us the neg. real root) is 2 the quotient of P(x) divided by x 8x 17 : x 3 x 2 8x 17 x 3 5x 2 7x 51 x 3 5x 2 7x 51 0 The third root is x = -3 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! The Rational Zero Theorem The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear somewhere in the list. The Rational Zero Theorem p q p n n-1 If f(x) a x a x … a1x a0 has integer coefficients and n n-1 q (where is reduced) is a rational zero, then p is a factor of the constant term a0 and q is a factor of the leading coefficient an. EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f(x) 15x3 14x2 3x – 2. Solution The constant term is –2 and the leading coefficient is 15. Factors of the constant term, 2 Factors of the leading coefficient, 15 1, 2 1, 3, 5, 15 Possible rational zeros 1, 2, 13 , 23 , Divide 1 and 2 by 1. Divide 1 and 2 by 3. 15 , 52 , Divide 1 and 2 by 5. 1 , 2 15 15 Divide 1 and 2 by 15. There are 16 possible rational zeros. The actual solution set to f(x) 15x3 14x2 3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16 possible solutions. EXAMPLE: Solve: Solving a Polynomial Equation x4 6x2 8x + 24 0. Solution Because we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots. Factors of the constant term, 24 Factors of the leading coefficient, 1 1, 2 3, 4, 6, 8, 12, 24 1 1, 2 3, 4, 6, 8, 12, 24 Possible rational zeros EXAMPLE: Solve: Solving a Polynomial Equation x4 6x2 8x + 24 0. Solution The graph of f(x) x4 6x2 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. 2 1 1 0 6 8 24 2 4 4 24 2 2 12 The 0 zero remainder indicates that 2 is a root of x4 6x2 8x + 24 0. x-intercept: 2 EXAMPLE: Solve: Solving a Polynomial Equation x4 6x2 8x + 24 0. Solution Now we can rewrite the given equation in factored form. x4 6x2 8x + 24 0 (x – 2)(x3 2x2 2x 12) 0 x–20 or x3 2x2 2x 12 0 This is the given equation. This is the result obtained from the synthetic division. Set each factor equal to zero. Now we must continue by factoring x3 + 2x2 - 2x - 12 = 0 EXAMPLE: Solve: Solving a Polynomial Equation x4 6x2 8x + 24 0. Solution Because the graph turns around at 2, this means that 2 is a root of even multiplicity. Thus, 2 must also be a root of x3 2x2 2x 12 = 0. These are the coefficients of x3 2x2 2x 12 = 0. 2 1 1 x-intercept: 2 2 2 4 2 12 8 12 6 0 The zero remainder indicates that 2 is a root of x3 2x2 2x 12 = 0. EXAMPLE: Solve: Solution Solving a Polynomial Equation x4 6x2 8x + 24 0. Now we can solve the original equation as follows. x4 6x2 8x + 24 0 (x – 2)(x3 2x2 2x 12) 0 (x – 2)(x – 2)(x2 4x 6) 0 x – 2 0 or x – 2 0 or x2 4x 6 0 x2 x2 x2 4x 6 0 This is the given equation. This was obtained from the first synthetic division. This was obtained from the second synthetic division. Set each factor equal to zero. Solve. EXAMPLE: Solve: Solution Solving a Polynomial Equation x4 6x2 8x + 24 0. We can use the quadratic formula to solve x2 4x 6 0. b b2 4ac x 2a 4 42 4 1 6 2 1 4 8 2 4 2i 2 2 2 i 2 We use the quadratic formula because x2 4x 6 0 cannot be factored. Let a 1, b 4, and c 6. Multiply and subtract under the radical. 8 4(2)(1) 2i 2 Simplify. The solution set of the original equation is {2, 2 i 2, 2 i 2}.