7.4 and 7.5 Solving and Zeros of Polynomials

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Presentation transcript:

7.4 and 7.5 Solving and Zeros of Polynomials Day 1

As with some quadratic equations, factoring a polynomial equation is one way to find its real roots. Recall the Zero Product Property. You can find the roots, or solutions, of the polynomial equation P(x) = 0 by setting each factor equal to 0 and solving for x.

Fundamental Theorem of Algebra Every polynomial function with degree n greater than or equal to 1 has exactly n complex zeros, including multiplicities

The polynomial 3x5 + 18x4 + 27x3 = 0 has two multiple roots, 0 and –3 The polynomial 3x5 + 18x4 + 27x3 = 0 has two multiple roots, 0 and –3. The root 0 is a factor three times because 3x3 = 0. The multiplicity of root r is the number of times that x – r is a factor of P(x). When a real root has even multiplicity, the graph of y = P(x) touches the x-axis but does not cross it. When a real root has odd multiplicity greater than 1, the graph “bends” as it crosses the x-axis.

You cannot always determine the multiplicity of a root from a graph You cannot always determine the multiplicity of a root from a graph. It is easiest to determine multiplicity when the polynomial is in factored form.

Example 1A: Using Factoring to Solve Polynomial Equations Solve the polynomial equation by factoring. 4x6 + 4x5 – 24x4 = 0 4x4(x2 + x – 6) = 0 Factor out the GCF, 4x4. 4x4(x + 3)(x – 2) = 0 Factor the quadratic. 4x4 = 0 or (x + 3) = 0 or (x – 2) = 0 Set each factor equal to 0. x = 0, x = –3, x = 2 Solve for x. The roots are 0, –3, and 2.

Example 1A Continued Check Use a graph. The roots appear to be located at x = 0, x = –3, and x = 2. 

Example 1B: Using Factoring to Solve Polynomial Equations Solve the polynomial equation by factoring. x4 + 25 = 26x2 x4 – 26 x2 + 25 = 0 Set the equation equal to 0. Factor the trinomial in quadratic form. (x2 – 25)(x2 – 1) = 0 (x – 5)(x + 5)(x – 1)(x + 1) Factor the difference of two squares. x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0 x = 5, x = –5, x = 1 or x = –1 Solve for x. The roots are 5, –5, 1, and –1.

Example 2A: Identifying Multiplicity Identify the roots of each equation. State the multiplicity of each root. x3 + 6x2 + 12x + 8 = 0 Since this is difficult to factor, use a graph. A calculator graph shows a bend near (–2, 0). 

Check to see if that is true! The root –2 has a multiplicity of 3. Therefore x + 2 is a factor three times. Check to see if that is true! (x + 2)(x + 2)(x + 2)= x3 + 6x2 + 12x + 8 

Example 2B: Identifying Multiplicity Identify the roots of each equation. State the multiplicity of each root. x4 + 8x3 + 18x2 – 27 = 0 A calculator graph shows a bend near (–3, 0) and crosses at (1, 0). The root 1 has a multiplicity of 1. The root –3 has a multiplicity of 3. Therefore (x – 1) is a factor once, and (x + 3) is a factor three times. Check: (x – 1)(x + 3)(x + 3)(x + 3)=x4 + 8x3 + 18x2 – 27 

Location Principle If P is a polynomial function and P(x1) and P(x2) have opposite signs, then there is a real number r between x1 and x2 that is a zero of P, that is, P(r) = 0

Lesson Quiz Identify the roots of each equation. State the multiplicity of each root. 0 and 2 each with multiplicity 2 1. 5x4 – 20x3 + 20x2 = 0 2. x3 – 12x2 + 48x – 64 = 0 4 with multiplicity 3 –3, 3, 1 3. x3 + 9 = x2 + 9x

Homework 7.4 ws