Finding Real Roots of Polynomial Equations 3-5

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Presentation transcript:

Finding Real Roots of Polynomial Equations 3-5 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

Warm Up Factor completely. 1. 2y3 + 4y2 – 30y 2y(y – 3)(y + 5) 2. 3x4 – 6x2 – 24 3(x – 2)(x + 2)(x2 + 2) Solve each equation. 3. x2 – 9 = 0 x = – 3, 3 4. x3 + 3x2 – 4x = 0 x = –4, 0, 1

Objectives Identify the multiplicity of roots. Use the Rational Root Theorem and the irrational Root Theorem to solve polynomial equations.

Vocabulary multiplicity

In Lesson 6-4, you used several methods for factoring polynomials In Lesson 6-4, you used several methods for factoring polynomials. As with some quadratic equations, factoring a polynomial equation is one way to find its real roots. Recall the Zero Product Property from Lesson 5-3. 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.

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 x = 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 x = 5, –5, 1, and –1.

Check It Out! Example 1a Solve the polynomial equation by factoring. 2x6 – 10x5 – 12x4 = 0

Check It Out! Example 1b Solve the polynomial equation by factoring. x3 – 2x2 – 25x = –50

Sometimes a polynomial equation has a factor that appears more than once. This creates a multiple root. Consider 3x5 + 18x4 + 27x3 = 0 , which factors to This equation has two multiple roots, x = 0 and –3. For example, the root x = 0 is a factor three times. The multiplicity of root x = 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 2A: Identifying Multiplicity Identify the roots of each equation. State the multiplicity of each root. x3 + 6x2 + 12x + 8 = 0 x3 + 6x2 + 12x + 8 = (x + 2)(x + 2)(x + 2) x + 2 is a factor three times. The root x = –2 has a multiplicity of 3. Check Use a graph. A calculator graph shows a bend near (–2, 0). 

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

Check It Out! Example 2a Identify the roots of each equation. State the multiplicity of each root. x4 – 8x3 + 24x2 – 32x + 16 = 0

Check It Out! Example 2b Identify the roots of each equation. State the multiplicity of each root. 2x6 – 22x5 + 48x4 + 72x3 = 0

Not all polynomials are factorable, but the Rational Root Theorem can help you find all possible rational roots of a polynomial equation.

Example 3: Marketing Application The design of a box specifies that its length is 4 inches greater than its width. The height is 1 inch less than the width. The volume of the box is 12 cubic inches. What is the width of the box? Step 1 Write an equation to model the volume of the box. Let x represent the width in inches. Then the length is x + 4, and the height is x – 1. x(x + 4)(x – 1) = 12 V = lwh. x3 + 3x2 – 4x = 12 Multiply the left side. x3 + 3x2 – 4x – 12 = 0 Set the equation equal to 0.

Example 3 Continued Step 2 Use the Rational Root Theorem to identify all possible rational roots. Factors of –12: ±1, ±2, ±3, ±4, ±6, ±12

Example 3 Continued Step 4 Factor the polynomial. The synthetic substitution of 2 results in a remainder of 0, so 2 is a root and the polynomial in factored form is (x – 2)(x2 + 5x + 6). (x – 2)(x2 + 5x + 6) = 0 Set the equation equal to 0. (x – 2)(x + 2)(x + 3) = 0 Factor x2 + 5x + 6. Set each factor equal to 0, and solve. x = 2, x = –2, or x = –3 The width must be positive, so the width should be 2 inches.

Polynomial equations may also have irrational roots.

The Irrational Root Theorem say that irrational roots come in conjugate pairs. For example, if you know that 1 + is a root of x3 – x2 – 3x – 1 = 0, then you know that 1 – is also a root. Recall that the real numbers are made up of the rational and irrational numbers. You can use the Rational Root Theorem and the Irrational Root Theorem together to find all of the real roots of P(x) = 0.

Example 4: Identifying All of the Real Roots of a Polynomial Equation Identify all the real roots of 2x3 – 9x2 + 2 = 0. Step 1 Use the Rational Root Theorem to identify possible rational roots. ±1, ±2 = ±1, ±2, ± . 1 2 p = 2 and q = 2 Step 2 Graph y = 2x3 – 9x2 + 2 to find the x-intercepts. The x-intercepts are located at or near x = –0.449, 0.5, and 4.449. The x-intercepts –0.449 and 4.449 do not correspond to any of the possible rational roots.

Step 3 Test the possible rational root . Example 4 Continued 1 2 Step 3 Test the possible rational root . 1 2 Test . The remainder is 0, so (x – ) is a factor. 1 2 2 –9 0 2 1 –4 –2 2 –8 –4 The polynomial factors into (x – )(2x2 – 8x – 4). 1 2 Step 4 Solve 2x2 – 8x – 4 = 0 to find the remaining roots. 2(x2 – 4x – 2) = 0 Factor out the GCF, 2 Use the quadratic formula to identify the irrational roots. 4± 16+8 2 6 2 x = = ±

( ) Example 4 Continued The fully factored equation is 1 2 x – x – 2 + 6 x – 2 – 6 = 0 2 æ é ç ë è ö ÷ ø ù û The roots are , , and . 1 2 2 6 + -

Check It Out! Example 4 Identify all the real roots of 2x3 – 3x2 –10x – 4 = 0.

Check It Out! Example 4 Continued

Check It Out! Example 4 Continued The fully factored equation is ( ) 1 2 x + x – 1+ 5 x – 1– 5 2 æ ù ç û è ö ÷ ø é ë The roots are – , , and . 1 2 1 5 + -

Homework pp. 186-188 15-22, 24-27, 29-32, 40-47