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Warm Up Factor completely. 1. 2 y 3 + 4 y 2 – 30 y. 2 y ( y – 3)( y + 5) . 2. 3 x 4 – 6 x 2 – 24. 3( x – 2)( x + 2)( x 2 + 2). Solve each equation. 3. x 2 – 9 = 0. x = – 3, 3. 4. x 3 + 3 x 2 – 4 x = 0. x = –4, 0, 1. Objectives.
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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 Use the Rational Root Theorem to solve polynomial equations.
Turning Points The vertexof a parabola marks the turning point of the graph of a quadratic function. • A turning point is a point at which the function values “turn” from increasing to decreasing or vice versa.
Turning Points The y-coordinate of a parabola’s turning point marks the absolute minimum or maximum of the function since there are no other points above or below it.
2nd degree polynomial For a polynomial function of degree n, how many tuning points will it have? n = 2 What is the domain of each function?
3rd degree polynomial For a polynomial function of degree n, how many tuning points will it have? n = 3 What is the domain of each function?
4th degree polynomial For a polynomial function of degree n, how many tuning points will it have? n = 4 What is the domain of each function?
Turning Points Let f be a polynomial function of degree n. • Then f has at mostn – 1 turning points • If f has n distinct real roots, then f has exactlyn – 1 turning points
Degree = # zeros This is real and imaginary combined
Solve the polynomial equation. 4x6 + 4x5 – 24x4 = 0 Step 1: Factor out any common factors 4x4(x2 + x – 6) = 0 Factor out the GCF, 4x4. Step 2: Factor the quadratic 4x4(x+ 3)(x – 2) = 0 Factor the quadratic. Step 3: Solve for x Set each factor equal to 0. 4x4 = 0 or (x+ 3) = 0 or (x – 2) = 0 x = 0, x = –3, x = 2 Solve for x. The roots are 0, –3, and 2.
x4 + 25 = 26x2 x4 – 26 x2 + 25 = 0 Rewrite in standard form. (x2 – 25)(x2 – 1) = 0 Factor the trinomial in quadratic form. Factor the difference of two squares. (x– 5)(x + 5)(x – 1)(x + 1) x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0 Solve for x. x = 5, x = –5, x = 1 or x = –1 The roots are 5, –5, 1, and –1.
Solve the polynomial equation by factoring. 2x6 – 10x5 – 12x4 = 0 2x4(x2 – 5x – 6) = 0 Factor out the GCF, 2x4. 2x4(x– 6)(x + 1) = 0 Factor the quadratic. 2x4 = 0 or (x– 6) = 0 or (x + 1) = 0 Set each factor equal to 0. x = 0, x = 6, x = –1 Solve for x. The roots are 0, 6, and –1.
Solve the polynomial equation by factoring. x3 – 2x2 – 25x = –50 Set the equation equal to 0. x3 – 2x2 – 25x + 50= 0 Group. (x3 – 2x2 )-(25x + 50)= 0 Factor. x2(x – 2) -25(x – 2)= 0 (x2-25)(x – 2)= 0 (x + 5)(x – 5)(x – 2) = 0 x = –5, x = 2, or x = 5 Solve for x. The roots are –5, 2, and 5.
The multiplicity refers to repeated factors. 2x4(x– 6)(x + 1) = 0 Notice that the x term is repeated 4 times That means 4 of my 6 factors are 0 Even multiplicity means the graph sits on that number Odd multiplicity means it bends at that number.
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.
Not all polynomials are factorable, but the Rational Root Theorem can help you find all possible rational roots of a polynomial equation.
A shipping crate must hold 12 cubic feet. The width should be 1 foot less than the length, and the height should be 4 feet greater than the length. What should the length of the crate be? Step 1 Write an equation to model the volume of the box. V = lwh Let x represent the length. Then the width is x – 1, and the height is x + 4. x(x – 1)(x + 4) = 12 V = lwh. x3 + 3x2 – 4x = 12 Multiply the left side. x3 + 3x2 – 4x– 12 = 0 Set the equation equal to 0.
p q x3 + 3x2 – 4x– 12 = 0 Step 2Use the Rational Root Theorem to identify all possible rational roots. Factors of –12: ±1, ±2, ±3, ±4, ±6, ±12 Step 3Test the possible roots to find one that is actually a root. The width must be positive, so try only positive rational roots. Use synthetic substitution
Step 4 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 length must be positive, so the length should be 2 feet.
Lesson Quiz Solve by factoring. 1. x3 + 9 = x2 + 9x –3, 3, 1 Identify the roots of each equation. State the multiplicity of each root. 0 and 2 each with multiplicity 2 2. 5x4– 20x3 + 20x2 = 0 3. x3 – 12x2+48x – 64 = 0 4 with multiplicity 3 4. A box is 2 inches longer than its height. The width is 2 inches less than the height. The volume of the box is 15 cubic inches. How tall is the box? 3 in.