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Warm Up Factor completely.

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.

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  1. 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

  2. Objectives Use the Rational Root Theorem to solve polynomial equations.

  3. 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.

  4. 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.

  5. 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?

  6. 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?

  7. 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?

  8. 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

  9. Degree = # zeros This is real and imaginary combined

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. 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.

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

  17. 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.

  18. 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

  19. 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.

  20. 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.

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