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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Do Now: Aim: How do we solve polynomial equations using factoring? Write the expression (x + 1)(x + 2)(x + 3) as a polynomial in standard form. (x + 1)(x + 2)(x + 3) FOIL first two factors (x 2 + 3x + 2)(x + 3) x 3 + 3x 2 + 3x 2 + 9x + 2x + 6 Multiply by distribution of resulting factors Combine like terms x 3 + 6x x + 6 cubic expression is standard form

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Solving Polynomials by Factoring Factor: 2x x x GCF Factor trinomial 2x(x 2 + 5x + 6x) 2x(x + 3)(x + 2) Solve: 2x x x = 0 2x(x 2 + 5x + 6x) = 0 2x(x + 3)(x + 2) = 0 Set factors equal to zero (Zero Product Property) 2x = 0 (x + 3) = 0 (x + 2) = 0 x = 0, -2, -3

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Graphical Solutions Solve: 2x x x = 0 2x(x + 3)(x + 2) = 0 x = 0, -2, -3 ax 3 + bx 2 + cx + d = y Cubic equation in Standard form y-intercept

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Model Problem Find the zeros of y = (x – 2)(x + 1)(x + 3) 0 = (x – 2)(x + 1)(x + 3) Set factors equal to zero (x – 2) = 0 (x + 1) = 0 (x + 3) = 0 Zeros/roots/x-intercepts are found at y = 0 (x-axis) x = -3, -1, 2 y-intercept? (-2)(1)(3) = -6 is the product of last terms of binomial factors y = x 3 + 2x 2 – 5x – 6

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Synthetic Division & Factors Divide x 3 – x by x + 1 using synthetic division x 2 – 2x quotient remainder Since there is no remainder x + 1 is a factor of x 3 – x Remainder Theorem (x 2 – 2x + 2)(x + 1) = x 3 – x = 0 x 3 – x Solve: (x 2 – 2x + 2)(x + 1) = 0 Set factors equal to zero (x 2 – 2x + 2) = 0 (x + 1) = 0 x = -1 Quadratic Formula x = 1 ± i 2 2

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Synthetic Division, Factors and Graphing = 0 x 3 – x Solve: (x 2 – 2x + 2)(x + 1) = 0 Set factors equal to zero (x 2 – 2x + 2) = 0 (x + 1) = 0 x = -1 Quadratic Formula x = 1 ± i

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Model Problem Use synthetic division to show that x + 3 is a factor of y = 2x x x + 9 then solve 2x x x + 9 = x 2 + 5x x x x + 9 = 0 (2x 2 + 5x + 3)(x + 3) = 0 (2x + 3)(x + 1)(x + 3) = 0 (2x + 3) = 0 (x + 1) = 0 (x + 3) = 0 x = (-3, -3/2, -1)

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Factor by Grouping x 3 – 2x 2 – 3x + 6 Group terms (x 3 – 2x 2 ) – (3x – 6) Factor Groups x 2 (x – 2) – 3(x – 2) Distributive Property (x 2 - 3)(x – 2)

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Model Problem 4x 3 – 6x x – 15 Group terms (4x 3 – 6x 2 ) + (10x – 15) Factor Groups 2x 2 (2x – 3) + 5(2x – 3) Distributive Property (2x 2 + 5)(2x – 3) Factor: x 3 – 2x 2 – 4x + 8 (x 3 – 2x 2 ) – (4x – 8) x 2 (x – 2) + 4(x – 2) (x 2 + 4)(x – 2) Factor:

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Model Problem x 3 – 3x 2 – 3x + 9 = 0 x 2 (x – 3) – 3(x – 3) = 0 Group (x 2 – 3)(x – 3) = 0 Factor (x – 3) = 0 x = 3 (x 2 – 3) = 0 x = (x 3 – 3x 2 ) – (3x – 9) = 0 Solve:

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Solving Quadratic Form Equations x 4 – 3x = 0 u 2 – 3u + 2 = 0 Quadratic form (x 2 – 1)(x 2 – 2) = 0 Factor Substitute (x – 1)(x + 1)(x 2 – 2) = 0 (x – 1) = 0 x = 1 (x + 1) = 0 x = -1 (x 2 + 2) = 0 x = u = x 2 u 2 – 3u + 2 = 0 (u – 1)(u – 2) = 0 Factor Set factors = 0 & solve for x

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Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Model Problem Solve: x 4 – x 2 = 12

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