5.5B – Synthetic Division. If I tell you that (x – 1) is a factor of the equation above, what does that tell you? x = 1 is a solution.

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Presentation transcript:

5.5B – Synthetic Division

If I tell you that (x – 1) is a factor of the equation above, what does that tell you? x = 1 is a solution

To divide by (x – n): n To divide by (x + n): –n–n

n Ax 4 + Bx 3 + Cx 2 + Dx + E  (x – n) ABCDE A nAnA B + (n  A) + If a term is missing you must have a place holder of 0 for it! Ax 3 + ….

Divide by using synthetic division. 1.(x 3 – 3x 2 + 8x – 5)  (x – 1) 1 1–38–5 1 1 – x 2 – 2x x – 1

Divide by using synthetic division. 2.(x 4 – 7x 2 + 9x – 10)  (x – 2) 2 10– 79– –3 –6 3 6 –4 x 3 + 2x 2 – 3x –4. x – 2

Divide by using synthetic division. 3.(2x 3 – 4x + 5)  (x + 4) – – –112 –107 2x 2 – 8x –107. x + 4

Factor completely, using the sum of cubes. 4. f(x) = x (a + b)(a 2 – ab + b 2 ) a = x b = 2 – –8 0 (x + 2)x 2 – 2x + 4( ) So, (x + 2) is a factor

Given polynomial f(x) and a factor of f(x), factor f(x) completely. 5. f(x) = x 3 – 3x 2 – 16x – 12; x + 1 –1 1–3–16–12 1 –1 –4 4 – (x + 1) x x –6 2 (x + 1)(x – 6)(x + 2) x 2 – 4x – 12( )

Given polynomial f(x) and a factor of f(x), factor f(x) completely. 6. f(x) = 4x x 3 – 8x x – 21; x + 7 –7 426–839 4 –28 – –42 –3 4x 3 – 2x 2 + 6x – 3 (x + 7) – ( ) 2x22x2 (2x – 1) + 3 (2x 2 + 3)(2x – 1)(x + 7)

Given polynomial function f and a zero of f, find the other zeros. 7. f(x) = 3x 3 – 16x 2 – 103x + 36; –4 –4 3–16– –12 – –36 0 (x + 4) 3x3x x –1 –9 (x + 4)(3x – 1)(x – 9) = 0 3x 2 – 28x + 9( ) This means x + 4 is a factor… x =–4, 1/3, 9

Given polynomial function f and a zero of f, find the other zeros. 8. f(x) = x 3 – 9x 2 – 5x + 45; 9 9 1–9– –5 –45 0 (x – 9) (x – 9)(x 2 – 5)= 0 x 2 – 5( ) This means x – 9 is a factor… x – 9 = 0 x 2 – 5 = 0 x 2 = 5 x = 9

You are given an expression for the volume of the rectangular prism. Find an expression for the missing dimension. 9. V = 2x 3 – 11x x – –3 –12 –2 –8 0 (x + 2)2x 2 – 3x – 2( ) 2x2x x 1 –2 (x + 2)(2x + 1)(x – 2) x – 2