 Unit 3 Practice Test Review. 1a) List all possible rational zeros of this polynomial: 5x 4 – 31x 3 + 11x 2 – 31x + 6 p  1, 2, 3, 6 q  1, 5 p  1, 2,

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Unit 3 Practice Test Review

1a) List all possible rational zeros of this polynomial: 5x 4 – 31x 3 + 11x 2 – 31x + 6 p  1, 2, 3, 6 q  1, 5 p  1, 2, 3, 6, 1/5, 2/5, 3/5, 6/5 q

1b) List all possible rational zeros of this polynomial: -2x 3 + 5x 2 + 6x + 18 p  1, 2, 3, 6, 9, 18 q  1, 2 p  1, 2, 3, 6, 9, 18, 1/2, 3/2,9/2 q

2) Use Synthetic Division to determine if x + 1 is a factor of x 3 + 5x 2 +7x + 9 SHOW WORK NEATLY AND EXPLAIN YOUR ANSWER -1  1 5 7 9 -1 - 4 - 3 1 4 3 6 No when synthetically dividing with -1, the remainder is 6 not 0; so x + 1 is not a factor.

3)Determine if 1 – 2i is a zero x 3 + x 2 – x+ 15 SHOW WORK NEATLY AND EXPLAIN YOUR ANSWER 1 – 2i| 1 1 -1 15 1 – 2i 1 2 – 2i (1 – 2i )(2 – 2i ) 2 – 2i – 4i + 4i 2 -2 – 6i 1 – 2i| 1 1 -1 15 1 – 2i -2 – 6i 1 2 – 2i -3 – 6i (1 – 2i )(-3 – 6i) -3 – 6i + 6i + 12i 2 -3 – 12 = -15 1 – 2i| 1 1 -1 15 1 – 2i -2 – 6i -15 1 2 – 2i -3 – 6i 0 Yes, when synthetically dividing with 1 – 2i, the remainder is zero; so 1 – 2i is a zero.

4) If -1/3 is a zero of h(x) = 3x 3 – 2x 2 – 61x – 20, find the other zeros. 3x 3 – 2x 2 – 61x – 20 -1/3 | 3 -2 -61 -20 -1 1 20 3 -3 -60 0 3x 2 – 3x – 60 = 0 3(x 2 – x – 20) = 0 3(x – 5)(x + 4) x = 5, -4

5) x 4 + 2x 3 – 4x –4; -1 + i is a zero -1 + i| 1 2 0 -4 -4 -1 + i 1 1 +i (-1 + i )(1 + i ) -1 – i + i + i 2 -2 -1 + i| 1 2 0 -4 -4 -1 + i -2 2 – 2i 4 1 1 +I -2 -2 – 2i (-1 + i )(-2 – 2i) 2 + 2 i - 2 i - 2 i 2 2 + 2 = 4 -1 - i| 1 1 + i -2 -2 – 2i 0 -1 – i 0 2 + i 1 0 -2 0 x 2 – 2 = 0 x = ±  2

6A) Graph (x + 2 )(x – 3 ) ( x – 2 ) ( x + 2) x 4 – x 3 – 10x 2 + 4x + 24 -2 | 1 -1 -10 4 24 ↓ -2 6 8 -24 1 -3 -4 12 0 x 3 – 3 x 2 – 4x + 12 = 0 x 2 ( x – 3 ) – 4 (x – 3 ) ( x – 3 ) (x 2 – 4 ) (x + 2 )(x – 3 ) ( x – 2 ) ( x + 2) y x 5 5 -5

6B) Graph x 4 – 20x 2 + 64 (x 2 – 4)(x 2 – 16) (x – 2)(x + 2)(x – 4)(x + 4) x = 2, -2, 4, -4 y x 5 5 -5

1)Find all zeros and factor 5x 4 – 31x 3 + 11x 2 – 31x + 6 From 1a p  1, 2, 3, 6, 1/5, 2/5, 3/5, 6/5 q 6  5 -31 11 -31 6 30 -6 30 -6 1/5  5 -1 5 -1  0  1 0 1 5 0 5  0  5x 2 + 5 = 0 5(x 2 + 1) = 0 x 2 = -1 x = ± i x = 6, 1/5, ± i 5(x + i)(x – i)(x – 6)(x – 1/5)

1)Find all zeros and factor 5x 4 – 31x 3 + 11x 2 – 31x + 6 6  5 -31 11 -31 6 30 -6 30 -6 5 -1 5 -1  0  5x 3 – 1x 2 + 5x – 1 x 2 (5x – 1) + 1(5x – 1) (5x – 1)(x 2 + 1) x 2 + 1 = 0 x 2 = -1 x = ± i x = 6, 1/5, ± i 5(x + i)(x – i)(x – 6)(x – 1/5) 6  5 -31 11 -31 6 30 -6 30 -6 1/5  5 -1 5 -1  0  1 0 1 5 0 5  0  5x 2 + 5 = 0 5(x 2 + 1) = 0 x 2 = -1 x = ± i x = 6, 1/5, ± i 5(x + i)(x – i)(x – 6)(x – 1/5)

2) Find all zeros and factor 18x 3 + 3x 2 – 7x – 2 p  1, 2 q  1, 2, 3, 6, 9, 18 p  1, 2, ½, 1/3, 2/3, 1/6, q 1/9, 2/9, 1/18 18x 3 + 3x 2 – 7x – 2 -½ | 18 3 -7 -2 ↓ -9 3 2 18 -6 -4 0 18x 2 – 6x – 4 = 0 2(9x 2 – 3x – 2) = 0 2(3x + 1)(3x – 2) x = -1/2, -1/3, 2/3 2(3x + 1)(3x – 2)(x + ½)

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