Find the zero for P(x) = 4x^3 + 8x^2 - 3x - 1.?

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Find the zero for P(x) = 4x^3 + 8x^2 - 3x - 1.?

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2007-10-14 14:42:50 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

3 answers

P(x) = 4x³ + 8x² - 3x - 1

Using the remainder theorem and synthetic division,
1/2 is a zero

4x³ + 8x² - 3x - 1 = (2x - 1)(2x² + 5x + 1)

(2x - 1)(2x² + 5x + 1) = 0

Using quadratic equation for the other 2 zeroes:

x = [-5 ± √(17)]/4

So this has 3 zeroes

at 1/2 and [-5 ± √(17)]/4

2007-10-14 14:53:01 · answer #1 · answered by Marvin 4 · 0⤊ 0⤋

1/2, -5/4 + sqrt(17)/4, -5/4 - sqrt(17)/4

2007-10-14 21:45:46 · answer #2 · answered by Mαtt 6 · 0⤊ 0⤋

2007-10-14 21:47:34 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋