guadractic form factor 4x^4-5x^2-9?

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2007-03-11 09:14:10 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

(4x^2 -9)(x^2+1)

(2x-3)(2x+3)(x^2+1)

2007-03-11 09:23:34 · answer #1 · answered by bignose68 4 · 0⤊ 0⤋

4x^4-5x^2-9 use Y = x^2 and obtain a quadratic 4Y^2 - 5Y - 9

2007-03-11 16:25:08 · answer #2 · answered by physicist 4 · 0⤊ 0⤋

4x^4-5x^2-9
Let z=x^2
Then 4z^2 -5z -9 = 0
So z = -1 or 2 1/4 [by quadratic formula]
So x^2 = -1 --> x = +/- i
x^2 = 9/4 --> x = +/- 3/2

2007-03-11 16:25:48 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋

(4x^2-9)(x^2+1)
(2x-3)(2x+3)(x^2+1)
For real numbers x^2 + 1 does not break down

2007-03-11 16:26:50 · answer #4 · answered by lizzie 3 · 0⤊ 0⤋

(4x^2 - 9)(x^2 + 1)
(2x + 3)(2x - 3)(x^2 + 1)

2007-03-11 16:23:35 · answer #5 · answered by Newbody 4 · 0⤊ 0⤋