2006-09-03 18:07:01 · 7 answers · asked by elbenito2513 2 in Science & Mathematics ➔ Mathematics
The Fundamental Theorem of Algebra tells you that every polynomial with complex coefficients can be factored into linear factors.
In this case, it is hard. You could use the cubic formula to do it by hand or a computer algebra system like Maple or Mathematica. In this case, the exact factorization is
(1 - 4/((-15 + I*Sqrt[31])/2)^(1/3) -
((-15 + I*Sqrt[31])/2)^(1/3) + x)*
(1 + (2*(1 + I*Sqrt[3]))/((-15 + I*Sqrt[31])/2)^(1/3) +
((1 - I*Sqrt[3])*((-15 + I*Sqrt[31])/2)^(1/3))/2 + x)*
(1 + (2*(1 - I*Sqrt[3]))/((-15 + I*Sqrt[31])/2)^(1/3) +
((1 + I*Sqrt[3])*((-15 + I*Sqrt[31])/2)^(1/3))/2 + x)
And approximation of this is
(-1.395426034725887 + x)*
(-0.5765347336712092 + x)*
(4.971960768397096 + x)
2006-09-04 00:09:22 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Maybe you can do this:
x^3 + 3x^2 - 9x + 4
=> x(x^2 + 3x - 9) + 4
=> x[x^2 + (3/2)^2 - (3/2)^2 - 9] + 4 (completing the square)
=> x[(x + 3/2)^2 - 27/4] + 4
2006-09-03 18:17:00 · answer #2 · answered by e.z p.z 2 · 0⤊ 0⤋
um
2006-09-03 19:01:24 · answer #3 · answered by Jason J 2 · 0⤊ 0⤋
No answer.
2006-09-03 18:19:38 · answer #4 · answered by michael2003c2003 5 · 0⤊ 1⤋
He's right, no answer.
2006-09-03 18:12:33 · answer #5 · answered by paintballer391 2 · 0⤊ 1⤋
no answer
2006-09-03 18:11:55 · answer #6 · answered by pro 2 · 0⤊ 1⤋
you can't
2006-09-03 18:31:43 · answer #7 · answered by Yaritza 3 · 0⤊ 0⤋