Factorise the following
x⁴+3x+2
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Factorise the following x⁴+3x+2
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Answer:
As it has no rational roots (±1 and ±2 would be the only possibilities, and they won’t do), try factoring it as a product of two quadratic polynomials :
x⁴ + 3x -2 = (x²+ax+b)(x²-ax+c).
[The linear terms must have opposite coefficients for the cubic term of the product to vanish.]
Equating coefficients, you get b+c = a², a(c-b) = 3 and bc = -2. The second equality yields c-b = 3/a, which, together with the first one, gives us c = (a² + 3/a)/2 and b= (a² - 3/a)/2; then the third equality yields a⁴ - 9/a² = -8, whence a⁶ + 8a² -9 = 0. This has the obvious solution a=1, whence b=-1 and c=2.
Thus x⁴ + 3x -2 = (x² + x - 1)(x² - x + 2).
The two quadratic factors can be further factored in the usual way, yielding
x⁴ + 3x -2 = (x + (1+√5)/2)(x + (1-√5)/2)(x - (1+i√7)/2)(x - (1-i√7)/2)