(x2 + y2)(x + y)(x - y) = x4 - y4.
It is: 1(x4+4y8) and can't be factored any further
The factorisation is as follows: 2x2 + 7x - 9 = (x - 1)(2x + 9).
If that's +28, the answer is (x - 4)
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(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
2x2 x 3x2 = 6 x4 (2x)2 x (3x)2 = 36 x4
2-3+9
x4 / 2x4 396(x2 + y2) / 396(2x2 + 2y2)
Quartic Binomial
264
No solution.
x^2(x + 2) - 1(x+2) (x+2)(x-1)(x+1)
∫(4x3 - 2x2 + x - 1) dx You can integrate this by taking the antiderivative of each term. Each of these terms is in the format axn, the antiderivative of which is axn-1/n: = ∫(4x3)dx - ∫(2x2)dx + ∫(x)dx - ∫(1)dx = x4 - 2x3/3 + x2/2 - x + C
There is no maximum since a quartic with a positive leading coefficient increases without limit. The minimum is approximately -4.4940
x4 + 3x3 - x2 - 9x - 6 = 0 x4 + x3 + 2x3 + 2x2 - 3x2 - 3x - 6x - 6 = 0 x3(x + 1) + 2x2(x + 1) - 3x(x + 1) - 6(x + 1) = 0 (x + 1)(x3 + 2x2 - 3x - 6) = 0 (x + 1)[x2(x + 2) - 3(x + 2)] = 0 (x + 1)(x + 2)(x2 - 3) = 0 So x + 1 = 0 so that x = -1 or x + 2 = 0 so that x = -2 or x2 - 3 = 0 so that x = +/- sqrt(3)
Yes, 2x2=4, 5x2=10.