(x2 + 4 - 2sqrt(3)) (x2 + 4 + 2sqrt(3))
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2
2x3 + 8x2 + 3x + 12 = (2x3 + 8x2) + (3x + 12) = 2x2(x + 4) + 3(x + 4) = (x + 4)(2x2 + 3) Since you have asked this question I am assuming that you are not yet at a level where you might want (2x2 + 3) factorised into its imaginary factors.
x4 +x2 =x2 (x2+1)
8x2 8x2
if 4 * 7 = x * 4, then x = (4 * 7) / 4 = 7
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2
Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4
4 x4 x4=64 + 13=77- - 4=81+(14x14=196)=277
2x3 + 8x2 + 3x + 12 = (2x3 + 8x2) + (3x + 12) = 2x2(x + 4) + 3(x + 4) = (x + 4)(2x2 + 3) Since you have asked this question I am assuming that you are not yet at a level where you might want (2x2 + 3) factorised into its imaginary factors.
(2x - 8)(-4x + 4) = -8x2 + 40x - 32
3x3 + 6x2 + x + 2 3x2 (x + 2) + (x + 2) (x + 2) (3x2 + 1) OR 3x3 + x + 6x2 + 2x (3x2 + 1) + 2 (3x2 + 1) (3x2 + 1) (x + 2) Check: (x + 2) (3x2 + 1) = 3x3 + x + 6x2 + 2 = 3x3 + 6x2 + x + 2 x3 - 3x2 - 4x + 12 x2 (x - 3) - 4 (x - 3) (x - 3) (x2 - 4) (x - 3) (x + 2) (x - 2) Check: (x - 3) (x + 2) (x - 2) = (x - 3) (x2 - 4) = x3 - 4x - 3x2 + 12= x3 - 3x2 - 4x + 12=== === x5 - x4 + 8x3 - 8x2 + 16x - 16 x4 (x - 1) + 8x2 (x - 1) + 16 (x - 1) (x - 1) (x4 + 8x2 + 16) (x - 1) (x2 + 4)(x2 + 4) (x - 1) (x2 + 4)2 Real Solution (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) (x - 1) (x + 2i)2 (x - 2i)2 Check: (x - 1) (x + 2i)2 (x - 2i)2 = (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) = (x - 1) (x2 + 2xi - 2xi - 4i2) (x2 + 2xi - 2xi - 4i2) = (x - 1) (x2 - 4(-1)) (x2 - 4(-1)) = (x - 1) (x2 + 4) (x2 + 4) = (x - 1) (x4 + 4x2 + 4x2 +16) = (x - 1) (x4 + 8x2 + 16) = x5 + 8x3 +16x - x4 - 8x2 - 16 = x5 - x4 + 8x3 - 8x2 + 16x - 16
x^4 - x^2 plus 2 has no real rational roots.x^4 - x^2 minus 2 factors to (x^2 + 1)(x^2 - 2)
(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)
The browser used by this site for posting questions is rubbish. We have complained about it but the powers that be are not doing anything about it. The consequence is that most mathematical symbols are not accepted and so all that I can see is of your question is:"If 4 plus x4 3 plus yy then x4 3y?"It is difficult to be sure, but making an informed guess as to what might be missing, I would say that the answer is NO. There is no solution for y.
It is: 25-4(8*6) = -167
x4 +x2 =x2 (x2+1)
x4 - 4x3 - 12x2 -32x + 64 (x - 4)(x + 2)(x + 2)(x - 4)