(x + y)2 = x2 + 2xy + y2
So x2 + y2 = (x + y)2 - 2xy = a2 - 2b
Then (x2 + y2)2 = x4 + 2x2y2 + y4
So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2
= a4 - 4a2b + 4b2 - 2b2
= a4 - 4a2b + 2b2
N = x4 x8 = x4+4 = (x4)2 = N2
No.
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
28 is.
yes
N = x4 x8 = x4+4 = (x4)2 = N2
Not necessarily.
No.
x^4-x^3+x
4X + 2x = 1. Where x = 0.166666666666666666666666666666666666666 recurring.
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
28 is.
7
yes
If the number is x, then the number multiplied by itself three times would be x4 ... so if the "given number" is y, then y = x4 and thus x is related to the "given number by x = y1/4 i.e. the quartic or fourth root of 4.
If there is a remainder then it's not a factor otherwise yes
Yes.