x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
(x - 1)(x2 + x - 1)
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
It is a polynomial of the fourth degree in X.
x3 + 1 = x3 + x2 - x2 - x + x + 1 = x2(x + 1) - x(x + 1) +1(x + 1) = (x + 1)(x2 - x + 1)
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
10x + 5 unless you know what x & y are * * * * * Perhaps this answer will be of help: x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). If you are willing to go to complex roots, then, x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). = (x + 1)(x + i√5)(x - i√5); in which case, x = -1 or ±i√5.
(x + 1)(x2 - x + 1)
x3 + 2x2 - 19x - 20 = x3 + x2 + x2 + x - 20x - 20 = x2(x + 1) - x(x + 1) - 20(x + 1) = (x + 1)(x2 + x - 20) = (x + 1)(x - 4)(x + 5)
x3 + 4x2 + x + 4 = (x + 4)(x2 + 1)
x3 + x = x2(x + 1)
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
(x - 1)(x2 + x - 1)
2x2+7/x1