x3 + x2 - 8x - 6 = 0
x3 + 3x2 - 2x2 - 6x - 2x - 6 = 0
x2(x + 3) - 2x(x + 3) - 2(x + 3) = 0
(x + 3)(x2 - 2x - 2) = 0
The second bracket cannot be factorised, so applying the quadratic formula to it gives the answers as
So x = -3 or x = 1 +/-sqrt(3)
that is,
x = -3, x = -0.732051 or x = 2.732051
cube root of 216 = 6 > X= 6
y2=x3+3x2
2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7
5x3 = 320 x3 = 320/5 = 64 So x = cuberoot(64) = 4
[ x3 - 4x2 + 7x ] / (x) = x2 - 4x + 7
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
IMPOSSIBLE
cube root of 216 = 6 > X= 6
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.
No.
y2=x3+3x2
It is x = -5
No, it is not.
(xn+2-1)/(x2-1)
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
Since that isn't an equation, there is really nothing to solve.