-x^6(64x^3 - 27)
If that were z^2 - 16z + 64, it would factor very neatly to (z - 8)(z - 8) or (z - 8)^2 As it is, you could use z(z^2 - 16) + 64 or the much more cumbersome -1/27 (3 z+2 2^(2/3) (3 (9-sqrt(69)))^(1/3)+4 3^(2/3) (2/(9-sqrt(69)))^(1/3)) (-9 z^2+(12 3^(2/3) (2/(9-sqrt(69)))^(1/3)+6 2^(2/3) (3 (9-sqrt(69)))^(1/3)) z-8 2^(1/3) (3 (9-sqrt(69)))^(2/3)-48 3^(1/3) (2/(9-sqrt(69)))^(2/3)+48)
There are no real solutions to this equation. Applying the quadratic formula, we find two imaginary solutions.x = 2.315032397181517ix = -2.315032397181517iwhere i is the square root of negative one.If that were 64x3, it would have been (4x + 7)(16x2 - 28x + 49)
(x+2)6 (x2 + 4x + 4)(x+2)4 (x2 + 4x + 4)(x2 + 4x + 4)(x+2)2 (x2 + 4x + 4)(x2 + 4x + 4)(x2 + 4x + 4) (x4 + 4x3 + 4x2 + 4x3 + 16x2 + 16x + 4x2 + 16x + 16)(x2 + 4x + 4) (x4 + 8x3 + 24x2 + 32x + 16)(x2 + 4x + 4) (x6 + 4x5 + 4x4 + 8x5 + 32x4 + 32x3 + 24x4 + 96x3 + 96x2 + 32x3 + 128x2 +128x + 16x2 + 64x + 64) (x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x +64) the coefficeints are 1, 12, 60, 160, 240, and 192
In case you meant 64x^3 - 27y^3, that factors to (4x - 3y)(16x^2 + 12xy + 9y^2)
(4x - 3y)(16x^2 + 12xy + 9y^2)
64x2-80x+25 = (8x-5)(8x-5) when factored
(4x - 3)(16x^2 + 12x + 9)
x5 - 1024 = (x - 4)(x4 + 4x3 + 16x2 + 64x + 256)
The GCF is 8x^3
(8x - 21)(8x + 21)
4. 12x^2-64x-256=4(3x^2-16x-64)
the GCF is 8x
(11x + 9)(x + 5)
(4x-8)2 = (4x-8)(4x-8) = 16x2-32x-32x+64 = 16x2-64x+64
(8x + 5y)(64x^2 - 40xy + 25y^2)