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How do you factor z squared minus 64x squared?
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)
Which of the following is a factor of 64x3 plus 343?
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)
What are the coefficients of the binomial x plus 2 to the power of 6?
(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
