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.
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x3 + 12x2 - 5x = x(x2 + 12x - 5) = x(x + 6 - √41)(x + 6 + √41)
(x + 4) / (x3 - 11x + 20) = (x + 4) / (x2 + 4x - 5)(x + 4) = 1 / (x2 + 4x - 5) = 1 / (x + 5)(x - 1), where x ≠ -4
x2 - 5x = 0x (x - 5) = 0x = 0andx = 5
x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
(x - 3)*(x + 5)*(x - 1) = 0 (x2 - 3x + 5x - 15)*(x - 1) = 0 (x2 + 2x - 15)*(x - 1) = 0 (x3 + 2x2 - 15x - x2 - 2x + 15) = 0 ie x3 + x2 - 17x + 15 = 0