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.
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
(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
x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
x2 - 5x = 0x (x - 5) = 0x = 0andx = 5
(x - 1)(x^2 - 5)
x3 + 12x2 - 5x = x(x2 + 12x - 5) = x(x + 6 - √41)(x + 6 + √41)
(x + 5)(x^2 - 5x + 25)
x3 - 4x2 + 5x - 20 = x2*(x - 4) + 5*(x - 4) = (x2 + 5)*(x - 4)
x2 + 5x + 25
x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)
5(x2 + x + 1)
Difference of two cubes: a3 - b3 = (a-b)(a2+ab+b2) 125 - x3 = 53 - x3 = (5 - x)(52 + 5x + x2) = (5 - x)(25 + 5x + x2)
x2 + 3x - 7 = 5x + 8; thus, x2 + 3x - 7 - 5x - 8 = x2 - 2x - 15 = (x - 5)(x + 3) = 0. Therefore the solution is x = 5 or -3.
x(x+5)
No, it is not.
x3 - 6x2 + 7x - 2 = x3 - x2 - 5x2 + 5x + 2x - 2 = x2(x - 1) - 5x(x - 1) + 2(x - 1) = (x - 1)(x2 - 5x + 2) = (x - 1){x - 0.5*[5 - sqrt(25 - 8)]}){x + 0.5*[5 - sqrt(25 - 8)]} = (x - 1)(x - 0.4384)(x - 4.5616) so that the roots are x = 1 x = 0.4384 and x = 4.5616