That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (3 plus or minus the square root of 2) divided by 7.
x = 0.6306019374818707
x = 0.2265409196609864
x3 - 7x2 + 6x =x (x2 - 7x + 6) =x (x - 6) (x - 1)
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
(7x2 + 9)(4x - 1)
(6x + 1)(x + 13)
(6x + 1)(x + 2)
(6x - 1)(6x - 1)
7x2 + 7x - 14 = 7(x2 + x - 2) = 7(x + 2)(x - 1)
(-3x+1)(-2x+1)
x2 + 6x + 5 can be factored into (x+1) (x+5)
(2x+4)(x+1)
6(x + 1)(x + 1)
Fill in the missing powers of x in the dividend by having a coefficient of zero (0); then use long division: _________________ 6x³ - 3x² - 2x + 1 ________--------------------------------- 2x + 1 | 12x⁴ + 0x³ - 7x² + 0x + 1 _________ 12x⁴ + 6x³ _________ ------------- _______________ - 6x³ - 7x² _______________ - 6x³ - 3x² _______________ ------------- ____________________ - 4x² + 0x ____________________ - 4x² - 2x ____________________ ------------ ____________________________ 2x + 1 ____________________________ 2x + 1 ____________________________ -------- __________________________________0 ____________________________ ===== → (12x⁴ - 7x² + 1) ÷ (2x + 1) = 6x³ - 3x² - 2x + 1 with no remainder.