In the polynomial;
an xn + an-1 xn-1 + · · · + a1 x + a0 = 0
where the coefficients an ,an-1 , . . . ,a1 ,a0 are integers.
If a rational number c/d, is factored to its lowest terms, is a solution to the equation, then c is a factor of the constant term of the polynomial and d is a factor of the leading coefficient of the polynomial.
ex.
find solutions to the equation:
6x3- 13x2- 13x + 30 = 0
c = constant term = 30
factors of 30 are: ±1,±2,±3,±5,±6,±10,±15,±30
d = leading coefficient = 6
factors of 6 are: ±1,±2,±3,±6
use synthetic division to find a factor for the polynomial:
NOTE:
(x - c/d) is factor of the polynomial if the remainder is zero.
try the integers first;
1 is not a factor since the remainder is not 0.
-1 is not a factor since the remainder is not 0.
2 is a factor since the remainder is 0.
(x-2)(6x2- x - 15) = 0
6x2- x - 15 can be factored by using the conventional factoring techniques.
(x-2)(6x2-10x + 9x - 15) = 0
(x-2)[2x(3x - 5) + 3(3x - 5)] = 0
(x-2)(2x + 3)(3x - 5) = 0
x = 2 or x = -3/2 or x = 5/3
ex.
6x4- x3- 21x2- 6x + 8 = 0
c = 8; ±1,±2,±4,±8
d = 6; ±1,±2,±3,±6
try the integers;
1 is not a factor since the remainder is not 0.
factor out 6x3- 7x2- 14x + 8 by using the same process as above.
c = 8; ±1,±2,±4,±8
d = 6; ±1,±2,±3,±6
try the integers;
(x + 1)(x - 2)(6x2+ 5x -4) = 0
factor 6x2+ 5x -4.
(x + 1)(x - 2)(6x2 + 5x - 4) = 0
(x + 1)(x - 2)(6x2 - 3x + 8x - 4) = 0
(x + 1)(x - 2)[3x(2x - 1) + 4(2x - 1)] = 0
(x + 1)(x - 2)(2x - 1)(3x + 4) = 0
6x4- x3- 21x2- 6x + 8 = (x + 1)(x - 2)(2x - 1)(3x + 4) = 0
x = -1 or x = 2 or x = 1/2 or x = -4/3
