Question

$x^{4}- 2 x^{3}- 7 x^{2} + 8x+12$
factorize using factor theorem or vanishing method.

Answer 1

take the factors of 12 as +_1 +_2 +_3  +_4 +_5 +_6..........
put the value of x as-1 and then solve it which will leave 0 as remainder now through long division method solve it. take that long term as divident and x+1 as divisor and solve it this will give u p(x) = $x^{4} -2x^{3} -7x^{2} + 8x +12$
                            = $(x+1)(x^{3} - 3x^{2} -4x +12)$ now take the commons
                            = $(x+1){ x^{2}(x - 3) - 4(x - 3)$ 
                            = $(x + 1)(x - 3)(x^{2} - 4)$
(x+1)(x-3)(x² - 4 ) are the factors...

Answer 2

x^4 - 2x^3 - 7x^2 + 8x + 12 is zero at x = 2 hence (x - 2) is a factor of this equation
x^3(x -2) - 0x^2(x - 2)  - 7x(x  - 2) - 6(x - 2)
(x - 2)(x^3 - 7x - 6)
x^3 - 7x - 6  is zero at x = -2 , hence (x + 2) is a factor of this equation
x^2(x + 2) - 2x(x + 2) - 3(x + 2)
(x + 2)(x^2 - 2x - 3)
(x + 2)(x^2 - 3x + x - 3)
(x + 2)(x -3)(x + 1)
hence factor of given equation is (x - 2)(x + 2)(x - 3)(x + 1)