Answer:
(x + 1)(x + 2)(2x + 1)(x - 3).
Step-by-step explanation:
From the leading and last coefficients, 2 and -6 the values +/- 1, +/-2 , +/- 3 could be roots of this equation ( well, up to 4 of them).
Try x = 1:
f(1) = 2(1)^4 + 1^3 - 14(1)^2 - 19(1) - 6 = -36 so its not 1.
f(-1) = 2 - 1 - 14 + 19 - 6 = 0 do x = -1 is a root.
Therefore (x + 1) is a factor (By the factor theorem)
Try x = 2:
f(2) = 32 + 8 -56 - 38 - 6 = -6o so x =2 not a root.
f(-2) = 32 - 8 - 56 + 38 - 6 = 0 so x = -2 is a root and (x + 2) is a factor.
(x + 1)(x + 2) = x^2 + 3x + 2.
Long division:
2x^2 - 5x - 3 <-----------Quotient
--------------------------------------------
x^2 + 3x + 2 ) 2x^4 + x^3 -14x^2 - 19x - 6
2x^4 + 6x^3 + 4x^2
-5x^3 - 18x^2 - 19x
-5x^3 - 15x^2 - 10x
-3x^2 - 9x - 6
-3x^2 - 9x - 6
.......................
2x^2 - 5x - 3 = (2x + 1)(x - 3)
So Finally, the factors are (x + 1)(x + 2)(2x + 1)(x - 3).
