factorise 2x^4 - 9x^3 + 5x^2 - 3x - 4?

Question

use any method you know. I greatly appreciate the whole answers.

Answer 1

Let f(x) = 2x^4 - 9x^3 + 5 x^2 - 3x -4

By trail and error f(4) = 2*(256) - 9*(64) + 5(16) -3*(4) -4
= 512 - 576 + 80 - 12 - 4
= 0
Hence (x-4) is a factor of f(x) =0 (Mostly it is better first to try all the multiples of the constant term in the expression)

Now by long division,
we get,
f(x) = (x-4)(2x^3 - x^2 +x +1) = 2(x-4)(x^3 - 0.5x^2 + 0.5 x + 0.5)
= 2(x-4)g(x)


For g(x)= 0
Substituting x= - 0.5
we get,
g(x) = (-.125)-(.125) + (-.25) +0.5
+0
Hence again by long division we get
g(x) = (x-0.5) ( x^2 - x +1)

Therefore,

f(x) = (x-4)(2x+1)(x^2 -x+1)

x^2 -x +1 cannot be further factorized as discriminant = (-1)^2 -4
= -3 <0
Hence we get unreal (complex or imaginary) roots for x^2 -x +1.

Answer 2

2x^4 - 9x^3 + 5x^2 - 3x - 4 = (x - 4)(2x + 1)(x^2 - x + 1)

i used www.quickmath.com

Answer 3

2(x - 4)(x + 0.5)(x^2 - x + 1)