How do you factor 4x^4 - 24x^3 + 35x^2 + 6x - 9 ?

Answer 1

4x^4 - 24x^3 + 35x^2 + 6x - 9 = 4x^3 (x-6) + (7x-3) (5x+3)

Answer 2

Trial and error...it is a (Ax²+Bx+C)(Dx²+Ex+F) factor

ADx^4+AEx³+AFx²+BDx³+BEx²+BFx+CDx²+CEx+CF
ADx^4 + (AE+BD)x³+(AF+BE+CD)x²+(BF+CE)x+CF

The posiblities for A and D or 2*2 or 4*1.
So plug in 2X2 for A and D
4x^4+(2E+2B)x³+(2F+BE+2C)x²+(BF+CE)x+CF
Then CF has to equal nine so it can be 3*-3, 9*-1, or -9*1.
It would be easiest if it was 3*-3 so that is the one to try first:
if C=-3 and F=3
4x^4+(2E+2B)x³+(6+BE-6)x²+(3B-3E)x-9
this leave BE = +35 so it can be 1*35, -1*-35, 7*5 or -7*-5
4x^4+(2E+2B)x³+(6+BE-6)x²+(3B-3E)x-9
2E+2B=-24, -7 or -5 works
3B-3E=6
-21+15=-6 or
-15+21=6 so B has to be -5 and E -7.
Therefore
(2x²-5x-3)(2x²-7x+3)
Then these can get factored down farther.

If the 2*2 wouldn't of worked out, it would of required going back to the start and using the 1*4, but it is always best to start with the easiest factors.