Okay, so I need to find all the integer roots of the polynomial equation \(x^4 + 5x^3 + 9x^2 - x - 14 = 0\). Hmm, let's start by recalling that integer roots of a polynomial are factors of the constant term divided by factors of the leading coefficient. In this case, the constant term is -14 and the leading coefficient is 1. So the possible integer roots are the divisors of -14. Let me list them out: ±1, ±2, ±7, ±14. 

Alright, so I should test these possible roots by substituting them into the polynomial and seeing if the result is zero. Let me start with x = 1. Plugging in 1: \(1 + 5 + 9 - 1 -14 = (1+5) + (9-1) -14 = 6 + 8 -14 = 14 -14 = 0\). Oh, so x=1 is a root! That's good. 

Now that I found x=1 is a root, I can factor the polynomial by (x - 1). To do that, I should perform polynomial division or use synthetic division. Let me try synthetic division with x=1. 

Set up the coefficients: 1 (x^4), 5 (x^3), 9 (x^2), -1 (x), -14 (constant). 

Bring down the 1. Multiply by 1: 1*1=1. Add to next coefficient: 5+1=6. Multiply by 1: 6*1=6. Add to next coefficient: 9+6=15. Multiply by 1: 15*1=15. Add to next coefficient: -1+15=14. Multiply by 1: 14*1=14. Add to last coefficient: -14+14=0. 

So the result of the division is \(x^3 + 6x^2 + 15x + 14\). So the polynomial factors as (x - 1)(x^3 + 6x^2 + 15x + 14) = 0. 

Now I need to find the integer roots of the cubic polynomial \(x^3 + 6x^2 + 15x + 14\). Again, using the Rational Root Theorem, possible integer roots are factors of 14 divided by 1, so ±1, ±2, ±7, ±14. Let me test these.

First, x = -1: (-1)^3 + 6*(-1)^2 + 15*(-1) +14 = -1 + 6*1 + (-15) +14 = -1 +6 -15 +14 = ( -1 +6 ) + ( -15 +14 ) = 5 -1 = 4 ≠ 0. Not a root.

x = -2: (-2)^3 +6*(-2)^2 +15*(-2) +14 = -8 +6*4 + (-30) +14 = -8 +24 -30 +14 = (-8 +24) + (-30 +14) = 16 -16 = 0. Oh, x=-2 is a root. 

So now factor out (x + 2) from the cubic. Let's use synthetic division again. Coefficients: 1 (x^3), 6 (x^2), 15 (x), 14 (constant). 

Using x = -2:

Bring down 1. Multiply by -2: 1*(-2) = -2. Add to next coefficient: 6 + (-2) = 4. Multiply by -2: 4*(-2) = -8. Add to next coefficient:15 + (-8)=7. Multiply by -2:7*(-2)=-14. Add to last coefficient:14 + (-14)=0. 

So the cubic factors into (x + 2)(x^2 +4x +7). Therefore, the original polynomial is now factored as (x -1)(x +2)(x^2 +4x +7). 

Now we need to find integer roots of the quadratic \(x^2 +4x +7\). Let's check the discriminant: \(16 - 4*1*7 = 16 -28 = -12\). Since the discriminant is negative, there are no real roots, let alone integer ones. 

Therefore, the only integer roots of the original equation are x=1 and x=-2. Wait, but let me double-check. Maybe I made a mistake in the synthetic division steps. Let me verify the cubic division again. 

When dividing the cubic by (x + 2), the coefficients after division should be 1, 4, 7. Let me check by expanding (x +2)(x^2 +4x +7). Multiply out: x*(x^2 +4x +7) +2*(x^2 +4x +7) = x^3 +4x^2 +7x +2x^2 +8x +14 = x^3 +6x^2 +15x +14. Yes, that's correct. So the cubic factors correctly. 

Therefore, the integer roots are indeed 1 and -2. 

But wait, let me check if there are any other possible integer roots that I might have missed. For the quadratic, even though the discriminant is negative, maybe there was a mistake in the initial possible roots. Let me check x=2 in the original polynomial. Plugging x=2: 16 + 40 + 36 -2 -14 = 16+40=56, 56+36=92, 92-2=90, 90-14=76 ≠0. Not a root. x=7: 7^4 is 2401, which is way too big, but let's compute: 2401 +5*343 +9*49 -7 -14. 2401 +1715=4116, 4116 +441=4557, 4557 -7=4550, 4550 -14=4536 ≠0. x=14 would be even larger. x=-1: we checked earlier, gives 4. x=-7: Let me compute. (-7)^4 +5*(-7)^3 +9*(-7)^2 -(-7) -14. 2401 +5*(-343) +9*49 +7 -14. 2401 -1715 +441 +7 -14. 2401-1715=686, 686+441=1127, 1127+7=1134, 1134-14=1120≠0. So none of the other possible integer roots work. 

Therefore, the only integer roots are x=1 and x=-2. 

But wait, let me check the original polynomial again with x=1 and x=-2 to make sure. 

For x=1: 1 +5 +9 -1 -14 = (1+5) + (9-1) -14 = 6 +8 -14 = 14 -14 =0. Correct. 

For x=-2: (-2)^4 +5*(-2)^3 +9*(-2)^2 -(-2) -14. 16 +5*(-8) +9*4 +2 -14. 16 -40 +36 +2 -14. (16-40)= -24, (-24 +36)=12, (12 +2)=14, (14-14)=0. Correct. 

So yeah, both are roots. 

Therefore, the integer roots are 1 and -2.
Therefore, the integer roots of the equation \(x^4 + 5x^3 + 9x^2 - x - 14 = 0\) are $\boxed{1}$ and $\boxed{-2}$.
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