<Question>
Find all the integer roots of
\[x^4 + 5x^3 + 9x^2 - x - 14 = 0.\]Enter all the integer roots, separated by commas.
</Question>

<think>
<Plan_1>
Okay, so I need to find all the integer roots of the polynomial equation x⁴ + 5x³ + 9x² - x - 14 = 0. Hmm, let's start by recalling how to find integer roots of a polynomial. I remember something about the Rational Root Theorem. Let me think... The Rational Root Theorem states that any possible rational root, expressed in lowest terms p/q, p is a factor of the constant term, and q is a factor of the leading coefficient. Since the leading coefficient here is 1 (because the x⁴ term has coefficient 1), that means any rational root is just an integer that divides the constant term, which is -14.

So, the possible integer roots are the divisors of -14.
Let me list those out: ±1, ±2, ±7, ±14.

Now, I need to test these values one by one to see which ones satisfy the equation. Let's start with x = 1.
Plugging into the equation: 1⁴ + 5*1³ + 9*1² - 1 - 14 = 1 + 5 + 9 - 1 - 14.

<prompt_1.1> Evaluate the polynomial at x = 1 and determine if it is a root. </prompt_1.1>

Next, let's try x = -1.

Plugging in x = -1: (-1)⁴ + 5*(-1)³ + 9*(-1)² - (-1) - 14.
That's 1 + 5*(-1) + 9*1 +1 -14. Calculating step by step: 1 -5 +9 +1 -14.
So, 1 -5 is -4, plus 9 is 5, plus 1 is 6, minus 14 is -8. Not zero, so -1 isn't a root.

Moving on to x = 2.

<prompt_1.2> Evaluate the polynomial at x = 2 and determine if it is a root. </prompt_1.2>

Okay, x = -2. Let's compute that.

<prompt_1.3> Evaluate the polynomial at x = -2 and determine if it is a root. </prompt_1.3>

Next, x = 7. Hmm, 7 might be a bit large, but let's try.

<prompt_1.4> Evaluate the polynomial at x = 7 and determine if it is a root. </prompt_1.4>

x = -7.

<prompt_1.5> Evaluate the polynomial at x = -7 and determine if it is a root. </prompt_1.5>

Finally, x =14. That's a big number, but let's check.

<prompt_1.6> Evaluate the polynomial at x = 14 and determine if it is a root. </prompt_1.6>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's compute that: 1 + 5 is 6, plus 9 is 15. Then 15 - 1 is 14, and 14 -14 is 0. Wait, that equals zero! So, x = 1 is a root. Great, so (x - 1) is a factor of the polynomial.
</execution_1.1>

<execution_1.2>
Plugging in x = 2: 2⁴ + 5*2³ + 9*2² - 2 - 14. Calculating each term: 16 + 5*8 + 9*4 -2 -14. So, 16 + 40 + 36 -2 -14. Adding those up: 16+40=56, 56+36=92, 92-2=90, 90-14=76. Not zero.
</execution_1.2>

<execution_1.3>
Plugging in x = -2: (-2)⁴ + 5*(-2)³ + 9*(-2)² - (-2) -14. Calculating each term: 16 + 5*(-8) + 9*4 +2 -14. So, 16 -40 +36 +2 -14. Let's compute step by step: 16 -40 = -24, +36 = 12, +2 =14, -14=0. Wait, that adds up to zero! So x = -2 is a root.
</execution_1.3>

<execution_1.4>
Plugging into the equation: 7⁴ + 5*7³ + 9*7² -7 -14. Let's compute each term:

7⁴ = 7*7*7*7 = 49*49 = 2401

5*7³ = 5*343 = 1715

9*7² = 9*49 = 441

Then subtract 7 and 14: -7 -14 = -21

Adding all together: 2401 + 1715 = 4116; 4116 + 441 = 4557; 4557 -21 = 4536. Definitely not zero.
</execution_1.4>

<execution_1.5>
Plugging into the equation: 7⁴ + 5*7³ + 9*7² -7 -14. Let's compute each term:
7⁴ is 2401, 5*7³ is 5*343=1715, 9*7²=9*49=441, then -7 and -14. Adding them up: 2401 +1715 = 4116, +441 = 4557, -7 = 4550, -14 = 4536. Definitely not zero.
</execution_1.5>

<execution_1.6>
Plugging in x =14: 14⁴ +5*14³ +9*14² -14 -14. That's going to be a huge number, probably not zero. Let me compute it roughly. 14² is 196, 14³ is 2744, 14⁴ is 38416. So, 38416 +5*2744 +9*196 -14 -14. 5*2744 is 13720, 9*196 is 1764. So total is 38416 +13720 = 52136, plus 1764 is 53900, minus 28 is 53872. Definitely not zero. So, 14 isn't a root.
</execution_1.6>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.3:
So far, we have two integer roots: 1 and -2.
Now, since we found these roots, we can factor the polynomial by dividing it by (x - 1)(x + 2).
Let me confirm that. If x = 1 is a root, then (x -1) is a factor, and if x = -2 is a root, then (x +2) is a factor.
So, the polynomial can be factored as (x -1)(x +2) multiplied by a quadratic polynomial. Let's perform polynomial division to find that quadratic.

Alternatively, since we have two roots, we can factor them out step by step. Let's first divide the original polynomial by (x -1).

<prompt_2.1> Perform polynomial division to factor out (x - 1) from the original polynomial. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Using synthetic division for x =1:

Coefficients of the original polynomial: 1 (x⁴), 5 (x³), 9 (x²), 0 (x), -14 (constant). Wait, hold on, the original polynomial is x⁴ +5x³ +9x² -x -14. So, the coefficients are 1, 5, 9, -1, -14. So, during synthetic division with x=1:

Bring down the 1. Multiply by 1: 1. Add to next coefficient: 5 +1=6. Multiply by 1:6. Add to next coefficient:9 +6=15. Multiply by1:15. Add to next coefficient: -1 +15=14. Multiply by1:14. Add to last coefficient: -14 +14=0. So, the result is x³ +6x² +15x +14. So, the polynomial is (x -1)(x³ +6x² +15x +14).
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now, we can take this cubic polynomial and try to factor it. Since we already found x = -2 is a root, let's use synthetic division again on the cubic with x = -2.

<prompt_3.1> Perform synthetic division on the cubic polynomial to factor out (x + 2). </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Coefficients of the cubic: 1 (x³),6 (x²),15 (x),14 (constant). Using synthetic division for x = -2:

Bring down the 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 result is x² +4x +7. Therefore, the original polynomial factors into (x -1)(x +2)(x² +4x +7).
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Now, we need to check if the quadratic x² +4x +7 has any integer roots. Let's compute its discriminant: D = 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 a second, but let me double-check my synthetic division steps to make sure I didn't make any mistakes.

<prompt_4.1> Double-check the synthetic division steps for both x = 1 and x = -2. </prompt_4.1>

The quadratic factor is x² +4x +7. Let me check if I can factor this further or if I made a mistake in the division.

<prompt_4.2> Verify the factorization by expanding (x - 1)(x + 2)(x² + 4x + 7) and check if it matches the original polynomial. </prompt_4.2>

But just to be thorough, let's check if there's another possible integer root. Wait, we already listed all possible integer roots as divisors of 14, which are ±1, ±2, ±7, ±14.
We tested all of them except for maybe x=7 and x=14 again, but they were way too big. But just to confirm, let me check x=7 again.

<prompt_4.3> Re-evaluate the potential integer roots x = 7 and x = 14 to ensure they are not roots. </prompt_4.3>

Alternatively, maybe there's a mistake in the synthetic division steps. Let me recheck them.

<prompt_4.4> Recheck the synthetic division steps for x = 1 and x = -2 to ensure accuracy. </prompt_4.4>

Alternatively, perhaps there's a calculation error when plugging in the roots. Let me verify x =1 again.

<prompt_4.5> Verify the calculations for x = 1 and x = -2 as roots and confirm the factorization. </prompt_4.5>

The quadratic factor is x² +4x +7. Let me check if this could be factored over integers.

<prompt_4.6> Verify if the quadratic factor x² + 4x + 7 can be factored over integers and confirm the integer roots. </prompt_4.6>

But just to make sure, perhaps there's a different approach. For example, factoring the original polynomial into quadratics. Let me try that.

<prompt_4.7> Attempt to factor the original polynomial into quadratics and confirm the factorization. </prompt_4.7>
</Plan_4>

<Execution_4>
<execution_4.1>
First, synthetic division for x=1:

Coefficients: 1,5,9,-1,-14

Bring down 1. Multiply by1:1. Add to 5:6. Multiply by1:6. Add to 9:15. Multiply by1:15. Add to -1:14. Multiply by1:14. Add to -14:0. Correct. So, cubic is x³ +6x² +15x +14.
</execution_4.1>

<execution_4.2>
If I multiply (x -1)(x +2)(x² +4x +7), do I get back the original polynomial?

First, multiply (x -1)(x +2): that's x² +2x -x -2 = x² +x -2.

Then multiply by x² +4x +7:

(x² +x -2)(x² +4x +7).

Multiply term by term:

x²*(x² +4x +7) = x⁴ +4x³ +7x²

x*(x² +4x +7) = x³ +4x² +7x

-2*(x² +4x +7) = -2x² -8x -14

Now, add them all together:

x⁴ +4x³ +7x² +x³ +4x² +7x -2x² -8x -14

Combine like terms:

x⁴ + (4x³ +x³) + (7x² +4x² -2x²) + (7x -8x) + (-14)

Which is x⁴ +5x³ +9x² -x -14.
</execution_4.2>

<execution_4.3>
Wait, we already checked x=7 and x=14, and they didn't work. So, seems like only 1 and -2 are integer roots. Therefore, the answer should be 1 and -2.
</execution_4.3>

<execution_4.4>
First, synthetic division for x =1:

Coefficients: 1,5,9,-1,-14

Bring down 1.

1*1 =1; 5+1=6

6*1=6; 9+6=15

15*1=15; -1+15=14

14*1=14; -14 +14=0. Correct. So the cubic is x³ +6x² +15x +14.

Then synthetic division for x = -2:

1,6,15,14

Bring down 1.

1*(-2) = -2; 6 + (-2) =4

4*(-2) = -8; 15 + (-8) =7

7*(-2) = -14; 14 + (-14)=0. Correct. So quadratic is x² +4x +7.
</execution_4.4>

<execution_4.5>
Plugging x=1 into the original equation: 1 +5 +9 -1 -14. 1+5=6, 6+9=15, 15-1=14, 14-14=0. Correct. Then x=-2: 16 -40 +36 +2 -14. 16-40=-24, +36=12, +2=14, -14=0. Correct. So those roots are confirmed.
</execution_4.5>

<execution_4.6>
The discriminant is 16 -28 = -12, so no real roots. Therefore, the only integer roots are 1 and -2. Therefore, the answer should be x=1 and x=-2.
</execution_4.6>

<execution_4.7>
Suppose the quartic factors into (x² + ax + b)(x² + cx + d). Then expanding:

x⁴ + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + bd.

Comparing coefficients with original polynomial x⁴ +5x³ +9x² -x -14.

So, we have:

1. a + c =5

2. ac + b + d =9

3. ad + bc = -1

4. bd = -14

We need to find integers a, b, c, d satisfying these equations. Since bd = -14, possible integer pairs (b,d) are (1,-14), (-1,14), (2,-7), (-2,7), (7,-2), (-7,2), (14,-1), (-14,1).

Let's try different pairs.

First pair: b=2, d=-7. Then bd=2*(-7)=-14. Now, equation 3: ad + bc = a*(-7) + c*2 = -7a +2c = -1. From equation 1: a + c=5. Let me solve for c=5 -a. Substitute into equation 3: -7a +2*(5 -a)= -7a +10 -2a= -9a +10= -1. So, -9a= -11 => a=11/9, not integer. Discard.

Next pair: b=-2, d=7. Then equation 3: a*7 + c*(-2)=7a -2c= -1. From a + c=5, so c=5 -a. Substitute into equation 3: 7a -2*(5 -a)=7a -10 +2a=9a -10= -1 => 9a=9 => a=1. Then c=5 -1=4. Now check equation 2: ac + b + d=1*4 + (-2) +7=4 -2 +7=9. Which matches. So, yes, this works. Thus, the quartic factors into (x² +ax +b)(x² +cx +d)= (x² +1x -2)(x² +4x +7). Which matches our earlier factorization: (x -1)(x +2)(x² +4x +7). Because (x² +x -2) factors further into (x +2)(x -1). So that's consistent.

Therefore, the quadratic x² +4x +7 has no real roots, so the only integer roots are x=1 and x=-2.
</execution_4.7>
</Execution_4>

<Plan_5>
Based on execution_4.7:
**Final Answer**
The integer roots are \boxed{1} and \boxed{-2}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the integer roots of the given equation are $\boxed{1}$ and $\boxed{-2}$.
</Final_answer>
</think>