<Question>
Find the product of the $y$-coordinates of all the distinct solutions $(x,y)$ for the two equations $y=x^2-8$ and $y^2=-5x+44$.
</Question>

<think>
<Plan_1>
Okay, so I need to find the product of the y-coordinates of all the distinct solutions (x, y) for the two equations y = x² - 8 and y² = -5x + 44. Hmm, let's start by understanding what the problem is asking. We have two equations: one is a parabola opening upwards, and the other is a parabola but in terms of y², which might open to the left or right. The question is about finding all the points (x, y) that satisfy both equations simultaneously and then multiplying their y-coordinates together.
First, since both equations are equal to y or y², maybe I can substitute one into the other. The first equation gives y in terms of x, so substituting y = x² - 8 into the second equation y² = -5x + 44 should work.
Let me try that.

So substituting, we get (x² - 8)² = -5x + 44.

Now, expanding the left side: (x² - 8)² is x⁴ - 16x² + 64. So the equation becomes x⁴ - 16x² + 64 = -5x + 44.

<prompt_1.1> Expand and simplify the equation (x² - 8)² = -5x + 44 to form a polynomial equation. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's bring all terms to one side: x⁴ - 16x² + 64 + 5x - 44 = 0. Simplify the constants: 64 - 44 is 20. So the equation is x⁴ - 16x² + 5x + 20 = 0.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, we have a quartic equation: x⁴ - 16x² + 5x + 20 = 0. Solving quartic equations can be tricky, but maybe this factors into simpler polynomials. Let's try to factor it.

Looking for rational roots using the Rational Root Theorem. Possible rational roots are factors of 20 over factors of 1, so ±1, ±2, ±4, ±5, ±10, ±20.

Let me test x=1: 1 -16 +5 +20 = 1 -16 is -15, +5 is -10, +20 is 10 ≠ 0.

<prompt_2.1> Test possible rational roots using the Rational Root Theorem and factor the polynomial if possible. </prompt_2.1>
-----

But wait, let me check if there are other solutions. Suppose there are multiple y's for each x. Wait, but each x gives a unique y from the first equation, so each solution (x, y) has a unique y once x is found. So the solutions are the intersections of the two curves, and each intersection point has one x and one y. So the total number of solutions is the number of real roots of the quartic.
Since it's a quartic, there can be up to 4 real roots. But depending on the curves, maybe there are fewer. However, the problem states "all the distinct solutions," so we need to consider all real solutions.
</Plan_2>

<Execution_2>
<execution_2.1>
x=2: 16 - 64 +10 +20 = 16-64=-48+10=-38+20=-18≠0.

x= -1: 1 -16 -5 +20 = 0. Wait, 1 -16 is -15, -5 is -20, +20 is 0. So x = -1 is a root.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Great, so (x + 1) is a factor. Let's perform polynomial division or use synthetic division.

Using synthetic division for root x = -1:

<prompt_3.1> Perform synthetic division using the root x = -1 and factor the polynomial. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Coefficients: 1 (x⁴), 0 (x³), -16 (x²), 5 (x), 20 (constant).

Bring down 1. Multiply by -1: -1. Add to next coefficient: 0 + (-1) = -1. Multiply by -1: 1. Add to next coefficient: -16 +1 = -15. Multiply by -1:15. Add to next coefficient:5 +15=20. Multiply by -1: -20. Add to last coefficient:20 + (-20)=0. So the quotient is x³ - x² -15x +20.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
So, we have (x + 1)(x³ - x² -15x +20) = 0. Now, let's factor the cubic. Again, try rational roots: possible roots are ±1, ±2, ±4, ±5, ±10, ±20.

<prompt_4.1> Factor the cubic polynomial by testing possible rational roots. </prompt_4.1>
-----

Now, moving on to the cubic factor: x³ - x² -15x +20. Let's check for roots again. Let's try x=1: 1 -1 -15 +20=5≠0. x=4: 64 -16 -60 +20= (64-16)=48, (48-60)=-12, (-12+20)=8≠0. x=5: 125 -25 -75 +20= (125-25)=100, (100-75)=25, 25+20=45≠0. x=2:8 -4 -30 +20= (8-4)=4, (4-30)=-26, (-26+20)=-6≠0. Hmm, maybe x= -2?

<prompt_4.2> Continue factoring the cubic polynomial by testing additional possible roots. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Testing x=1: 1 -1 -15 +20 = 5 ≠0. x= -1: -1 -1 +15 +20=33≠0. x=2: 8 -4 -30 +20= -6≠0. x=4: 64 -16 -60 +20=8≠0. x=5: 125 -25 -75 +20=45≠0. x= -2: -8 -4 +30 +20=38≠0. Wait, maybe x=5? Wait, tried that. How about x= -4? Maybe, but perhaps there's a mistake here. Wait, let's try x=4 again. 4³=64, 4²=16, so 64 -16 -60 +20= (64-16)=48, 48-60= -12, -12+20=8. Not zero. Hmm.
</execution_4.1>

<execution_4.2>
Wait, tried x=2 earlier. Maybe x= -4? Let's check x= -4: (-64) -16 -(-60) +20= (-64-16)= -80 +60= -20 +20=0. Wait, x= -4: (-4)^3= -64, -(-4)^2= -16, -15*(-4)=60, +20. So total: -64 -16 +60 +20= (-80) +80=0. So x= -4 is a root. Therefore, (x + 4) is a factor.

Using synthetic division for root x= -4 on the cubic x³ -x² -15x +20.

Coefficients: 1 (x³), -1 (x²), -15 (x), 20 (constant).

Bring down 1. Multiply by -4: -4. Add to next coefficient: -1 + (-4)= -5. Multiply by -4: 20. Add to next coefficient: -15 +20=5. Multiply by -4: -20. Add to last coefficient:20 + (-20)=0. So the quotient is x² -5x +5.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, maybe x=5?

<prompt_5.1> Verify if the cubic polynomial has any real roots using graphical reasoning. </prompt_5.1>
-----

Based on execution_4.2:
Therefore, the cubic factors into (x +4)(x² -5x +5). So the original quartic factors into (x +1)(x +4)(x² -5x +5)=0.

Now, solving x² -5x +5=0 using quadratic formula: x=(5 ±√(25 -20))/2= (5 ±√5)/2. So the roots are x= -1, x= -4, x=(5 +√5)/2, and x=(5 -√5)/2.

So all the x-coordinates of the solutions are these four. Now, we need to find the corresponding y-coordinates for each x using y =x² -8.

<prompt_5.2> Find the corresponding y-coordinates for each x using the equation y = x² - 8. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
Wait, x=5 gives 125 -25 -75 +20=45. x= -1 gives -1 -1 +15 +20=33. Hmm. Maybe x=10? That seems too big. Wait, maybe there's a mistake here. Alternatively, maybe the cubic doesn't factor further, but that seems unlikely. Let me check again.
</execution_5.1>

<execution_5.2>
Let's compute each:

1. x = -1: y = (-1)^2 -8 =1 -8= -7.
2. x = -4: y = (-4)^2 -8=16 -8=8.
3. x=(5 +√5)/2: Let's compute x² first. [(5 +√5)/2]^2 = (25 +10√5 +5)/4 = (30 +10√5)/4 = (15 +5√5)/2. Then y = (15 +5√5)/2 -8 = (15 +5√5 -16)/2 = (-1 +5√5)/2.
4. x=(5 -√5)/2: Similarly, x² = [(5 -√5)/2]^2 = (25 -10√5 +5)/4 = (30 -10√5)/4 = (15 -5√5)/2. Then y = (15 -5√5)/2 -8 = (15 -5√5 -16)/2 = (-1 -5√5)/2.

So the y-coordinates are: -7, 8, (-1 +5√5)/2, and (-1 -5√5)/2.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
Now, the problem asks for the product of these y-coordinates. Let's compute that.

First, multiply the two irrational y-coordinates:

[(-1 +5√5)/2] * [(-1 -5√5)/2] = [(-1)^2 - (5√5)^2]/(2*2) = [1 - 25*5]/4 = [1 -125]/4 = (-124)/4 = -31.

Then, multiply this result with the other two y-coordinates: -7 and 8. So total product is (-31) * (-7) * 8.

<prompt_6.1> Calculate the product of all y-coordinates. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
First, (-31)*(-7) = 217. Then 217*8= 1736.

Wait, let me check that again. 217*8: 200*8=1600, 17*8=136, so 1600+136=1736. So the product is 1736.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Hmm, but let me verify if this is correct because quartic equations can sometimes have multiple roots, and maybe I missed something. Alternatively, perhaps there's a smarter way to compute the product without finding each root.

<prompt_7.1> Verify the product of the y-coordinates using Vieta's formula. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Wait, the original quartic equation in x was x⁴ -16x² +5x +20=0. When we substituted y =x² -8 into y² = -5x +44, we got x⁴ -16x² +5x +20=0. Then, we found the roots x=-1, x=-4, and the two roots from the quadratic. Then, for each x, we computed y. Then multiplied all four y's.

Alternatively, maybe there's a way to relate the product of the y's to the coefficients of the equation. Let's think. If we have four y's: y1, y2, y3, y4. Then the product y1*y2*y3*y4 is equal to the product of the roots of some polynomial equation in y. But how to relate that?

But each y is related to x via y =x² -8. So perhaps if we can find a polynomial equation in y whose roots are the y's corresponding to the x's of the original system. Then, the product of the roots of that polynomial would be the product we need. Let's try that approach.

Starting with the two equations:

1. y =x² -8 => x² = y +8.
2. y² = -5x +44.

From the first equation, x = ±√(y +8). But substituting this into the second equation would lead to y² = -5*(±√(y +8)) +44. That introduces square roots, which might complicate things. Alternatively, maybe square both sides to eliminate the square root, but that might get messy.

Alternatively, perhaps express x from the second equation in terms of y². From the second equation: x = (44 - y²)/5. Then substitute this into the first equation: y =x² -8. So substituting x gives y = [(44 - y²)/5]^2 -8.

Let's compute that:

y = [(44 - y²)^2]/25 -8.

Multiply both sides by 25 to eliminate the denominator:

25y = (44 - y²)^2 - 200.

Expand the right side: (44 - y²)^2 = 44² - 88y² + y⁴ = 1936 -88y² + y⁴. So:

25y = 1936 -88y² + y⁴ -200.

Simplify the constants: 1936 -200 = 1736. So:

25y = y⁴ -88y² +1736.

Bring all terms to one side:

y⁴ -88y² -25y +1736 =0.

So the quartic equation in y is y⁴ -88y² -25y +1736=0. The roots of this equation are the y-coordinates of the solutions. Then, the product of the roots of this quartic equation is the constant term (if the leading coefficient is 1) multiplied by (-1)^4=1. So the product is 1736. Wait, but let's check.

Wait, for a quartic equation y⁴ + a y³ + b y² + c y + d =0, the product of the roots is d if the leading coefficient is 1. Wait, no: actually, for a quartic equation y⁴ + a y³ + b y² + c y + d =0, the product of the roots is d (since the product is (-1)^4 * (d/a_coeff)), but since a_coeff=1 here, it's d. Wait, but in our case, the equation is y⁴ -88y² -25y +1736=0. So a=0 (coefficient of y³), b=-88, c=-25, d=1736. Therefore, the product of the roots is d=1736. Therefore, the product of all y-coordinates is 1736, which matches our previous calculation. So that's a good consistency check.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
But wait, but in our previous step-by-step, we had four y's: -7, 8, (-1 +5√5)/2, and (-1 -5√5)/2. Multiplying all four gives 1736.
Let me confirm that.

<prompt_8.1> Verify the product using the roots found earlier. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
Multiply the two irrational terms: [(-1 +5√5)/2]*[(-1 -5√5)/2] = [(-1)^2 - (5√5)^2]/4 = (1 - 25*5)/4 = (1 -125)/4 = (-124)/4 = -31. Then, multiply by -7 and 8: (-31)*(-7)=217, 217*8=1736. Yes, correct.

Alternatively, using Vieta's formula, since the quartic in y is y⁴ -88y² -25y +1736=0, the product of the roots is 1736. So that's a much quicker way. Therefore, the answer is 1736.
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
But let me think again: when we substituted x = (44 - y²)/5 into y =x² -8, we derived the quartic equation in y. Each solution y corresponds to an x via x=(44 - y²)/5. But we need to ensure that for each y, there exists an x such that both equations are satisfied. However, since we derived the equation by substitution, all roots of the quartic in y should correspond to valid solutions (x,y) of the original system. Therefore, the four y's are exactly the y-coordinates of the distinct solutions, and their product is 1736.

But just to be thorough, let's check if each y indeed corresponds to a valid x and that there are four solutions.

<prompt_9.1> Verify the validity of each y and ensure four solutions exist. </prompt_9.1>
</Plan_9>

<Execution_9>
<execution_9.1>
For y=-7: x=(44 - (-7)^2)/5=(44 -49)/5=(-5)/5=-1. Which matches our earlier x=-1. Then, y=-7, which is correct.

For y=8: x=(44 -8^2)/5=(44 -64)/5=(-20)/5=-4. Which matches x=-4, y=8.

For y=(-1 +5√5)/2: Let's compute x=(44 - y²)/5. Let's see, from earlier, when x=(5 +√5)/2, y=x² -8. Let's compute y²: [(-1 +5√5)/2]^2 = (1 -10√5 +125)/4 = (126 -10√5)/4. Then, -5x +44= -5*(5 +√5)/2 +44= (-25 -5√5)/2 +44= (-25 -5√5 +88)/2= (63 -5√5)/2. But y² is (126 -10√5)/4 = (63 -5√5)/2, which matches. So yes, correct.

Similarly for the other root. Therefore, all four y's correspond to valid solutions. Therefore, the product is indeed 1736.
</execution_9.1>
</Execution_9>

<Plan_10>
Based on execution_9.1:
Alternatively, if there were extraneous solutions introduced during substitution, but in this case, since we substituted one equation into another, the solutions should be valid. Therefore, the answer is 1736.

But wait, the problem says "distinct solutions". Let me check if any of these solutions are the same. For example, is there a case where two different x's give the same y? For instance, the two x's (5 +√5)/2 and (5 -√5)/2 give two different y's, as we saw. Similarly, the other y's are distinct. So all four y's are distinct. Therefore, all four solutions are distinct, so the product is 1736.

But let me confirm once again. The quartic equation in y is y⁴ -88y² -25y +1736=0, and the product of its roots is 1736, as the constant term is 1736 and the leading coefficient is 1. Therefore, the product is 1736.

**Final Answer**
The product of the y-coordinates is \boxed{1736}.
</Plan_10>

<Execution_10>
No prompt to execute.
</Execution_10>

<Final_answer>
Therefore, the product of the $y$-coordinates of all the distinct solutions $(x,y)$ for the given equations is $\boxed{1736}$.
</Final_answer>
</think>