<Question>
The product of $3t^2+5t+a$ and $4t^2+bt-2$ is $12t^4+26t^3-8t^2-16t+6$. What is $a+b$?
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the values of 'a' and 'b' such that when I multiply the two quadratic polynomials \(3t^2 + 5t + a\) and \(4t^2 + bt - 2\), the result is the given quartic polynomial \(12t^4 + 26t^3 -8t^2 -16t +6\). Then, I can add 'a' and 'b' to get the answer. Hmm, how do I approach this?

Well, usually when you multiply two polynomials, you use the distributive property, right? So maybe I should multiply these two quadratics out and then compare the coefficients with the given product. That way, I can set up equations for 'a' and 'b'.

Let me start by expanding \((3t^2 + 5t + a)(4t^2 + bt - 2)\). I'll do it term by term.

<prompt_1.1> Expand the polynomial multiplication of \(3t^2 + 5t + a\) and \(4t^2 + bt - 2\) term by term. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, multiply \(3t^2\) by each term in the second polynomial:

\(3t^2 \times 4t^2 = 12t^4\)

\(3t^2 \times bt = 3b t^3\)

\(3t^2 \times (-2) = -6t^2\)

Next, multiply \(5t\) by each term in the second polynomial:

\(5t \times 4t^2 = 20t^3\)

\(5t \times bt = 5b t^2\)

\(5t \times (-2) = -10t\)

Then, multiply 'a' by each term in the second polynomial:

\(a \times 4t^2 = 4a t^2\)

\(a \times bt = ab t\)

\(a \times (-2) = -2a\)

Now, let me combine all these terms:

\(12t^4 + 3b t^3 + (-6t^2) + 20t^3 + 5b t^2 + (-10t) + 4a t^2 + ab t + (-2a)\)

Now, I need to combine like terms. Let's group them by the power of t:

- \(t^4\): 12t^4
- \(t^3\): 3b t^3 + 20t^3 = (3b + 20) t^3
- \(t^2\): -6t^2 + 5b t^2 + 4a t^2 = (-6 + 5b + 4a) t^2
- \(t\): -10t + ab t = (-10 + ab) t
- Constants: -2a
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
According to the problem, this product should equal \(12t^4 + 26t^3 -8t^2 -16t +6\). So, the coefficients of corresponding powers of t must be equal. Let's write down the equations:

<prompt_2.1> Set up equations for the coefficients of the expanded polynomial by comparing them to the given product. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
1. Coefficient of \(t^4\): 12 = 12. That checks out, so no equation here.
2. Coefficient of \(t^3\): 3b + 20 = 26
3. Coefficient of \(t^2\): -6 + 5b + 4a = -8
4. Coefficient of \(t\): -10 + ab = -16
5. Constant term: -2a = 6

Okay, so now I have a system of equations here. Let's solve them step by step.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Starting with the constant term equation (equation 5):

-2a = 6

Divide both sides by -2:

<prompt_3.1> Solve the system of equations to find the values of \(a\) and \(b\). </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
a = 6 / (-2) = -3

So, a = -3.

Now, let's move to the equation for \(t^3\) (equation 2):

3b + 20 = 26

Subtract 20 from both sides:

3b = 26 - 20 = 6

Divide by 3:

b = 6 / 3 = 2

So, b = 2.

Wait, but we also have equation 4 (coefficient of t):

-10 + ab = -16

Let's check if this holds with a = -3 and b = 2.

Compute ab:

ab = (-3)(2) = -6

Then, -10 + (-6) = -16.

Yes, that matches the equation. So that's consistent. So all equations are satisfied with a = -3 and b = 2.

Therefore, a + b = -3 + 2 = -1.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me double-check to make sure I didn't make any mistakes in the multiplication or combining terms.

Let me redo the expansion with a = -3 and b = 2 to see if we indeed get the given polynomial.

<prompt_4.1> Redo the polynomial multiplication with \(a = -3\) and \(b = 2\) to verify the result. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Original polynomials:

First: 3t² + 5t + a = 3t² + 5t -3

Second: 4t² + bt -2 = 4t² + 2t -2

Multiply them:

(3t² +5t -3)(4t² +2t -2)

Multiply each term:

3t²*4t² = 12t⁴

3t²*2t = 6t³

3t²*(-2) = -6t²

5t*4t² = 20t³

5t*2t =10t²

5t*(-2) = -10t

-3*4t² = -12t²

-3*2t = -6t

-3*(-2) =6

Now, combine all terms:

12t⁴ +6t³ -6t² +20t³ +10t² -10t -12t² -6t +6

Combine like terms:

t⁴: 12t⁴

t³: 6t³ +20t³ =26t³

t²: -6t² +10t² -12t² = (-6 +10 -12)t² = (-8)t²

t: -10t -6t = -16t

Constants: +6

So, the product is 12t⁴ +26t³ -8t² -16t +6, which matches exactly the given polynomial. Therefore, the values of a = -3 and b = 2 are correct, so a + b = -1.
</execution_4.1>
</Execution_4>

<Final_answer>
Therefore, the sum of $a$ and $b$ is $\boxed{-1}$.
</Final_answer>
</think>