Okay, let's see. I need to find the values of 'a' and 'b' such that when I multiply the two polynomials 3t² + 5t + a and 4t² + bt - 2, the result is 12t⁴ + 26t³ -8t² -16t +6. Then, once I have 'a' and 'b', I can add them together to get a + b. So, the key here is to perform the multiplication of the two polynomials and then equate the coefficients of corresponding powers of t to the given product. Then, solve the resulting system of equations for 'a' and 'b'.

First, let me recall how to multiply two polynomials. Each term in the first polynomial must be multiplied by each term in the second polynomial, and then like terms are combined. So, let's start by expanding the product (3t² + 5t + a)(4t² + bt - 2).

Let me write down each term's multiplication step by step.

First, multiply 3t² by each term in the second polynomial:
- 3t² * 4t² = 12t⁴
- 3t² * bt = 3b t³
- 3t² * (-2) = -6t²

Next, multiply 5t by each term in the second polynomial:
- 5t * 4t² = 20t³
- 5t * bt = 5b t²
- 5t * (-2) = -10t

Then, multiply 'a' by each term in the second polynomial:
- a * 4t² = 4a t²
- a * bt = ab t
- a * (-2) = -2a

Now, let's write all these terms out:

12t⁴ + 3b t³ -6t² + 20t³ + 5b t² -10t + 4a t² + ab t -2a

Now, combine like terms:

- The t⁴ term: 12t⁴
- The t³ terms: 3b t³ + 20t³ = (3b + 20) t³
- The t² terms: -6t² + 5b t² + 4a t² = (-6 + 5b + 4a) t²
- The t terms: -10t + ab t = (ab -10) t
- The constant term: -2a

So, the expanded polynomial is:

12t⁴ + (3b + 20) t³ + (-6 + 5b + 4a) t² + (ab -10) t -2a

Now, this must equal the given polynomial:

12t⁴ +26t³ -8t² -16t +6

Therefore, the coefficients of corresponding powers of t must be equal. Let's write down the equations:

1. Coefficient of t⁴: 12 = 12 (which is already satisfied, so no new information here)
2. Coefficient of t³: 3b + 20 = 26
3. Coefficient of t²: -6 + 5b + 4a = -8
4. Coefficient of t: ab -10 = -16
5. Constant term: -2a = 6

So, now we have a system of equations:

From equation 5: -2a = 6 → a = 6 / (-2) = -3

So, a = -3

Then, let's check equation 2: 3b + 20 = 26 → 3b = 26 -20 =6 → b=6/3=2

So, b=2

Now, let's verify these values in equations 3 and 4.

Equation 3: -6 +5b +4a = -6 +5*(2) +4*(-3) = -6 +10 -12 = (-6 -12) +10 = -18 +10 = -8. Which matches the given coefficient.

Equation 4: ab -10 = (-3)(2) -10 = -6 -10 = -16. Which also matches.

So, everything checks out. Therefore, a = -3 and b = 2. Therefore, a + b = -3 + 2 = -1.

Wait, but let me double-check the calculations step by step to make sure I didn't make any arithmetic errors.

First, equation 5: -2a =6 → a= -3. Correct.

Equation 2: 3b +20 =26 → 3b=6 → b=2. Correct.

Equation 3: -6 +5b +4a. Plugging b=2 and a=-3:

-6 +5*2 +4*(-3) = -6 +10 -12. 10 -6 is 4, 4 -12 is -8. Correct.

Equation 4: ab -10. a=-3, b=2: (-3)(2)= -6. -6 -10= -16. Correct.

So, all equations are satisfied. Therefore, a + b = -3 +2 = -1.

I think that's the answer. But let me just go through the multiplication again to be thorough.

Multiply (3t² +5t -3) by (4t² +2t -2).

First 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, combining all terms:

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

Combine like terms:

12t⁴

6t³ +20t³=26t³

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

-10t -6t= -16t

+6

So, the product is 12t⁴ +26t³ -8t² -16t +6, which matches the given polynomial. So, everything is correct.

Therefore, a + b = -3 +2 = -1.
Therefore, the final answer is $\boxed{-1}$.
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