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$(3t^2 + 5t + a ) (4t^2 + bt - 2) = 12t^4 +26t^3 - 8t^2 - 16t + 6$

What is a + b?

The answer would be:
$6=-2a$
$a=-3$
$-16t=(5t\cdot-2)+a\cdot bt\Longrightarrow-16t=-10t+(-3)bt\Longrightarrow b=2$
$a+b=-1$

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SolveAlways[] makes quick work of your problem:

SolveAlways[(3 t^2 + 5 t + a) (4 t^2 + b t - 2) == 12 t^4 + 26 t^3 - 8 t^2 - 16 t + 6, t]
   {{a -> -3, b -> 2}}
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One way could be

ClearAll[t,a,b}
z1 = CoefficientList[(3*t^2 + 5*t + a)*(4*t^2 + b*t - 2), t];
z2 = CoefficientList[12*t^4 + 26*t^3 - 8*t^2 - 16*t + 6, t];
eqs = Thread[z1 == z2];
Solve[eqs,{a,b}]

Mathematica graphics

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try this

Reduce[ForAll[t, (3*t^2 + 5*t + a)*(4*t^2 + b*t - 2) == 12*t^4 + 26*t^3 - 8*t^2 - 16*t + 6], a]

b == 2 && a == -3

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