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

 Jul 24, 2018
 #1
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PLEASE post  readable questions.

 Jul 24, 2018
 #2
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The product of $3t^2+5t+a$ and $4t^2+bt-2$ is $12t^4+26t^3-8t^2-16t+6$. What is $a+b$?

 

\(\begin{array}{|rcll|} \hline && (3t^2+5t+a)(4t^2+bt-2) \\ &=& 12t^4+3bt^3-6t^2+20t^3+5bt^2-10t+4at^2+abt-2a \\ &=& 12t^4+(3b+20)t^3 +(-6+5b+4a)t^2 +(-10+ab)t -2a \\ \hline \text{compare}&=&12t^4+26t^3-8t^2-16t+6 \\\\ -2a &=& 6 \quad | \quad : (-2) \\ \mathbf{a} & \mathbf{=} & \mathbf{-3} \\\\ 3b+20 &=& 26 \\ 3b &=& 6 \\ \mathbf{b} & \mathbf{=} & \mathbf{2} \\\\ a+b &=& -3+2 \\ \mathbf{a+b} & \mathbf{=} & \mathbf{-1} \\ \hline \end{array} \)

 

 

laugh

 Jul 25, 2018

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