# Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform Lleft{3e^{-4t}-t^{2}+6t-9right}

Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform
$L\left\{3{e}^{-4t}-{t}^{2}+6t-9\right\}$
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$\text{Step 1}$
$L\left\{3{e}^{-4t}-{t}^{2}+6t-9\right\}$
$L\left({e}^{at}\right)=\frac{1}{s-a}$
$L\left({t}^{n}\right)=\frac{n!}{{s}^{n+1}}$
$\text{Step 2}$
$L\left(3{e}^{-4t}-{t}^{2}+6t-9\right)$
$=3\left(\frac{1}{s+4}\right)-\frac{2!}{{s}^{2+1}}+6\left(\frac{1!}{{s}^{1+1}}\right)-9\left(\frac{1}{s}\right)$
$=\frac{3}{s+4}-\frac{2}{{s}^{3}}+\frac{6}{{s}^{2}}-\frac{9}{s}$
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