# find the laplace transform of this equation: $$\displaystyle{4}-{4}{t}+{2}{t}^{{2}}$$ What

find the laplace transform of this equation:
$4-4t+2{t}^{2}$
What I am doing:
$\frac{4}{s}-\frac{4}{{s}^{2}}+\frac{4}{{s}^{3}}$
$\frac{4{s}^{2}-4s}{{s}^{3}}+\frac{4}{{s}^{3}}$
$\frac{4{s}^{2}-4s+4}{s3}$
But I am getting the wrong answer, can you please tell me what I am doing wrong?
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Gonarsu2dw8
Using the linearity of the Laplace transform, we have
$L\left(4-4t+2{t}^{2}\right)=4L\left(1\right)-4L\left(t\right)+2L\left({t}^{2}\right)=\frac{4}{s}-\frac{4}{{s}^{2}}+\frac{4}{{s}^{3}}$
The lowest common denominator is ${s}^{3}$, thus
$\frac{4}{s}-\frac{4}{{s}^{2}}+\frac{4}{{s}^{3}}=\frac{4{s}^{2}}{{s}^{3}}-\frac{4s}{{s}^{3}}+\frac{4}{{s}^{3}}=\frac{4{s}^{2}-4s+4}{{s}^{3}}=\frac{4\left({s}^{2}-s+1\right)}{{s}^{3}}$