$\displaystyle t^2+4t+3=A(1+t^2)+(Mt+N)2t$

I got that $\displaystyle A=3$

But how do I get $\displaystyle Mt$and$\displaystyle N$

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- Dec 10th 2006, 06:00 AMtotalnewbieHow to find Mt and N
$\displaystyle t^2+4t+3=A(1+t^2)+(Mt+N)2t$

I got that $\displaystyle A=3$

But how do I get $\displaystyle Mt$and$\displaystyle N$ - Dec 10th 2006, 06:38 AMThePerfectHacker
One way is to expand,

Expand,

$\displaystyle t^2+4t+3=(A+2M)t^2+(2N)t+(A)$

Corresponding parts are equal,

$\displaystyle \left\{ \begin{array}{cccc}A&+2M& \,&=1 \\ \,&\,&N&=4\\ A&\,&\,& = 3 \end{array} \right\}$

You can solve the first equation since you know the third.

Thus,

$\displaystyle \left\{ \begin{array}{c}A=3\\M=-1\\N=4 \end{array} \right\}$