If 3x2+x+1x-14=ax-1+bx-12+cx-13+dx-14then the determinator of abcd

3x2+x+1(x-1)4x=1=y ⇒ x=y+1 3(y+1)2+y+1+1y4 =3y2+7y+5y4 =3y2+7y3+5y4a=0 b=3, c=7, d=50375=0-21=-21

If 3x2+x+1x-14=ax-1+bx-12+cx-13+dx-14then the determinator of abcd

3x2+x+1(x-1)4x=1=y ⇒ x=y+1 3(y+1)2+y+1+1y4 =3y2+7y+5y4 =3y2+7y3+5y4a=0 b=3, c=7, d=50375=0-21=-21

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$If \frac{3 x^{2} + x + 1}{{(x - 1)}^{4}} = \frac{a}{x - 1} + \frac{b}{{(x - 1)}^{2}} + \frac{c}{{(x - 1)}^{3}} + \frac{d}{{(x - 1)}^{4}} then the determinator of [\begin{matrix} a & b \\ c & d \end{matrix}]$

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Detailed Solution

$\begin{array}{rcl} \frac{3 x^{2} + x + 1}{{(x - 1)}^{4}} \\ x & = & 1 = y \Rightarrow x = y + 1 \\ \frac{3 {(y + 1)}^{2} + y + 1 + 1}{y^{4}} \\ = & \frac{3 y^{2} + 7 y + 5}{y^{4}} \\ = & \frac{3}{y^{2}} + \frac{7}{y^{3}} + \frac{5}{y^{4}} \\ a & = & 0 b = 3, c = 7, d = 5 \\ |\begin{matrix} 0 & 3 \\ 7 & 5 \end{matrix}| & = & 0 - 21 = - 21 \end{array}$