If $x^3-4x^2 + 19 = 6 ( x-1)$, then what is the value of $[x^2 + (\frac{1}{( x-4)}]$.


Given:  Expression: $x^3 – 4x^2 + 19 = 6 (x – 1)$.

To do: To find the value of $[x^2+(\frac{1}{( x-4)}]$.

Solution:

Given expression: $x^3-4x^2+19=6( x-1)$

$\Rightarrow x^3-4x^2+1+18=6x-6$

$\Rightarrow x^3-4x^2+1=6x-24$

$\Rightarrow x^3-4x^2+1=6(x-4)$ ...... $( i)$

$x^2+\frac{1}{( x-4)}=\frac{x^2( x-4)+1}{( x-4)}=\frac{( x^3-4x^2+1)}{( x-4)}$

From equation $( i)$

$\frac{( x^3-4x^2+1)}{( x-4)}=\frac{6( x-4)}{( x-4)}=6$     

$\therefore x^2+\frac{1}{( x-4)}=6$

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Updated on: 10-Oct-2022

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