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Kindly can someone please help me solve this particular question from the start till the end. I don't know how to solve partial fractions with improper fractions. Please show me step-by-step working to this.

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

Please help.

Your assistance is much appreciated. Many thanks.

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HINT:

Observe that the highest power of $x$ are $4,3$ in the numerator & the denominator respectively.

Using Partial Fraction Decomposition,

$$\frac{2x^4-2x^3+x}{(2x-1)^2(x-2)}=Ax+B+\frac C{2x-1}+\frac D{(2x-1)^2}+\frac E{x-2}$$

More generally for

$\displaystyle\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0}=\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{\prod (d_ix-c_i)^{n_i}}$

where $m\ge n,$

we can write $\displaystyle\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0}$

$$=e_{m-n}x^{m-n}+e_{m-n-1}x^{m-n-1}+\cdots+e_1x+e_0+\sum\left(\frac {f_{i1}}{d_ix-c_i}+\frac {f_{i2}}{(d_ix-c_i)^2}+\cdots+\frac {f_{i(n_i-1)}}{(d_ix-c_i)^{n_i-1}}+\frac{f_{in_i}}{(d_ix-c_i)^{n_i}}\right)$$

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use long division to find A and B THEN equate the remainder over the denominatore with the sum of partial fractions

add the parial fractions and equate the numeratores of both sides then substitute x=2 then x=1/2 and x=0 you will get A=1/2 B=1 C=-1/4 D= -1/4 E=2

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