alleryBefore splitting into partial fractions, $\dfrac{x^2-2x+1}{(x+1)^2}$ is first written as:
A$1-\frac{2x}{(x+1)^2}$
B$1+\frac{4x}{(x+1)^2}$
C$\frac{4x}{(x+1)^2}-1$
D$1-\frac{4x}{(x+1)^2}$
Answer & Solution
Correct answer: D. $1-\frac{4x}{(x+1)^2}$
1. Numerator and denominator have the same degree, so divide before splitting.
2. Expand the denominator: $(x+1)^2=x^2+2x+1$.
3. Subtract it from the numerator: $(x^2-2x+1)-(x^2+2x+1)=-4x$.
4. So $\dfrac{x^2-2x+1}{(x+1)^2}=1+\dfrac{-4x}{(x+1)^2}=1-\dfrac{4x}{(x+1)^2}$.
5. The $-2x$ version subtracts only the middle term rather than the whole denominator.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 6: Algebra_
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