ACSEE (Form 6) Algebra — practice questions

Explore 2 free Algebra questions in ACSEE (Form 6) Advanced Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Algebra questions and exercises with answers

Showing questions 1–2 of 2.

Expanded by the binomial theorem in ascending powers of $x$ up to $x^3$, $\dfrac{1}{(4-x)^2}$ is:Read the answer and worked solution →Before splitting into partial fractions, $\dfrac{x^2-2x+1}{(x+1)^2}$ is first written as:Read the answer and worked solution →

Worked practice questions

Expanded by the binomial theorem in ascending powers of $x$ up to $x^3$, $\dfrac{1}{(4-x)^2}$ is:

$\frac{1}{16}+\frac{x}{16}+\frac{3x^2}{256}+\frac{x^3}{256}$
$\frac{1}{16}+\frac{x}{32}+\frac{3x^2}{256}+\frac{x^3}{256}$
$\frac{1}{4}+\frac{x}{32}+\frac{3x^2}{256}+\frac{x^3}{256}$
$\frac{1}{16}-\frac{x}{32}+\frac{3x^2}{256}-\frac{x^3}{256}$

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Correct answer: B. $\frac{1}{16}+\frac{x}{32}+\frac{3x^2}{256}+\frac{x^3}{256}$

1. Factor the 4 out: $\dfrac{1}{(4-x)^2}=\dfrac{1}{16}\left(1-\dfrac{x}{4}\right)^{-2}$. 2. Apply $(1+u)^n=1+nu+\dfrac{n(n-1)u^2}{2!}+\dfrac{n(n-1)(n-2)u^3}{3!}$ with $n=-2$ and $u=-\dfrac{x}{4}$. 3. The bracket becomes $1+\dfrac{x}{2}+\dfrac{3x^2}{16}+\dfrac{x^3}{16}$ — every sign is positive because $n$ is negative and $u$ is negative. 4. Multiply through by $\dfrac{1}{16}$. 5. This gives $\dfrac{1}{16}+\dfrac{x}{32}+\dfrac{3x^2}{256}+\dfrac{x^3}{256}$. The alternating-sign option forgets that the two negatives cancel. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 6: Algebra_

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Before splitting into partial fractions, $\dfrac{x^2-2x+1}{(x+1)^2}$ is first written as:

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

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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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