MYP 3 · Maths
LAWS OF ALGEBRA
73 questions across 8 sub-topics
Use the Sub-Topic filter above to focus on one.
Index laws
Expansion laws
The zero index law
The negative index law
The distributive law
The product (a + b)(c + d)
Perfect square expansion
Difference of two squares
Index laws 10 questions
QUESTION 1
4 marks
Criterion A
Medium
Simplify each expression:
a.
$x^5 \times x^3$
[1] b.
$\dfrac{y^9}{y^4}$
[1] c.
$(a^3)^4$
[1] d.
$\dfrac{m^6 \times m^2}{m^3}$
[1] Show complete worked solution
(a)
$$x^8$$
(b)
$$y^5$$
(c)
$$a^{12}$$
(d)
$$m^5$$
QUESTION 2
3 marks
Criterion A
Medium
Simplify $\dfrac{p^4 \times p^7}{p^5 \times p^2}$, showing each index law used.
a.
Simplify the numerator and denominator separately first.
[2] b.
Complete the simplification.
[1] Show complete worked solution
(a)
Numerator: $p^{4+7}=p^{11}$. Denominator: $p^{5+2}=p^7$.
(b)
$$\frac{p^{11}}{p^7}=p^4$$
QUESTION 3
4 marks
Criterion B
Medium
Investigate what happens when you raise a power to another power, e.g. $(x^2)^3$.
a.
Expand $(x^2)^3$ fully (writing out all factors) to find the simplified result.
[2] b.
Repeat for $(x^3)^2$ and $(x^4)^2$. State the general rule connecting the two exponents to the final exponent.
[2] Show complete worked solution
(a)
$(x^2)^3=x^2\times x^2\times x^2 = x^{2+2+2}=x^6$
(b)
$(x^3)^2=x^6$. $(x^4)^2=x^8$. Rule: $(x^m)^n = x^{mn}$ — multiply the exponents.
QUESTION 4
3 marks
Criterion C
Medium
A student simplifies $x^5 \times x^3$ as $x^{15}$ (multiplying the exponents instead of adding).
a.
Explain the error, and verify the correct answer using $x=2$.
[3] Show complete worked solution
(a)
The student confused the multiplication rule ($x^m\times x^n=x^{m+n}$, ADD exponents) with the power-of-a-power rule ($(x^m)^n=x^{mn}$, MULTIPLY exponents) — these are different situations. Correct: $x^5\times x^3=x^8$. Check with $x=2$: $2^5\times2^3=32\times8=256$, and $2^8=256$ (while $2^{15}=32768$, clearly different).
QUESTION 5
3 marks
Criterion D
Medium
A cube-shaped storage container has side length $x^2$ cm.
a.
Write and simplify an expression for the volume of the cube.
[2] b.
If $x=2$, find the numerical volume.
[1] Show complete worked solution
(a)
$$V=(x^2)^3=x^6 \text{ cm}^3$$
(b)
$$V=2^6=64 \text{ cm}^3$$
QUESTION 6
7 marks
Criterion A
Medium
Simplify each expression using the index laws:
a.
$\dfrac{x^8 \times x^3}{x^5}$
[2] b.
$(x^4)^2 \div x^3$
[2] c.
$\dfrac{(x^3)^2 \times x}{x^4}$
[3] Show complete worked solution
(a)
$$\frac{x^{11}}{x^5}=x^6$$
(b)
$$x^8 \div x^3 = x^5$$
(c)
$$\frac{x^6\times x}{x^4}=\frac{x^7}{x^4}=x^3$$
QUESTION 7
5 marks
Criterion A
Hard
A cell culture's population model uses $P(t)=P_0 \times 2^{t/3}$, where $P_0$ is the initial population and $t$ is time in hours (the population doubles every 3 hours).
a.
If $P_0=500$, find the population after 9 hours.
[2] b.
Find, using index laws, an expression for the RATIO $\dfrac{P(12)}{P(6)}$ in simplest index form, and explain what this ratio represents physically.
[3] Show complete worked solution
(a)
$$P(9)=500\times2^{9/3}=500\times2^3=500\times8=4000$$
(b)
$$\frac{P(12)}{P(6)} = \frac{P_0\times2^{12/3}}{P_0\times2^{6/3}} = \frac{2^4}{2^2}=2^2=4$$ This represents how many times LARGER the population is at 12 hours compared to 6 hours — since 12 hours is 6 hours (2 doubling periods) later than 6 hours, the population has doubled twice, i.e. quadrupled.
QUESTION 8
6 marks
Criterion B
Hard
Investigate the pattern in $x^n \div x^n$ for different values of $n$, connecting it to the zero index law.
a.
Using the DIVISION index law ($x^m\div x^n=x^{m-n}$), simplify $x^5\div x^5$, $x^{10}\div x^{10}$, and $x^1\div x^1$.
[3] b.
Since ANY nonzero number divided by itself directly equals 1, and the index law gives $x^0$ for all these cases, what must $x^0$ always equal? Explain why this reasoning works for ANY exponent $n$, not just the specific ones tested.
[3] Show complete worked solution
(a)
$x^5\div x^5=x^0$. $x^{10}\div x^{10}=x^0$. $x^1\div x^1=x^0$ — all give $x^0$, regardless of the exponent value.
(b)
$x^0$ must always equal 1 (for $x\ne0$), since $x^n\div x^n$ is ALWAYS exactly 1 by direct calculation (anything divided by itself is 1), regardless of what $n$ is — and the index law ALWAYS gives $x^{n-n}=x^0$ for this same calculation. Since both methods calculate the identical quantity $x^n\div x^n$ for ANY $n$, they must always agree, proving $x^0=1$ universally (for nonzero $x$), not just for the specific cases tested.
QUESTION 9
3 marks
Criterion C
Medium
A student simplifies $\dfrac{x^9}{x^3}$ as $x^3$ (dividing the exponents instead of subtracting them).
a.
Explain the error, and verify the correct answer using $x=2$.
[3] Show complete worked solution
(a)
The division index law requires SUBTRACTING exponents ($x^m\div x^n=x^{m-n}$), not dividing them. Correct: $x^9\div x^3=x^6$. Verify with $x=2$: $2^9\div2^3=512\div8=64$, and $2^6=64$ (while the student's $2^3=8\ne64$).
QUESTION 10
6 marks
Criterion D
Hard
A data centre's storage doubles capacity every time a new server rack is added, following $S(n)=1024\times2^n$ GB, where $n$ is the number of additional racks installed beyond the base configuration.
a.
Find the storage capacity after adding 5 additional racks.
[2] b.
The company needs at least 500,000 GB of storage. Using index laws (not trial and error), find the minimum number of additional racks needed, and discuss ONE practical limitation of this doubling model that would make it unrealistic to keep adding racks indefinitely (e.g. physical space, cost, power requirements).
[4] Show complete worked solution
(a)
$$S(5)=1024\times2^5=1024\times32=32768 \text{ GB}$$
(b)
Need $1024\times2^n\ge500000 \Rightarrow 2^n\ge488.3 \Rightarrow n\ge9$ (since $2^8=256<488.3$ but $2^9=512\ge488.3$). So 9 additional racks are needed. Limitation: doubling capacity with every added rack assumes UNLIMITED physical space, power supply, and cooling capacity — in reality, data centres have finite floor space and power infrastructure, meaning this exponential doubling model would eventually hit a hard physical or financial limit long before storage needs could keep scaling this way indefinitely.
Expansion laws 9 questions
QUESTION 1
6 marks
Criterion A
Medium
Expand and simplify each expression:
a.
$3(2x+5)$
[1] b.
$(x+4)(x+2)$
[2] c.
$(x+3)^2$
[2] d.
$(x+5)(x-5)$
[1] Show complete worked solution
(a)
$$6x+15$$
(b)
$$x^2+2x+4x+8=x^2+6x+8$$
(c)
$$x^2+6x+9$$
(d)
$$x^2-25$$
QUESTION 2
6 marks
Criterion B
Medium
Investigate which type of expansion (distributive, binomial product, perfect square, or difference of squares) applies to different expression shapes.
a.
Classify each of $2(x+3)$, $(x+1)(x+4)$, $(x+2)^2$, $(x+6)(x-6)$ by expansion type, then expand each.
[4] b.
Describe one visual clue in each expression's shape that helps you identify which expansion rule to use.
[2] Show complete worked solution
(a)
$2(x+3)$: distributive law $\to 2x+6$. $(x+1)(x+4)$: general binomial product $\to x^2+5x+4$. $(x+2)^2$: perfect square $\to x^2+4x+4$. $(x+6)(x-6)$: difference of squares $\to x^2-36$.
(b)
A single term outside brackets: distributive law. Two DIFFERENT binomials multiplied: general product (FOIL). The SAME bracket squared: perfect square. Two IDENTICAL binomials but with opposite signs (one $+$, one $-$): difference of squares.
QUESTION 3
5 marks
Criterion C
Medium
A student expands $(x+3)(x+3)$ using the general product method and separately notices it 'looks like' $(x+3)^2$, but isn't sure if they should get the same answer either way.
a.
Expand $(x+3)(x+3)$ using the general FOIL method, and expand $(x+3)^2$ using the perfect square rule. Compare the results.
[3] b.
Explain why this makes sense — what does squaring a bracket actually mean?
[2] Show complete worked solution
(a)
$(x+3)(x+3)=x^2+3x+3x+9=x^2+6x+9$. $(x+3)^2=x^2+2(3)x+9=x^2+6x+9$. Identical results.
(b)
$(x+3)^2$ means $(x+3)\times(x+3)$ by definition of squaring — it's literally the same multiplication as $(x+3)(x+3)$, just written with exponent notation, so both methods must always give the same result.
QUESTION 4
5 marks
Criterion D
Medium
A rectangular garden has length $(x+7)$ m and width $(x+2)$ m.
a.
Write and expand an expression for the area of the garden.
[3] b.
If $x=5$, find the numerical area, and verify it against the original (unexpanded) dimensions.
[2] Show complete worked solution
(a)
$$A=(x+7)(x+2)=x^2+2x+7x+14=x^2+9x+14$$
(b)
Using expanded form: $A=25+45+14=84$ m$^2$. Verify with original dimensions: length $=12$, width $=7$, area $=12\times7=84$ m$^2$.
QUESTION 5
4 marks
Criterion A
Medium
Expand and fully simplify: $$4(x+3) - 2(x-5)$$
a.
Expand each bracket separately.
[2] b.
Combine, taking care with the SUBTRACTION of the second bracket's expansion.
[2] Show complete worked solution
(a)
$4(x+3)=4x+12$. $2(x-5)=2x-10$.
(b)
$$4x+12-(2x-10) = 4x+12-2x+10 = 2x+22$$
QUESTION 6
7 marks
Criterion B
Hard
Investigate the pattern in expanding $(x+a)(x-a)$, $(x+a)^2$, and general $(x+a)(x+b)$, comparing their MIDDLE terms.
a.
Expand $(x+5)(x-5)$, $(x+5)^2$, and $(x+5)(x+3)$.
[4] b.
For $(x+5)(x-5)$, explain why there is NO middle ($x$) term, connecting this to the general rule $(x+a)(x-b)$ having middle term coefficient $(a-b)$ specifically here with $a=b=5$.
[3] Show complete worked solution
(a)
$(x+5)(x-5)=x^2-25$ (no middle term). $(x+5)^2=x^2+10x+25$. $(x+5)(x+3)=x^2+8x+15$.
(b)
In general, $(x+a)(x-b)$ expands to $x^2+(a-b)x-ab$ — the middle term's coefficient is $(a-b)$. Here $a=b=5$, so $(a-b)=5-5=0$, meaning the middle term coefficient is exactly 0, which is why it disappears entirely — this is the SPECIFIC case (difference of squares) where the two constants are equal in size but opposite in sign.
QUESTION 7
7 marks
Criterion B
Hard
Investigate whether $(x+a)(x+b)(x+c)$ (a product of THREE binomials) follows a similarly predictable pattern to the two-binomial case.
a.
Fully expand $(x+1)(x+2)(x+3)$, showing your working (expand two brackets first, then multiply by the third).
[4] b.
Compare the coefficients ($1, 6, 11, 6$) to the numbers $1,2,3$ (the constants used). State what you notice about the coefficient of $x^2$ (which is 6) in relation to $1+2+3$, and the constant term (6) in relation to $1\times2\times3$.
[3] Show complete worked solution
(a)
$(x+1)(x+2)=x^2+3x+2$. Then $(x^2+3x+2)(x+3) = x^3+3x^2+3x^2+9x+2x+6 = x^3+6x^2+11x+6$.
(b)
Coefficient of $x^2$ is 6, which equals $1+2+3=6$ (the SUM of the constants). The constant term is also 6, which equals $1\times2\times3=6$ (the PRODUCT of the constants) — these connections extend the pattern seen in two-binomial products, where the coefficient of $x$ was the sum $(a+b)$ and the constant was the product $(ab)$.
QUESTION 8
5 marks
Criterion C
Medium
A student expands $3(2x-1) - (x+4)$ as $6x-3-x+4$ (forgetting that subtracting the SECOND bracket means every term inside it changes sign).
a.
Identify precisely where the error occurs, and give the fully correct expansion.
[3] b.
Verify the correct answer using $x=2$.
[2] Show complete worked solution
(a)
The student correctly expanded $3(2x-1)=6x-3$, but then subtracted the bracket $(x+4)$ INCORRECTLY as $-x+4$ instead of $-x-4$ (both terms inside the bracket must have their sign flipped when subtracting the whole bracket). Correct: $6x-3-x-4=5x-7$.
(b)
Original: $3(2(2)-1)-(2+4)=3(3)-6=9-6=3$. Correct formula $5x-7$ at $x=2$: $10-7=3$ (student's wrong version $6x-3-x+4=5x+1$ at $x=2$ gives $11\ne3$).
QUESTION 9
7 marks
Criterion D
Hard
A garden has a rectangular lawn of length $(x+6)$m and width $(x+2)$m, surrounded by a path of UNIFORM width 1m on all sides.
a.
Write and expand an expression for the TOTAL area (lawn + path), noting the path adds 1m to EACH side, so total dimensions are $(x+8)$ by $(x+4)$.
[3] b.
Write and expand an expression for the area of the PATH alone (total area minus lawn area), and verify your simplified expression by evaluating both the total and lawn areas at $x=10$m, confirming the difference matches.
[4] Show complete worked solution
(a)
$$(x+8)(x+4) = x^2+12x+32$$
(b)
Lawn area: $(x+6)(x+2)=x^2+8x+12$. Path area: $(x^2+12x+32)-(x^2+8x+12)=4x+20$. Check at $x=10$: total $=(18)(14)=252$, lawn $=(16)(12)=192$, path $=252-192=60$. Formula: $4(10)+20=60$.
The zero index law 9 questions
QUESTION 1
4 marks
Criterion A
Medium
Evaluate each expression:
a.
$5^0$
[1] b.
$(-3)^0$
[1] c.
$4x^0$ (where $x\ne0$)
[1] d.
$(4x)^0$ (where $x\ne0$)
[1] Show complete worked solution
(a)
$$1$$
(b)
$$1$$
(c)
$$4 \times 1 = 4$$
(d)
$$1$$
QUESTION 2
3 marks
Criterion B
Medium
Investigate WHY any non-zero number raised to the power 0 equals 1, using the index law for division.
a.
Simplify $\dfrac{x^3}{x^3}$ two ways: (i) using the division index law $\dfrac{x^m}{x^n}=x^{m-n}$, and (ii) by direct cancellation (since anything divided by itself is 1).
[2] b.
Since both methods calculate the SAME expression, what must $x^0$ equal?
[1] Show complete worked solution
(a)
(i) $x^{3-3}=x^0$. (ii) $\frac{x^3}{x^3}=1$ directly, since any nonzero number divided by itself is 1.
(b)
Since both results describe $\frac{x^3}{x^3}$, we must have $x^0=1$.
QUESTION 3
4 marks
Criterion C
Medium
A student thinks $x^0=0$, reasoning that 'anything to the power of nothing should give nothing'.
a.
Explain why this reasoning is incorrect, using the division-based proof.
[2] b.
State the one exception where the zero index law does NOT apply, and explain why.
[2] Show complete worked solution
(a)
The 'power of nothing gives nothing' intuition is a misleading guess, not based on any actual rule. The real reason comes from the index laws: $\frac{x^n}{x^n}=x^{n-n}=x^0$, but also $\frac{x^n}{x^n}=1$ directly — so $x^0$ must equal 1, not 0.
(b)
The law does not apply when the base is 0 (i.e. $0^0$ is undefined) — this is a special case because $\frac{0^n}{0^n}$ involves division by zero, which is never allowed.
QUESTION 4
3 marks
Criterion D
Medium
A scientist writes a very small measurement as $3.2 \times 10^{-4}$ metres, and a colleague asks what $10^0$ would represent if the exponent were 0 instead.
a.
Evaluate $3.2 \times 10^0$.
[1] b.
Explain what this tells you about the role of $10^0$ in scientific notation generally.
[2] Show complete worked solution
(a)
$$3.2 \times 1 = 3.2$$
(b)
$10^0=1$ means that when the exponent is 0, the number is simply left unchanged (multiplying by 1 does nothing) — it represents 'no scaling at all', which is why $10^0$ acts as the 'neutral' starting point in the powers-of-ten scale.
QUESTION 5
5 marks
Criterion A
Medium
Evaluate each expression:
a.
$8^0 + 3^0$
[1] b.
$7x^0$ (where $x\ne0$)
[1] c.
$(7x)^0$ (where $x\ne0$)
[1] d.
$5^0 \times 5^3$
[2] Show complete worked solution
(a)
$$1+1=2$$
(b)
$$7\times1=7$$
(c)
$$1$$
(d)
$$1\times125=125$$
QUESTION 6
5 marks
Criterion B
Medium
Investigate why $(ab)^0=1$ but $ab^0$ (without brackets) can give a DIFFERENT result, using $a=3, b=5$.
a.
Evaluate $(3\times5)^0$ and $3\times5^0$ separately.
[2] b.
Explain why these give different results, referring to WHAT exactly the zero exponent applies to in each case.
[3] Show complete worked solution
(a)
$(3\times5)^0=15^0=1$. $3\times5^0=3\times1=3$.
(b)
In $(3\times5)^0$, the exponent 0 applies to the ENTIRE product $(3\times5)=15$, giving $15^0=1$. In $3\times5^0$ (no brackets), the exponent 0 applies ONLY to the 5, giving $5^0=1$, which is then multiplied by the separate factor of 3, giving $3\times1=3$ — brackets determine exactly what the exponent 'reaches', and this changes the result significantly.
QUESTION 7
5 marks
Criterion C
Hard
A student believes $0^0=1$ 'because everything to the power 0 is 1', while another believes $0^0=0$ 'because 0 to any power is 0'.
a.
Explain why BOTH students' reasoning, while each individually plausible-sounding, leads to a genuine mathematical CONTRADICTION when applied to the special case $0^0$.
[3] b.
State the correct convention taught at this level: is $x^0=1$ valid for ALL values of $x$, or does it require an exception?
[2] Show complete worked solution
(a)
The first student's rule ('anything to power 0 is 1') and the second student's rule ('0 to any power is 0') are both generally true patterns — but they directly CONTRADICT each other specifically at $0^0$, since it can't be BOTH 1 and 0 simultaneously. This is exactly why $0^0$ is typically treated as a special, undefined (or context-dependent) case in mathematics, rather than following either general pattern automatically.
(b)
The rule $x^0=1$ specifically requires $x\ne0$ — it is valid for every nonzero value of $x$, but the case $x=0$ (i.e. $0^0$) is excluded as a special undefined case, precisely because of the contradiction shown in part (a).
QUESTION 8
2 marks
Criterion C
Medium
A classmate simplifies $\dfrac{x^5}{x^5}$ using the division law and gets $x^0$, then writes: 'so the answer is $x^0$', without simplifying further.
a.
Explain why '$x^0$' is not fully simplified, and complete the simplification to a single number.
[2] Show complete worked solution
(a)
$x^0$ is not a FINAL simplified numerical answer — since $x\ne0$ is implied (as $\frac{x^5}{x^5}$ requires $x\ne0$ to be defined), $x^0=1$, and a fully simplified answer should state this final numerical value, not leave it in index form when it evaluates to a specific constant.
QUESTION 9
5 marks
Criterion D
Medium
A chemistry experiment involves a substance where the concentration formula includes a term $C_0 \times k^0$ at time zero (before any reaction has started), where $k$ is the reaction rate constant.
a.
Explain, using the zero index law, why the term $k^0$ always simplifies to 1 REGARDLESS of the specific numerical value of the reaction rate constant $k$ (as long as $k\ne0$).
[2] b.
Given this, simplify the full expression $C_0\times k^0$, and explain what this simplified result tells a chemist about the concentration at time zero, REGARDLESS of the specific reaction rate.
[3] Show complete worked solution
(a)
By the zero index law, ANY nonzero base raised to the power 0 equals 1 — this holds true no matter what specific value $k$ takes (whether $k=0.5$, $k=3.7$, or any other nonzero rate constant), so $k^0=1$ universally in this formula.
(b)
$C_0\times k^0=C_0\times1=C_0$. This tells the chemist that the concentration at time zero is simply $C_0$ (the initial concentration), completely UNAFFECTED by the reaction rate constant $k$ — which makes physical sense, since no reaction has had time to occur yet at $t=0$, so the rate at which the reaction proceeds shouldn't influence the STARTING concentration at all.
The negative index law 9 questions
QUESTION 1
6 marks
Criterion A
Medium
Rewrite each with a positive index, then evaluate:
a.
$2^{-3}$
[2] b.
$5^{-2}$
[2] c.
$4 \times 3^{-1}$
[2] Show complete worked solution
(a)
$$2^{-3}=\frac{1}{2^3}=\frac{1}{8}$$
(b)
$$5^{-2}=\frac{1}{5^2}=\frac{1}{25}$$
(c)
$$4\times\frac{1}{3}=\frac{4}{3}$$
QUESTION 2
4 marks
Criterion B
Medium
Investigate the connection between the negative index law and the division index law, using $\dfrac{x^2}{x^5}$.
a.
Simplify $\dfrac{x^2}{x^5}$ using the division law $\dfrac{x^m}{x^n}=x^{m-n}$.
[1] b.
Now simplify $\dfrac{x^2}{x^5}$ by cancelling common factors directly (writing out the $x$'s).
[2] c.
Since both methods calculate the same thing, what must $x^{-3}$ equal?
[1] Show complete worked solution
(a)
$$x^{2-5}=x^{-3}$$
(b)
$\dfrac{x\times x}{x\times x\times x\times x\times x} = \dfrac{1}{x\times x\times x}=\dfrac{1}{x^3}$
(c)
$$x^{-3}=\frac{1}{x^3}$$
QUESTION 3
4 marks
Criterion C
Medium
A student simplifies $2^{-3}$ as $-8$ (treating the negative sign as making the whole answer negative).
a.
Explain the error, and give the correct value.
[2] b.
Explain how $2^{-3}$ (positive result) is different from $-(2^3)$ (which genuinely does give a negative result).
[2] Show complete worked solution
(a)
A negative EXPONENT does not make the result negative — it means 'reciprocal' (1 over the positive-power version). Correct: $2^{-3}=\frac{1}{2^3}=\frac{1}{8}$, not $-8$.
(b)
$2^{-3}=\frac{1}{8}$ (positive, since the negative applies to the EXPONENT, meaning 'take the reciprocal'). $-(2^3)=-8$ (negative, since the minus sign applies to the whole result AFTER evaluating $2^3=8$). These are two completely different operations that happen to use similar-looking negative signs.
QUESTION 4
3 marks
Criterion D
Medium
In physics, the intensity of light follows an inverse-square law: intensity is proportional to $d^{-2}$, where $d$ is distance from the source.
a.
Rewrite $d^{-2}$ using a positive index (as a fraction).
[1] b.
Explain, using this rewritten form, why doubling the distance from a light source (i.e. $d\to2d$) reduces the intensity to $\frac{1}{4}$ of its original value, not $\frac{1}{2}$.
[2] Show complete worked solution
(a)
$$d^{-2}=\frac{1}{d^2}$$
(b)
Since intensity $\propto\frac{1}{d^2}$, doubling $d$ gives $\frac{1}{(2d)^2}=\frac{1}{4d^2}=\frac{1}{4}\times\frac{1}{d^2}$ — the intensity is divided by $2^2=4$, not just 2, because the distance is SQUARED in the denominator.
QUESTION 5
4 marks
Criterion A
Medium
Rewrite each with a positive index, then evaluate as a fraction:
a.
$6^{-2}$
[2] b.
$5 \times 2^{-3}$
[2] Show complete worked solution
(a)
$$\frac{1}{6^2}=\frac{1}{36}$$
(b)
$$5\times\frac{1}{8}=\frac{5}{8}$$
QUESTION 6
3 marks
Criterion A
Medium
Simplify $\dfrac{1}{5^{-2}}$ (a NEGATIVE index in the DENOMINATOR).
a.
Rewrite $5^{-2}$ with a positive index first, then simplify the overall fraction (dividing by a fraction).
[3] Show complete worked solution
(a)
$$\frac{1}{5^{-2}} = \frac{1}{\frac{1}{25}} = 1\div\frac{1}{25} = 1\times25=25=5^2$$
QUESTION 7
5 marks
Criterion B
Hard
Investigate the general rule for $\dfrac{1}{x^{-n}}$, based on the specific case $\dfrac{1}{5^{-2}}=5^2$ found above.
a.
Test the pattern with $\dfrac{1}{3^{-4}}$ — does it equal $3^4$? Verify numerically.
[2] b.
State the general rule for $\dfrac{1}{x^{-n}}$ in terms of $x^n$, and explain WHY 'a negative index in the denominator flips to become a positive index' using the definition of a negative index as a reciprocal.
[3] Show complete worked solution
(a)
$3^{-4}=\frac{1}{81}$. $\frac{1}{3^{-4}}=\frac{1}{\frac{1}{81}}=81=3^4$, confirming the pattern.
(b)
$$\frac{1}{x^{-n}} = x^n$$ Since $x^{-n}$ ALREADY means $\frac{1}{x^n}$ (its definition as a reciprocal), taking the reciprocal of THAT (i.e. $\frac{1}{x^{-n}}$) means finding the reciprocal of a reciprocal, which always returns to the original value — hence $\frac{1}{x^{-n}}=\frac{1}{\frac{1}{x^n}}=x^n$.
QUESTION 8
4 marks
Criterion C
Medium
A student simplifies $3^{-2}$ as $-9$ (treating the negative index as making the base negative before squaring).
a.
Explain the error, and give the correct value, distinguishing between $3^{-2}$, $(-3)^2$, and $-(3^2)$.
[4] Show complete worked solution
(a)
$3^{-2}$ means 'take the reciprocal of $3^2$', giving $\frac{1}{9}$ (positive, since it's a fraction, not related to sign at all). This is completely different from $(-3)^2=9$ (squaring a negative base gives positive) and $-(3^2)=-9$ (negating the result of squaring). The student incorrectly conflated 'negative index' with 'negative base' — these are unrelated concepts that happen to both involve the word 'negative'.
QUESTION 9
6 marks
Criterion D
Hard
The intensity of sound decreases with distance according to $I \propto d^{-2}$, similar to light intensity. A sound source has intensity 80 units at 1m distance.
a.
Using $I=\dfrac{80}{d^2}$, find the intensity at $d=4$m.
[2] b.
A safety guideline requires intensity below 2 units. Find the MINIMUM distance required to meet this guideline, and explain (using the inverse-square relationship) why doubling the required 'safe intensity' threshold (e.g. from 2 to 4 units) would NOT simply double the minimum safe distance.
[4] Show complete worked solution
(a)
$$I=\frac{80}{4^2}=\frac{80}{16}=5 \text{ units}$$
(b)
$\frac{80}{d^2}\le2 \Rightarrow d^2\ge40 \Rightarrow d\ge\sqrt{40}\approx6.32$m. Because intensity depends on $d^{-2}$ (an INVERSE SQUARE relationship, not a simple inverse), changing the threshold doesn't scale distance linearly — doubling the allowed intensity threshold to 4 would give $d^2\ge20 \Rightarrow d\ge\sqrt{20}\approx4.47$m, which is NOT half of 6.32m; because of the squared relationship, distance changes by a factor of $\sqrt{2}$ (not 2) when intensity threshold changes by a factor of 2.
The distributive law 8 questions
QUESTION 1
5 marks
Criterion A
Medium
Expand each expression:
a.
$4(x+6)$
[1] b.
$-3(2x-5)$
[1] c.
$x(x+7)$
[1] d.
$-2x(3x-4)$
[2] Show complete worked solution
(a)
$$4x+24$$
(b)
$$-6x+15$$
(c)
$$x^2+7x$$
(d)
$$-6x^2+8x$$
QUESTION 2
4 marks
Criterion B
Medium
Investigate why $a(b+c)=ab+ac$ is true, using areas of rectangles.
a.
A rectangle has width $a$ and length $(b+c)$, split into two smaller rectangles of length $b$ and $c$. Write an expression for the area of the WHOLE rectangle two ways: (i) directly as width times total length, and (ii) as the sum of the two smaller rectangles' areas.
[3] b.
Since both expressions describe the exact same total area, what does this prove about the distributive law?
[1] Show complete worked solution
(a)
(i) $a(b+c)$. (ii) $ab + ac$ (the two smaller rectangle areas added together).
(b)
Since both must be equal (same total area, calculated two ways), this proves $a(b+c)=ab+ac$.
QUESTION 3
4 marks
Criterion C
Medium
A student expands $5(2x+3)$ as $10x+3$ (forgetting to multiply the second term by 5).
a.
Explain the student's error, and give the correct expansion.
[2] b.
Verify the correct answer by substituting $x=2$ into both the original expression $5(2x+3)$ and your expanded answer.
[2] Show complete worked solution
(a)
The student only multiplied the FIRST term inside the brackets by 5, forgetting that the 5 must multiply EVERY term inside the brackets. Correct: $5(2x+3)=10x+15$.
(b)
Original: $5(2(2)+3)=5(7)=35$. Expanded: $10(2)+15=20+15=35$ — matches, confirming the correct expansion.
QUESTION 4
4 marks
Criterion D
Medium
A cinema charges $x$ dollars per adult ticket. A family buys 3 adult tickets and also pays a \$4 booking fee per ticket.
a.
Write an expression for the total cost, in the form $3(x+4)$, and expand it.
[2] b.
If tickets cost $x=\$12$ each, find the total cost using both the unexpanded and expanded forms, confirming they match.
[2] Show complete worked solution
(a)
$$3(x+4)=3x+12$$
(b)
Unexpanded: $3(12+4)=3(16)=48$. Expanded: $3(12)+12=36+12=48$.
QUESTION 5
2 marks
Criterion A
Easy
Expand: $$-4(3x-2)$$
a.
Expand carefully, paying attention to the negative sign outside the bracket.
[2] Show complete worked solution
(a)
$$-4(3x-2) = -12x+8$$
QUESTION 6
5 marks
Criterion B
Medium
Investigate the distributive law applied to THREE terms inside the bracket, e.g. $a(b+c+d)$, using $3(x+y+2)$.
a.
Expand $3(x+y+2)$, then verify by substituting $x=4,y=5$ into BOTH the original and expanded form.
[3] b.
State the general rule: does the distributive law work the same way for ANY number of terms inside the bracket (not just 2), and explain briefly why.
[2] Show complete worked solution
(a)
$3(x+y+2)=3x+3y+6$. Original at $x=4,y=5$: $3(4+5+2)=3(11)=33$. Expanded: $3(4)+3(5)+6=12+15+6=33$.
(b)
Yes — the distributive law extends to any number of terms: $a(b_1+b_2+\cdots+b_n)=ab_1+ab_2+\cdots+ab_n$. This works because multiplication distributes over EACH addition individually, regardless of how many terms are being added together inside the bracket.
QUESTION 7
5 marks
Criterion C
Medium
A student simplifies $x(x+7)-3(x-2)$ as $x^2+7x-3x-2$ (forgetting to multiply the $-3$ by BOTH terms in the second bracket).
a.
Explain the error precisely, and give the fully correct expansion.
[3] b.
Verify the correct answer using $x=3$.
[2] Show complete worked solution
(a)
The student correctly distributed $x(x+7)=x^2+7x$, but only multiplied $-3$ by the $x$ term (getting $-3x$) and forgot to multiply $-3$ by $-2$ as well. Correct: $x(x+7)-3(x-2)=x^2+7x-3x+6=x^2+4x+6$.
(b)
Original: $3(10)-3(1)=30-3=27$. Correct formula $x^2+4x+6$ at $x=3$: $9+12+6=27$.
QUESTION 8
3 marks
Criterion C
Medium
A classmate says the distributive law $a(b+c)=ab+ac$ 'only works for addition, never subtraction', after seeing it written that way in a textbook.
a.
Explain why the distributive law ALSO applies to subtraction (i.e. $a(b-c)=ab-ac$), by rewriting subtraction as 'adding a negative' and applying the SAME rule.
[3] Show complete worked solution
(a)
Subtraction can always be rewritten as addition of a negative: $b-c=b+(-c)$. Applying the distributive law to THIS form: $a(b+(-c))=ab+a(-c)=ab-ac$ — showing that the subtraction version is really just a special case of the addition version, using a negative term, so the SAME underlying rule covers both.
The product (a + b)(c + d) 8 questions
QUESTION 1
6 marks
Criterion A
Medium
Expand each product:
a.
$(x+3)(x+5)$
[2] b.
$(x-2)(x+7)$
[2] c.
$(2x+1)(x-4)$
[2] Show complete worked solution
(a)
$$x^2+5x+3x+15=x^2+8x+15$$
(b)
$$x^2+7x-2x-14=x^2+5x-14$$
(c)
$$2x^2-8x+x-4=2x^2-7x-4$$
QUESTION 2
6 marks
Criterion B
Medium
Investigate the pattern connecting the coefficients in $(x+a)(x+b)$ to its expanded form.
a.
Expand $(x+2)(x+5)$, $(x+3)(x+4)$, and $(x+1)(x+6)$.
[3] b.
In each case, compare the coefficient of $x$ and the constant term to the original numbers $a$ and $b$. State the general pattern for $(x+a)(x+b)$.
[3] Show complete worked solution
(a)
$(x+2)(x+5)=x^2+7x+10$. $(x+3)(x+4)=x^2+7x+12$. $(x+1)(x+6)=x^2+7x+6$.
(b)
In each expansion, the coefficient of $x$ equals $a+b$ (sum), and the constant term equals $a\times b$ (product). General rule: $(x+a)(x+b)=x^2+(a+b)x+ab$.
QUESTION 3
5 marks
Criterion C
Medium
A student expands $(x+4)(x+3)$ as $x^2+12$ (multiplying only the first terms and the last terms, skipping the 'outer' and 'inner' products).
a.
Explain the student's error, referring to the FOIL method (First, Outer, Inner, Last).
[3] b.
Verify the correct answer using $x=1$.
[2] Show complete worked solution
(a)
The student only calculated the First terms ($x\times x=x^2$) and Last terms ($4\times3=12$), but forgot the Outer ($x\times3=3x$) and Inner ($4\times x=4x$) products — all FOUR products must be included and combined. Correct: $x^2+3x+4x+12=x^2+7x+12$.
(b)
Original: $(1+4)(1+3)=5\times4=20$. Correct expansion: $1+7+12=20$. (Student's wrong answer: $1+12=13\ne20$.)
QUESTION 4
5 marks
Criterion D
Medium
A rectangular field has length $(x+15)$ m and width $(x+8)$ m, where $x$ represents an unknown base length.
a.
Write and expand an expression for the field's area.
[3] b.
If $x=10$ m, find the numerical area using your expanded expression, and verify against the original dimensions.
[2] Show complete worked solution
(a)
$$(x+15)(x+8)=x^2+8x+15x+120=x^2+23x+120$$
(b)
Expanded: $100+230+120=450$ m$^2$. Original: length$=25$, width$=18$, area$=25\times18=450$ m$^2$.
QUESTION 5
5 marks
Criterion A
Medium
Expand each product:
a.
$(x+9)(x+4)$
[2] b.
$(2x+3)(x-5)$
[3] Show complete worked solution
(a)
$$x^2+13x+36$$
(b)
$$2x^2-7x-15$$
QUESTION 6
3 marks
Criterion A
Medium
A rectangle's length is $(x+9)$cm and width is $(x+4)$cm.
a.
Write and expand an expression for the area, and find the numerical area when $x=6$cm.
[3] Show complete worked solution
(a)
$(x+9)(x+4)=x^2+13x+36$. At $x=6$: $36+78+36=150$cm² (check directly: $15\times10=150$).
QUESTION 7
7 marks
Criterion B
Hard
Investigate the pattern connecting the SIGN of the constant term in $(x+a)(x+b)$'s expansion to the signs of $a$ and $b$.
a.
Expand $(x+3)(x+5)$ (both positive), $(x-3)(x-5)$ (both negative), and $(x+3)(x-5)$ (one of each sign). State the sign of the CONSTANT term in each case.
[4] b.
Explain WHY the constant term is always positive when $a$ and $b$ have the SAME sign (both positive or both negative), but negative when they have DIFFERENT signs — referring to the fact that the constant term equals $a\times b$.
[3] Show complete worked solution
(a)
$(x+3)(x+5)=x^2+8x+15$ — constant POSITIVE. $(x-3)(x-5)=x^2-8x+15$ — constant POSITIVE. $(x+3)(x-5)=x^2-2x-15$ — constant NEGATIVE.
(b)
Since the constant term always equals $a\times b$ (the product of the two constants), and multiplying two numbers with the SAME sign always gives a POSITIVE result (positive×positive or negative×negative), while multiplying numbers with DIFFERENT signs always gives a NEGATIVE result — this directly explains the pattern observed: matching signs give a positive constant term, differing signs give a negative one.
QUESTION 8
6 marks
Criterion D
Hard
A construction company estimates the area needed for a rectangular foundation as $(x+12)(x+7)$ m², where $x$ represents a variable buffer zone width.
a.
Expand this expression.
[2] b.
If regulations require $x\ge3$ m and the site can accommodate at most $200\text{ m}^2$, determine the greatest whole-number value of $x$ that satisfies both constraints.
[4] Show complete worked solution
(a)
$$x^2+19x+84$$
(b)
The constraints are $x\ge3$ and $(x+12)(x+7)\le200$. Test whole-number values near the boundary: $$x=4:\ (16)(11)=176\le200,$$ $$x=5:\ (17)(12)=204>200.$$ Therefore the greatest whole-number value that satisfies both constraints is $$\boxed{x=4\text{ m}}.$$
Perfect square expansion 11 questions
QUESTION 1
6 marks
Criterion A
Medium
Expand each perfect square:
a.
$(x+6)^2$
[2] b.
$(x-4)^2$
[2] c.
$(3x+2)^2$
[2] Show complete worked solution
(a)
$$x^2+12x+36$$
(b)
$$x^2-8x+16$$
(c)
$$9x^2+12x+4$$
QUESTION 2
5 marks
Criterion B
Medium
Investigate the pattern in $(x+a)^2$ compared to $(x-a)^2$.
a.
Expand $(x+5)^2$ and $(x-5)^2$.
[2] b.
Compare the two expansions. What stays the same, and what changes?
[2] c.
State the general rule for $(x-a)^2$ compared to $(x+a)^2 = x^2+2ax+a^2$.
[1] Show complete worked solution
(a)
$(x+5)^2=x^2+10x+25$. $(x-5)^2=x^2-10x+25$.
(b)
The $x^2$ term and the constant term ($25$) are IDENTICAL in both. Only the sign of the middle term ($+10x$ vs $-10x$) changes.
(c)
$$(x-a)^2 = x^2-2ax+a^2$$ (only the middle term's sign flips)
QUESTION 3
5 marks
Criterion C
Medium
A student expands $(x+7)^2$ as $x^2+49$ (squaring each term separately, forgetting the middle term).
a.
Explain the error using the definition $(x+7)^2=(x+7)(x+7)$, and give the correct expansion.
[3] b.
Verify using $x=3$: evaluate both $(x+7)^2$ directly and the student's (incorrect) expansion.
[2] Show complete worked solution
(a)
$(x+7)^2$ means $(x+7)(x+7)$, which by FOIL gives FOUR terms ($x^2$, $7x$, $7x$, $49$), not just two. The student incorrectly assumed squaring 'distributes' over addition (i.e. $(a+b)^2=a^2+b^2$), which is a common but false shortcut. Correct: $x^2+14x+49$.
(b)
$(3+7)^2=10^2=100$. Student's version: $3^2+49=9+49=58\ne100$ — confirms the error.
QUESTION 4
5 marks
Criterion D
Medium
A square swimming pool has side length $(x+3)$ m, surrounded by a uniform paved border making the OUTER edge of the paving $(x+3+4)=(x+7)$ m per side (since the border is 2 m wide on each side).
a.
Write and expand an expression for the area of the pool.
[2] b.
Write and expand an expression for the TOTAL area (pool plus paving), then find the area of JUST the paving (border) in simplified form.
[3] Show complete worked solution
(a)
$$(x+3)^2=x^2+6x+9 \text{ m}^2$$
(b)
Total area $=(x+7)^2=x^2+14x+49$. Paving area $=(x^2+14x+49)-(x^2+6x+9)=8x+40$.
QUESTION 5
7 marks
Criterion A
Medium
Expand each perfect square:
a.
$(x+11)^2$
[2] b.
$(x-7)^2$
[2] c.
$(4x+1)^2$
[3] Show complete worked solution
(a)
$$x^2+22x+121$$
(b)
$$x^2-14x+49$$
(c)
$$16x^2+8x+1$$
QUESTION 6
5 marks
Criterion B
Hard
Investigate whether $(x+y)^2$ can EVER equal $x^2+y^2$ (i.e. whether the 'cross term' $2xy$ can ever vanish).
a.
Expand $(x+y)^2$ fully, identifying the cross term.
[2] b.
For the cross term $2xy$ to equal exactly 0 (making $(x+y)^2=x^2+y^2$ true), what must be true about $x$ or $y$? Test your conclusion with a specific example.
[3] Show complete worked solution
(a)
$$(x+y)^2 = x^2+2xy+y^2$$ — the cross term is $2xy$.
(b)
$2xy=0$ requires EITHER $x=0$ OR $y=0$ (or both) — since if neither is zero, their product can't be zero. Test: $x=0,y=7$: $(0+7)^2=49$, and $0^2+7^2=49$ — matches, confirming $(x+y)^2=x^2+y^2$ ONLY when at least one of $x,y$ is zero.
QUESTION 7
6 marks
Criterion B
Hard
Investigate the pattern in $(x+1)^2-(x-1)^2$ compared to $(x+2)^2-(x-2)^2$, looking for a general rule.
a.
Fully expand and simplify $(x+1)^2-(x-1)^2$.
[3] b.
Now do the same for $(x+2)^2-(x-2)^2$, and state the general pattern for $(x+a)^2-(x-a)^2$.
[3] Show complete worked solution
(a)
$(x^2+2x+1)-(x^2-2x+1)=4x$.
(b)
$(x^2+4x+4)-(x^2-4x+4)=8x$. General pattern: $$(x+a)^2-(x-a)^2=4ax$$ (verify: with $a=1$, $4(1)x=4x$; with $a=2$, $4(2)x=8x$).
QUESTION 8
5 marks
Criterion C
Medium
A student expands $(x-9)^2$ as $x^2-81$ (treating it like a difference of squares instead of a perfect square).
a.
Explain the error, clarifying the difference between $(x-9)^2$ and $(x+9)(x-9)$.
[3] b.
Verify the correct expansion of $(x-9)^2$ using $x=10$.
[2] Show complete worked solution
(a)
$(x-9)^2$ means $(x-9)(x-9)$ — the SAME bracket multiplied by itself, which expands to $x^2-18x+81$ (including a middle term). The student's answer, $x^2-81$, is actually the expansion of the DIFFERENT expression $(x+9)(x-9)$ (a difference of squares, two DIFFERENT brackets) — confusing a perfect square with a difference of squares is a common but significant error.
(b)
$(10-9)^2=1^2=1$. Correct formula $x^2-18x+81$ at $x=10$: $100-180+81=1$ (student's wrong version $x^2-81$ at $x=10$ gives $19\ne1$).
QUESTION 9
5 marks
Criterion C
Hard
A classmate says: 'since $(x+5)^2$ has a PLUS in the brackets, the middle term of its expansion must also be positive — and since $(x-5)^2$ has a MINUS, its middle term must be negative.' They conclude this rule always works, without checking $(-x+5)^2$.
a.
Test the classmate's rule using $(-x+5)^2$ — does the '+' inside this bracket predict a positive middle term as the classmate's rule suggests?
[3] b.
Explain why the classmate's simplified rule fails here, identifying what ACTUALLY determines the sign of the middle term (hint: think about the sign of the SECOND term specifically, not just whether there's a visible '+' or '-' symbol).
[2] Show complete worked solution
(a)
$(-x+5)^2 = (-x)^2+2(-x)(5)+5^2 = x^2-10x+25$ — the middle term is NEGATIVE, despite the bracket containing a '+' sign. This DISPROVES the classmate's simple rule.
(b)
The middle term's sign is determined by the sign of the SECOND term being squared (here, $+5$, combined with the NEGATIVE first term $-x$), not simply by which symbol appears in the bracket — the classmate's rule oversimplified by ignoring that the FIRST term itself ($-x$) can also carry a sign that affects the outcome; the correct approach is always to properly expand $(a+b)^2=a^2+2ab+b^2$ with the ACTUAL signed values of $a$ and $b$.
QUESTION 10
7 marks
Criterion D
Hard
A square painting's canvas has side length $(x+7)$cm, and is placed inside a square frame with a UNIFORM 3cm border on all sides (so the frame's outer side length is $(x+13)$cm).
a.
Write and expand expressions for the canvas area and the total (frame+canvas) area.
[4] b.
Find a simplified expression for the area of the FRAME alone (border only, excluding canvas), and evaluate it at $x=20$cm.
[3] Show complete worked solution
(a)
Canvas: $(x+7)^2=x^2+14x+49$. Total: $(x+13)^2=x^2+26x+169$.
(b)
Frame area $=(x^2+26x+169)-(x^2+14x+49)=12x+120$. At $x=20$: $12(20)+120=240+120=360$cm².
QUESTION 11
7 marks
Criterion D
Hard
A city's population model predicts population $P(t)=(x+t)^2$ thousand people, where $x$ is a base constant and $t$ is years elapsed.
a.
If $x=50$, expand $(50+t)^2$ to find $P(t)$ as a polynomial in $t$.
[2] b.
City planners want to know the population at $t=10$ years, and separately want to know how much of that population figure comes from the 'interaction' between the base population and time (the middle term), versus the base population itself and the pure time-squared growth. Break down $P(10)$ into its three components and their individual contributions.
[5] Show complete worked solution
(a)
$$P(t)=2500+100t+t^2$$
(b)
$P(10)=2500+1000+100=3600$ thousand. Component breakdown: base term ($2500$) $=69.4\%$ of total; middle/interaction term ($1000$) $=27.8\%$; time-squared term ($100$) $=2.8\%$. This shows the base population still dominates the total even after 10 years, though the interaction term contributes meaningfully, while the pure $t^2$ term remains relatively small at this stage.
Difference of two squares 9 questions
QUESTION 1
4 marks
Criterion A
Medium
Expand each expression:
a.
$(x+8)(x-8)$
[1] b.
$(2x+3)(2x-3)$
[2] c.
$(x-9)(x+9)$
[1] Show complete worked solution
(a)
$$x^2-64$$
(b)
$$4x^2-9$$
(c)
$$x^2-81$$
QUESTION 2
4 marks
Criterion B
Medium
Investigate why the 'middle terms' always cancel out in a difference-of-squares expansion, using $(x+a)(x-a)$.
a.
Fully expand $(x+a)(x-a)$ using FOIL, showing all four terms before simplifying.
[2] b.
Explain why the middle two terms always cancel, for ANY value of $a$.
[2] Show complete worked solution
(a)
$(x+a)(x-a) = x^2 -ax+ax-a^2$ (First: $x^2$, Outer: $-ax$, Inner: $ax$, Last: $-a^2$).
(b)
The Outer term is $-ax$ and the Inner term is $+ax$ — these are always exact opposites of each other (same size, opposite sign) regardless of what $a$ is, so they always sum to zero and cancel out, leaving just $x^2-a^2$.
QUESTION 3
5 marks
Criterion C
Medium
A student is confused why $(x+6)(x-6)$ expands to $x^2-36$ (no middle term), while $(x+6)(x+6)$ expands to $x^2+12x+36$ (WITH a middle term).
a.
Expand both expressions fully to confirm these results.
[3] b.
Explain the key difference between the two expressions that causes this different behaviour.
[2] Show complete worked solution
(a)
$(x+6)(x-6)=x^2-6x+6x-36=x^2-36$ (middle terms cancel). $(x+6)(x+6)=x^2+6x+6x+36=x^2+12x+36$ (middle terms ADD instead).
(b)
In $(x+6)(x-6)$, the second bracket has a MINUS sign, making the Outer and Inner terms opposite in sign (so they cancel). In $(x+6)(x+6)$, both brackets have a PLUS sign, so the Outer and Inner terms have the SAME sign (so they add together instead of cancelling).
QUESTION 4
5 marks
Criterion D
Medium
A mental-maths trick: to calculate $47\times53$ quickly, write it as $(50-3)(50+3)$.
a.
Use the difference-of-squares pattern to evaluate $(50-3)(50+3)$ without a calculator.
[2] b.
Verify this matches $47\times53$ by direct multiplication, and explain why this 'trick' works for any two numbers that are equally spaced above and below a round number.
[3] Show complete worked solution
(a)
$$(50-3)(50+3)=50^2-3^2=2500-9=2491$$
(b)
$47\times53=2491$ (matches). This works because any such pair can be written as $(n-a)(n+a)=n^2-a^2$, which is often much easier to calculate than direct multiplication when $n$ is a round number like 50.
QUESTION 5
6 marks
Criterion B
Hard
Investigate a mental-maths trick for multiplying numbers close to a round number, using $97\times103$.
a.
Write $97\times103$ in the form $(100-a)(100+a)$ for an appropriate value of $a$, then apply the difference-of-squares pattern to evaluate it without direct multiplication.
[3] b.
Verify this matches direct multiplication, then explain why this 'trick' only works efficiently when the two numbers are EQUALLY spaced above and below a convenient round number.
[3] Show complete worked solution
(a)
$97\times103=(100-3)(100+3)=100^2-3^2=10000-9=9991$.
(b)
Direct check: $97\times103=9991$. The trick relies specifically on the pattern $(n-a)(n+a)=n^2-a^2$, which requires the two numbers to be exactly $a$ below and $a$ above some central value $n$ — if the numbers aren't symmetric around a round number this way, they can't be written in this convenient $(n-a)(n+a)$ form, and the shortcut doesn't apply directly.
QUESTION 6
3 marks
Criterion C
Medium
A student expands $(5x-2)(5x+2)$ as $5x^2-4$ (forgetting to square the coefficient 5 as well as the $x$).
a.
Explain the error precisely, and give the correct expansion.
[3] Show complete worked solution
(a)
The student only squared the VARIABLE part ($x\to x^2$) but forgot the COEFFICIENT must also be squared: $(5x)^2=25x^2$, not $5x^2$. Correct: $(5x-2)(5x+2)=(5x)^2-2^2=25x^2-4$.
QUESTION 7
7 marks
Criterion D
Hard
A square garden plot has side length $(x+20)$m. A smaller square garden plot has side length $(x-20)$m, made from the same available materials budget scaled differently.
a.
Write and expand an expression for the DIFFERENCE in area between the two plots (larger minus smaller).
[3] b.
Alternatively, this difference could be found by factoring $(x+20)^2-(x-20)^2$ as a difference of squares FIRST: $[(x+20)+(x-20)][(x+20)-(x-20)]$. Simplify this factored form, and confirm it matches your answer to part (a).
[4] Show complete worked solution
(a)
$$(x+20)^2-(x-20)^2$$ Using the earlier pattern $(x+a)^2-(x-a)^2=4ax$ with $a=20$: $$=80x$$
(b)
$[(x+20)+(x-20)][(x+20)-(x-20)] = [2x][40] = 80x$ — matches part (a) exactly, confirming both approaches (expanding perfect squares directly, or factoring as a difference of squares first) give the same correct result.
QUESTION 8
7 marks
Criterion D
Hard
A manufacturer produces circular metal discs. The area difference between a disc of radius $(r+2)$cm and a smaller disc of radius $(r-2)$cm (using $\text{Area}=\pi r^2$, and treating the DIFFERENCE OF SQUARES pattern on the radii) needs to be calculated for material cost estimates.
a.
Write an expression for the area difference, $\pi(r+2)^2 - \pi(r-2)^2$, and simplify using the difference-of-squares / expansion pattern.
[4] b.
If $r=15$cm, find the exact area difference (in terms of $\pi$) and as a decimal correct to 1 decimal place.
[3] Show complete worked solution
(a)
$$\pi[(r+2)^2-(r-2)^2] = \pi[4(2)r] = 8\pi r$$ (using the pattern $(x+a)^2-(x-a)^2=4ax$ with $a=2$).
(b)
$8\pi(15)=120\pi \approx 376.99 \approx 377.0 \text{ cm}^2$.
QUESTION 9
6 marks
Criterion D
Medium
A solar-panel installation compares two square arrangements. Their side lengths are $(x+50)$ cm and $(x-50)$ cm, where $x$ is a whole number of centimetres and $x>50$.
a.
Write and simplify an expression for the difference in area between the two arrangements.
[3] b.
The larger arrangement must exceed the smaller one by at least $8000\text{ cm}^2$. Find the minimum permitted whole-number value of $x$, remembering that both side lengths must be positive.
[3] Show complete worked solution
(a)
$(x+50)^2-(x-50)^2 = 4(50)x = 200x$ (using the established pattern).
(b)
The area condition gives $200x\ge8000$, so $x\ge40$. The smaller side must also be positive: $x-50>0$, so $x>50$. Since $x$ is a whole number of centimetres, $$\boxed{x=51\text{ cm}}.$$