LAWS OF ALGEBRA

QUESTION 1 [6 marks] — Criterion A Medium

Expand and simplify each expression:

a. $3(2x+5)$

[1]

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$$6x+15$$

b. $(x+4)(x+2)$

[2]

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$$x^2+2x+4x+8=x^2+6x+8$$

c. $(x+3)^2$

[2]

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$$x^2+6x+9$$

d. $(x+5)(x-5)$

[1]

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$$x^2-25$$

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.

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$2(x+3)$: 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. Describe one visual clue in each expression's shape that helps you identify which expansion rule to use.

[2]

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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]

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$(x+3)(x+3)=x^2+3x+3x+9=x^2+6x+9$. $(x+3)^2=x^2+2(3)x+9=x^2+6x+9$. Identical results.

b. Explain why this makes sense — what does squaring a bracket actually mean?

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$(x+3)^2$ 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]

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$$A=(x+7)(x+2)=x^2+2x+7x+14=x^2+9x+14$$

b. If $x=5$, find the numerical area, and verify it against the original (unexpanded) dimensions.

[2]

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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]

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$4(x+3)=4x+12$. $2(x-5)=2x-10$.

b. Combine, taking care with the SUBTRACTION of the second bracket's expansion.

[2]

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$$4x+12-(2x-10) = 4x+12-2x+10 = 2x+22$$

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 $(x+5)(x+3)$.

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$(x+5)(x-5)=x^2-25$ (no middle term). $(x+5)^2=x^2+10x+25$. $(x+5)(x+3)=x^2+8x+15$.

b. 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]

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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]

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$(x+1)(x+2)=x^2+3x+2$. Then $(x^2+3x+2)(x+3) = x^3+3x^2+3x^2+9x+2x+6 = x^3+6x^2+11x+6$.

b. 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 $1\times2\times3$.

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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]

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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. Verify the correct answer using $x=2$.

[2]

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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]

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$$(x+8)(x+4) = x^2+12x+32$$

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.

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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: $4(10)+20=60$ ?.

Expansion laws