LAWS OF ALGEBRA

QUESTION 1 [6 marks] — Criterion A Medium

Expand each product:

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

[2]

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$$x^2+5x+3x+15=x^2+8x+15$$

b. $(x-2)(x+7)$

[2]

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$$x^2+7x-2x-14=x^2+5x-14$$

c. $(2x+1)(x-4)$

[2]

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$$2x^2-8x+x-4=2x^2-7x-4$$

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]

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$(x+2)(x+5)=x^2+7x+10$. $(x+3)(x+4)=x^2+7x+12$. $(x+1)(x+6)=x^2+7x+6$.

b. In each 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]

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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]

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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. Verify the correct answer using $x=1$.

[2]

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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]

Show Solution

$$(x+15)(x+8)=x^2+8x+15x+120=x^2+23x+120$$

b. If $x=10$ m, find the numerical area using your expanded expression, and verify against the original dimensions.

[2]

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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]

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$$x^2+13x+36$$

b. $(2x+3)(x-5)$

[3]

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$$2x^2-7x-15$$

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]

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$(x+9)(x+4)=x^2+13x+36$. 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]

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$(x+3)(x+5)=x^2+8x+15$ — constant POSITIVE. $(x-3)(x-5)=x^2-8x+15$ — constant POSITIVE. $(x+3)(x-5)=x^2-2x-15$ — constant NEGATIVE.

b. 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]

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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]

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$$x^2+19x+84$$

b. If regulations require the buffer zone $x$ to be at least 3m, and the company's site can accommodate AT MOST 200 m² for this foundation, determine the MAXIMUM value of $x$ (to the nearest whole metre) that satisfies BOTH constraints, using your expanded expression.

[4]

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Testing values: $x=3$: $9+57+84=150\le200$ ?. $x=6$: $36+114+84=234>200$ ?. $x=5$: $25+95+84=204>200$ ? (close). $x=4$: $16+76+84=176\le200$ ?. So the maximum whole-metre value satisfying both constraints ($x\ge3$ and area $\le200$) is $x=4$m.

The product (a + b)(c + d)