Algebra — Introduction & Intermediate

Q1 (8 marks)

Start: $ax^2 + bx + c = 0$, $a\neq 0$.

• Divide by $a$: $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$. (1)
• Move constant: $x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$. (1)
• Complete square — add $\left(\dfrac{b}{2a}\right)^2$ to both sides (half the linear coefficient, squared). (1) $x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}$
• LHS is a perfect square: $\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2}$. (1)
• Take square roots: $x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2-4ac}}{2a}$. (1)
• Isolate $x$: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. (2)
• Condition: real roots exist iff $b^2 - 4ac \ge 0$ (the discriminant is non-negative, else square root is imaginary). (1)

Q2 (10 marks)

(a) Remainder Theorem proof (4): By the division algorithm, $p(x) = (x-a)q(x) + r$ where $r$ is constant (degree of remainder < degree of divisor = 1). (2) Substitute $x = a$: $p(a) = (a-a)q(a) + r = 0 + r = r$. Hence remainder $= p(a)$. (2)

(b) Factor Theorem (2): $(x-a)$ is a factor $\iff$ remainder $= 0 \iff p(a) = 0$. (2)

(c) Application (4): $p(2) = 8 - 16 + 2 + 6 = 0$ → $(x-2)$ is a factor. (2) Divide: $x^3 - 4x^2 + x + 6 = (x-2)(x^2 - 2x - 3)$. (1) Factor quadratic: $x^2 - 2x - 3 = (x-3)(x+1)$. $p(x) = (x-2)(x-3)(x+1). \quad **(1)**$

Q3 (9 marks)

If $\alpha,\beta$ are roots, then $ax^2+bx+c = a(x-\alpha)(x-\beta)$. (1) Expand RHS: $a(x^2 - (\alpha+\beta)x + \alpha\beta) = ax^2 - a(\alpha+\beta)x + a\alpha\beta$. (2) Compare coefficients:

• $x$-term: $-a(\alpha+\beta) = b \Rightarrow \alpha+\beta = -\dfrac{b}{a}$. (2)
• constant: $a\alpha\beta = c \Rightarrow \alpha\beta = \dfrac{c}{a}$. (2)

For $2x^2 - 7x + 3 = 0$: $\alpha+\beta = 7/2$, $\alpha\beta = 3/2$. $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \tfrac{49}{4} - 3 = \tfrac{37}{4}. \quad **(2)**$

Q4 (8 marks)

$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. (2) Rearrange: $a^3 + b^3 = (a+b)^3 - 3a^2b - 3ab^2 = (a+b)^3 - 3ab(a+b)$. (2) Factor out $(a+b)$: $a^3+b^3 = (a+b)\big[(a+b)^2 - 3ab\big] = (a+b)(a^2 - ab + b^2). \quad **(2)**$

Apply with $a = 3x$, $b = 2$: $27x^3 + 8 = (3x+2)(9x^2 - 6x + 4). \quad **(2)**$

Q5 (8 marks)

Domain/logic: squaring maps $u = v$ and $u = -v$ to same $u^2=v^2$; hence a squared equation may satisfy solutions of the "wrong sign" branch — these are extraneous. Also require RHS $x-4 \ge 0$ and radicand $\ge 0$. (2)

Square both sides: $2x + 7 = (x-4)^2 = x^2 - 8x + 16$. (1) $\Rightarrow x^2 - 10x + 9 = 0 \Rightarrow (x-1)(x-9)=0 \Rightarrow x = 1$ or $x = 9$. (2)

Check $x=1$: $\sqrt{9}=3$; RHS $=1-4=-3$. $3 \ne -3$ → extraneous. (1) Check $x=9$: $\sqrt{25}=5$; RHS $=9-4=5$. ✓ valid. (1) Solution: $x = 9$. (1)

Q6 (7 marks)

$|3x-5| \le 7$ means the distance of $3x-5$ from 0 is at most 7, so it lies between $-7$ and $7$ simultaneously (AND — both bounds must hold). (2) $-7 \le 3x - 5 \le 7$ (1) Add 5: $-2 \le 3x \le 12$. (1) Divide by 3: $-\dfrac{2}{3} \le x \le 4$. (2) Number line: closed dots at $-\tfrac23$ and $4$, shaded segment between. (1)

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 -2/3           4

[
 {"claim":"p(x)=x^3-4x^2+x+6 factors as (x-2)(x-3)(x+1)","code":"x=symbols('x'); result = expand((x-2)*(x-3)*(x+1)) == expand(x**3-4*x**2+x+6)"},
 {"claim":"alpha^2+beta^2 = 37/4 for 2x^2-7x+3","code":"s=Rational(7,2); p=Rational(3,2); result = (s**2-2*p) == Rational(37,4)"},
 {"claim":"27x^3+8 = (3x+2)(9x^2-6x+4)","code":"x=symbols('x'); result = expand((3*x+2)*(9*x**2-6*x+4)) == expand(27*x**3+8)"},
 {"claim":"radical eqn solution x=9 valid, x=1 extraneous","code":"x=symbols('x'); sols=solve(Eq(2*x+7,(x-4)**2),x); valid=[s for s in sols if sqrt(2*s+7)==s-4]; result = valid==[9]"}
]