Maths — Answers & Corrections

Exercises 2.1, 3.3, 3.4, 3.5 · for Ethan · 2026-06-17

● Answer — Answer shown — check yours against it.

● Redo — Direction only — redo these yourself (no answer given).

Exercise 2.1 — Simultaneous Equations

Q1Answer$(3,9)$ and $(-2,4)$

Q2RedoYour factorisation $(2x+2)(x-4)=0$ is right — re-check the sign of the root from $2x+2=0$ (you wrote $x=1$).

Q3Redo$x^2+y^2=25$ is not $x+y=5$. Substitute $y=x-1$ into $x^2+y^2$, keep the squares, expand → a quadratic with two solutions.

Q4Answer$(1,4)$ and $(-2,-2)$

Q5Answer$(0,1)$ and $(\tfrac12,\tfrac12)$

Q6Answer$(2,1)$ and $(-1,-3)$

Q7Answer$(\tfrac32,4)$ and $(2,3)$

Q8Answer$(3,1)$ and $(9,7)$

Q9Answer$(1,3)$ and $(1.8,2.6)$

Q10Answer$\left(\sqrt{\tfrac35},\,2\sqrt{\tfrac35}\right)$ and its negative

Q11Answer$(1,2)$ and $(2,1)$

Q12Answer$(1,2)$ and $(4,-4)$

Q13Answer$(-\tfrac12,\tfrac16)$ and $(1,-\tfrac13)$

Q14RedoThis question is $x+y=4,\ x^2+y^2=10$ — you solved Q15 by mistake. Redo against the right pair.

Q15Answer$(1,3)$ and $(-1,-3)$

Q16Answer$(0,-\tfrac12)$ and $(-1,-1)$

Q17Answer$(-1,2)$ and $(\tfrac{15}{2},-\tfrac75)$

Q18Answer$(4,3)$ and $(-\tfrac32,-8)$

Q19RedoSkipped — find where $y=1-2x$ cuts $x^2+y^2=2$ (same substitution method).

Q20Answer$8.5$ and $2.5$

Q21Answersides $23$ cm and $17$ cm

Q22Answer$AB=6\sqrt5$

Q23RedoYou found the two points — the question wants the MIDPOINT (average them). Also re-check the first point's $x$.

Q24Answer$AB=5\sqrt2$

Q25Answer$P=(2,2)$

Q26Answer$y=-2x-1$

Exercise 3.3 — Factor Theorem & Polynomials

Q1aAnswer$f(4)=0$ → $x-4$ is a factor ✓

Q1bAnswer$f(-1)=0$ ✓

Q1cAnswer$f(2)=0$ ✓

Q1dAnswerroot $x=-\tfrac13$, $f(-\tfrac13)=0$ ✓

Q2aAnswer$a=29$

Q2bAnswer$a=546$

Q2cRedoSetup is right — combine the number terms and isolate $\tfrac{9a}{4}$ to finish.

Q3Answer$b=-2a-2$

Q4aAnswer$a=0,\ b=-19$

Q4bAnswer$a=2,\ b=38$

Q4cAnswer$a=-12,\ b=-7$

Q5Answer$a=-6$

Q6Answer$a=\tfrac{27}{2},\ b=\tfrac{11}{2}$

Q7aAnswer$p=1,\ q=9$

Q7bRedoA cubic has three linear factors and you have two — what is the third? (test $x=-3$, then state the factor theorem).

Q8aAnswer$f(-a)=0 \Rightarrow a^3-4a^2+3a=0$ ✓

Q8bAnswer$a=0,\ 1,\ 3$

Exercise 3.4 — Factorising & Solving Cubics

Q1Answer$(x-1)(2x-1)(x+1)$

Q2Answera) $(x-2)(x^2+4x+5)$; b) $(x-2)(x+2)(x+4)$; c) $x(x-6)(2x+3)$; d) $(x-7)(x-2)(x+1)$; e) $(x-4)(x-3)(2x+1)$; f) $(x-2)(x+3)(3x-1)$; g) $(x-2)(2x-1)(2x+1)$; h) $(x-3)(x+5)(2x-1)$

Q3Answera) $7,1,-5$; b) $3,2,1$; c) $-2,-4,\tfrac13$; d) $1,-4,\tfrac32$; e) $3,-2,\tfrac12$; f) $1,-4,-\tfrac12$; g) $-2,\tfrac12,-\tfrac32$; h) $3,-4,\tfrac52$

Q4Answera) $1,\ -3\pm\sqrt7$; b) $-3,\ \tfrac{-5\pm\sqrt{37}}{2}$; c) $2,\ -2\pm\sqrt3$; d) $1,-4,\tfrac32$

Q5Answer$-\tfrac12,\ -2\pm\sqrt{13}$

Q6Answer$-3,\ 0.54,\ -5.54$

Q7Answera) $f(2)=0$ ✓; b) only real root $x=2$ (since $x^2+x+1$ has no real roots)

Q8RedoYou listed the roots — now form $f(x)$: multiply the $(x-\text{root})$ factors and expand (leading coefficient 1).

Q9RedoSame, but leading coefficient 2 — use $(2x-1)$ for the root $\tfrac12$ before expanding.

Q10Redo$1\pm\sqrt2$ gives $x^2-2x-1$; multiply by $(x+3)$ and expand (result should be monic). Re-check.

Q11Redo$2\pm\sqrt3$ gives $x^2-4x+1$; times $(2x-1)$. Re-check the middle ($x$) term.

Q12Answer$a=2,\ \tfrac{1\pm\sqrt{33}}{8}$

Exercise 3.5 — Remainder Theorem

Q1Redo(a) $=5$, (c) $=76$, (d) $=2$ ✓. For (b), re-add $8-24+22-7$ carefully and re-check the remainder (no answer given).

Q2Answer$a=5,\ b=57,\ c=11$

Q3Answer$a=4,\ b=0$

Q4Answer$a=-6,\ b=-6$

Q5Answer$a=-8,\ b=15$; roots $3,\ \tfrac{-1\pm\sqrt{21}}{2}$

Q6Answer$a=-9,\ b=2$; remainder $5$

Q7Answer$a=-4,\ b=2$

Q8Answer$a=6,\ b=-3$

Q9Answer$b=12-2a$; then $a=5,\ b=2$

Q10Answer$k=5$; remainder $-72$

Q11Answer$a=-8,\ b=-5$; remainder $-30$

Q12Answer$k=32$

Q13Answer$3a^3-2a^2-18a-9=0$; $a=3,\ \tfrac{-7\pm\sqrt{13}}{6}$

Q14Answer$k=3$; remainder $-5$

Q15Answer$a=2,\ b=-5,\ c=7$

Q16RedoChallenge — with roots $1,k,k+1$ and $f(2)=20$, reach $k^2-3k-18=0$, then solve for $k$.