Sec 1 Maths: Basic Algebra — Expansion & Factorisation practice questions & worked solutions

(a) Using a common denominator of $12$:

$$\begin{aligned}&\frac{1.5(5x-7)}{6}-3(x-2)+\frac{7x}{4}\end{aligned}$$
$$\begin{aligned}&= \frac{3(5x-7)-36(x-2)+21x}{12}\end{aligned}$$
$$\begin{aligned}&= \frac{15x-21-36x+72+21x}{12}\end{aligned}$$
$$\begin{aligned}&= \frac{51}{12}\end{aligned}$$
$$\begin{aligned}&= \frac{17}{4}\end{aligned}$$

(b) $x$ dollars $=100x$ cents. Mass in grams:

$$\begin{aligned}\text{mass} &= \frac{100x}{w} \text{ g}\end{aligned}$$

(a) Common factor $5r$:

$$\begin{aligned}15r^{2}s-5pr &= 5r(3rs-p)\end{aligned}$$

(b) Common factor $(3c-5)$:

$$\begin{aligned}-2a(3c-5)+6b(3c-5) &= (3c-5)(-2a+6b)\end{aligned}$$
$$\begin{aligned}&= 2(3c-5)(3b-a)\end{aligned}$$

From the grid, $P(-2, 1)$, $Q(6, 1)$ and $R(4, 5)$.

(a) $PQ$ lies along the line $y = 1$ (horizontal), so

$$\begin{aligned}\text{gradient of } PQ = 0\end{aligned}$$

(b) For $PQRS$ a parallelogram, $\vec{PS} = \vec{QR}$:

$$\begin{aligned}S &= P + (R – Q) \\ S &= (-2 + (4 – 6),\; 1 + (5 – 1)) \\ S &= (-4, 5)\end{aligned}$$

(c) $x = 5$ is the vertical line through all points with $x$-coordinate $5$.

(a)

$$\begin{aligned}\frac{2}{3} – \frac{x-2}{6} – 8x &= \frac{4}{6} – \frac{x-2}{6} – \frac{48x}{6} \\ &= \frac{4 – (x-2) – 48x}{6} \\ &= \frac{4 – x + 2 – 48x}{6} \\ &= \frac{6 – 49x}{6}\end{aligned}$$

(b)

$$\begin{aligned}6\left[ 5(a+2b) – \frac{3}{2}(a+4b) \right] &= 30(a+2b) – 9(a+4b) \\ &= 30a + 60b – 9a – 36b \\ &= 21a + 24b\end{aligned}$$

(c)

$$\begin{aligned}28hk + 14h^2 – 4h(3 – k) &= 28hk + 14h^2 – 12h + 4hk \\ &= 14h^2 + 32hk – 12h \\ &= 2h(7h + 16k – 6)\end{aligned}$$

(a) Common factor $-6b^{2}c$:

$$\begin{aligned}-6b^{2}c – 18b^{2}c^{2} – 12b^{3}c &= -6b^{2}c(1 + 3c + 2b)\end{aligned}$$

(b) Common denominator $30$:

$$\begin{aligned}\frac{3 – 2x}{15} + \frac{2x + 5}{5} – \frac{x – 4}{10} &= \frac{2(3 – 2x)}{30} + \frac{6(2x + 5)}{30} – \frac{3(x – 4)}{30}\end{aligned}$$
$$\begin{aligned}&= \frac{(6 – 4x) + (12x + 30) – (3x – 12)}{30}\end{aligned}$$
$$\begin{aligned}&= \frac{6 – 4x + 12x + 30 – 3x + 12}{30}\end{aligned}$$
$$\begin{aligned}&= \frac{5x + 48}{30}\end{aligned}$$

Note $(q-2p)=-(2p-q)$, so take $(2p-q)$ as a common factor:

$$\begin{aligned}&2p-q+(2p-q)(p-3q)+(7p+5q)(q-2p)\\ &=(2p-q)\big[1+(p-3q)-(7p+5q)\big]\end{aligned}$$
$$\begin{aligned}&=(2p-q)(1+p-3q-7p-5q)\end{aligned}$$
$$\begin{aligned}&=(2p-q)(1-6p-8q)\end{aligned}$$

(a)

$$\begin{aligned}&\frac{3p – 2(q – 1)}{3} – \frac{4p – 5q + 1}{2} \\ &= \frac{2[3p – 2(q-1)] – 3(4p – 5q + 1)}{6} \\ &= \frac{2(3p – 2q + 2) – 3(4p – 5q + 1)}{6} \\ &= \frac{6p – 4q + 4 – 12p + 15q – 3}{6} \\ &= \frac{-6p + 11q + 1}{6} \\ &= \frac{11q – 6p + 1}{6}\end{aligned}$$

(b)

$$\begin{aligned}&2ay + 4(xy – ab) + 2x(y – 6b) \\ &= 2ay + 4xy – 4ab + 2xy – 12bx \\ &= 2ay + 6xy – 4ab – 12bx \\ &= 2y(a + 3x) – 4b(a + 3x) \\ &= 2(a + 3x)(y – 2b) \\ &= 2(y – 2b)(3x + a)\end{aligned}$$

(i) First $= x$, so

$$\begin{aligned}\text{second} &= 2x \\ \text{third} &= x – 6 \\ \text{fourth} &= \frac{2}{7}(x – 6)\end{aligned}$$

(ii)

$$\begin{aligned}\frac{x + 2x + (x – 6) + \frac{2}{7}(x – 6)}{4} &= 27 \\ x + 2x + x – 6 + \frac{2}{7}(x – 6) &= 108 \\ 7x + 14x + 7(x – 6) + 2(x – 6) &= 756 \\ 7x + 14x + 7x – 42 + 2x – 12 &= 756 \\ 30x – 54 &= 756 \\ 30x &= 810 \\ x &= 27\end{aligned}$$

(iii)

$$\begin{aligned}\text{first} + \text{third} &= x + (x – 6) \\ &= 27 + 21 \\ &= 48\end{aligned}$$

(a) Expand the inner brackets:

$$\begin{aligned}2h\left(\frac{1}{h}+4k\right) &= 2 + 8hk\\ 6k\left(2h-\frac{2}{3k}\right) &= 12hk – 4\end{aligned}$$

Inside the square bracket:

$$\begin{aligned}(2 + 8hk) – (12hk – 4) &= 6 – 4hk\end{aligned}$$

Multiply by $-\frac{1}{2}$:

$$\begin{aligned}-\frac{1}{2}(6 – 4hk) &= -3 + 2hk\\ &= 2hk – 3\end{aligned}$$

(b)

$$\begin{aligned}2q – p – (5pr – 10qr) &= 2q – p – 5pr + 10qr\\ &= (2q – p) + 5r(2q – p)\\ &= (2q – p)(1 + 5r)\\ &= -(p – 2q)(1 + 5r)\end{aligned}$$

(a)(i) Greatest $x – y$ uses largest $x$ and smallest $y$.

$$\begin{aligned}x – y &= 7 – (-5) \\ x – y &= 12\end{aligned}$$

(a)(ii) Since $-5 \le y \le 3$, the range includes $y = 0$, giving the smallest square.

$$y^2 = 0$$

(b) Round each value to one significant figure.

$$\begin{aligned}\frac{42100}{2.95 \times 996} &\approx \frac{40000}{3 \times 1000} \\ &\approx \frac{40000}{3000} \\ &\approx 13.3 \\ &\approx 10\end{aligned}$$

(c)(i)

$$\begin{aligned}6x – \frac{2(3 – 11x)}{5} &= \frac{30x – 2(3 – 11x)}{5} \\ &= \frac{30x – 6 + 22x}{5} \\ &= \frac{52x – 6}{5}\end{aligned}$$

(c)(ii)

$$\begin{aligned}\sqrt[3]{\frac{216v^9}{27u^{12}}} &= \frac{6v^3}{3u^4} \\ &= \frac{2v^3}{u^4}\end{aligned}$$
$$\begin{aligned}\frac{4v^2}{u^3} \div \frac{2v^3}{u^4} &= \frac{4v^2}{u^3} \times \frac{u^4}{2v^3} \\ &= \frac{2u}{v}\end{aligned}$$

(a)(i) $a^2$ is least when $a = 0$ (an allowed integer), giving $a^2 = 0$; $b$ least $= -5$.

$$\begin{aligned}0 + (-5)\end{aligned}$$
$$= -5$$

(a)(ii) Largest product from $a = -5$ and $b = -5$:

$$\begin{aligned}(-5)(-5)\end{aligned}$$
$$= 25$$

(b)(i)

$$\begin{aligned}\frac{9}{4x} \times \frac{16y}{3} \\ \frac{9 \times 16y}{4x \times 3}\end{aligned}$$
$$= \frac{12y}{x}$$

(b)(ii)

$$\begin{aligned}\frac{12}{6} – \frac{3 + w}{6} \\ \frac{12 – 3 – w}{6}\end{aligned}$$
$$= \frac{9 – w}{6}$$

(a)

$$\begin{aligned}28x-20y &= 9x-12y\\ 28x-9x &= 20y-12y\\ 19x &= 8y\\ \frac{x}{y} &= \frac{8}{19}\end{aligned}$$

(b)(i)

$$\begin{aligned}15a-9b-18bk+30ak &= 3(5a-3b)+6k(5a-3b)\\ &= 3(5a-3b)(1+2k)\end{aligned}$$

(b)(ii)

$$t^{2}+t=t(t+1)$$

$t$ and $t+1$ are consecutive integers, so one of them is even and the other is odd. The product of an even and an odd number is always even, hence $t^{2}+t$ is always even.

(a) Area of triangle $AOB=\dfrac{1}{2}\times6\times2=6$. Area of triangle $AOC$ is twice this $=12$.

$$\begin{aligned}\frac{1}{2}\times6\times OC &= 12\\ OC &= 4\end{aligned}$$

As $C$ is on the negative $y$-axis: $C=(0,-4)$.

(b) $B$ and $C$ both lie on the $y$-axis, so the line $BC$ is

$$x=0$$

(c) For parallelogram $ABCD$, $\vec{AD}=\vec{BC}$, so $D=A+C-B$:

$$\begin{aligned}D &= (6+0-0,\ 0+(-4)-2)\\ &= (6,-6)\end{aligned}$$

(a)

$$\begin{aligned}30abc + 3ab – 18bc = 3b(10ac + a – 6c)\end{aligned}$$

(b)

$$\begin{aligned}4y + 3(2z – 6) &= 4y + 6z – 18 \\ &= 2(2y + 3z – 9)\end{aligned}$$

(a)

$$\begin{aligned}3a \times 4b – 2c = 12ab – 2c\end{aligned}$$

(b)

$$\begin{aligned}p – 3(q – 2p) &= p – 3q + 6p \\ &= 7p – 3q\end{aligned}$$

(c)

$$\begin{aligned}\frac{5x – 1}{6} + \frac{7(x + 3)}{8} &= \frac{4(5x – 1)}{24} + \frac{21(x + 3)}{24} \\ &= \frac{20x – 4}{24} + \frac{21x + 63}{24} \\ &= \frac{20x – 4 + 21x + 63}{24} \\ &= \frac{41x + 59}{24}\end{aligned}$$

(a)

$$\begin{aligned}2a(2+b)-4(b+2) &= 2a(2+b)-4(2+b) \\ &= (2+b)(2a-4) \\ &= 2(2+b)(a-2)\end{aligned}$$

(b)

$$\begin{aligned}3x-\frac{1}{2}\left[4+5(2-4x)\right] &= 3x-\frac{1}{2}\left[4+10-20x\right] \\ &= 3x-\frac{1}{2}(14-20x) \\ &= 3x-7+10x \\ &= 13x-7\end{aligned}$$

(a) Substitute $x=3$:

$$\begin{aligned}4(3a-2)-1 &= a(3+6) \\ 12a-8-1 &= 9a \\ 12a-9 &= 9a \\ 3a &= 9 \\ a &= 3\end{aligned}$$

(b) Remaining $=2-\dfrac{3+4y}{21}-\dfrac{2(4-5y)}{7}$. Use common denominator $21$:

$$\begin{aligned}\text{Remaining} &= \frac{42}{21}-\frac{3+4y}{21}-\frac{6(4-5y)}{21} \\ &= \frac{42-(3+4y)-(24-30y)}{21} \\ &= \frac{42-3-4y-24+30y}{21} \\ &= \frac{15+26y}{21}\end{aligned}$$

The remaining piece is $\dfrac{15+26y}{21}$ m.

(a) Common factor $3a^2mn$:

$$3a^2mn(1 – 2am)$$

(b)

$$\begin{aligned}&\tfrac{1}{3}\left[12 – 6 + 3x\right] – 5x\\ &= \tfrac{1}{3}(6 + 3x) – 5x\\ &= 2 + x – 5x\\ &= 2 – 4x\end{aligned}$$

(a) Substitute $a=-2$:

$$-2\left[6 – (-2)(p+3)\right] = p\left[(-2)^2 + 4\right]$$
$$-2\left[6 + 2(p+3)\right] = 8p$$
$$-2(2p + 12) = 8p$$
$$-4p – 24 = 8p$$
$$-24 = 12p$$
$$p = -2$$

(b) In a rhombus, co-interior angles are supplementary:

$$\angle BCD = 180^\circ – 116^\circ = 64^\circ$$

The diagonal $CA$ bisects $\angle BCD$:

$$\angle ACD = \tfrac{1}{2}(64^\circ) = 32^\circ$$