Sec 2 Maths: Expansion & Factorisation of Quadratic Expressions, practice questions & worked solutions

(a)

$$\begin{aligned}3(3x-2)+1&=9x-6+1\\ &=9x-5\end{aligned}$$

(b)

Write $x$ as $\dfrac{3x}{3}$ so both terms share the denominator $3$:

$$\begin{aligned}\frac{2x}{3}+x&=\frac{2x}{3}+\frac{3x}{3}\\ &=\frac{5x}{3}\end{aligned}$$

(c)

The $-2$ multiplies both terms in the bracket, so $-2\times(-5)=+10$:

$$\begin{aligned}3x-2(x-5)&=3x-2x+10\\ &=x+10\end{aligned}$$

(a)

This is the difference of two squares, $(a+b)(a-b)=a^2-b^2$, with $a=3p$ and $b=4$:

$$\begin{aligned}(3p+4)(3p-4)&=(3p)^2-4^2\\ &=9p^2-16\end{aligned}$$

(b)

$$\begin{aligned}(2c-5)(c-3)&=2c(c-3)-5(c-3)\\ &=2c^2-6c-5c+15\\ &=2c^2-11c+15\end{aligned}$$

(a)

$$\begin{aligned}(2x+3y)(x+2y)&=2x^2+4xy+3xy+6y^2\\ &=2x^2+7xy+6y^2\end{aligned}$$

(b)

Expand $(x+y)(x-y)=x^2-y^2$ inside a bracket first, so the minus sign applies to both terms:

$$\begin{aligned}3x^2-(x+y)(x-y)&=3x^2-(x^2-y^2)\\ &=3x^2-x^2+y^2\\ &=2x^2+y^2\end{aligned}$$

(a)

$$\begin{aligned}(3x+4)(x-2)&=3x^2-6x+4x-8\\ &=3x^2-2x-8\end{aligned}$$

(b)

Take out the common factor $2$ first, then factorise the quadratic that remains:

$$\begin{aligned}6y^2+10y-4&=2(3y^2+5y-2)\\ &=2(3y-1)(y+2)\end{aligned}$$

Check: $(3y-1)(y+2)=3y^2+6y-y-2=3y^2+5y-2$.

Use $(a-b)^2=a^2-2ab+b^2$ with $a=3p$ and $b=5q$, then multiply by $2$:

$$\begin{aligned}2(3p-5q)^2&=2\left[(3p)^2-2(3p)(5q)+(5q)^2\right]\\ &=2(9p^2-30pq+25q^2)\\ &=18p^2-60pq+50q^2\end{aligned}$$

Expand $(2x-y)^2=4x^2-4xy+y^2$ first:

$$\begin{aligned}3(2x-y)^2-4(1+3x^2)&=3(4x^2-4xy+y^2)-4-12x^2\\ &=12x^2-12xy+3y^2-4-12x^2\\ &=3y^2-12xy-4\end{aligned}$$

Group the terms in pairs and take out the common factor of each pair:

$$\begin{aligned}4ab+6b+16a+24&=(4ab+6b)+(16a+24)\\ &=2b(2a+3)+8(2a+3)\\ &=(2a+3)(2b+8)\\ &=2(2a+3)(b+4)\end{aligned}$$

The bracket $(2b+8)$ still has a common factor $2$, so take it out to factorise completely.

(a)

Take out the common factor $3$, then use $a^2-b^2=(a+b)(a-b)$:

$$\begin{aligned}3x^2-27y^2&=3(x^2-9y^2)\\ &=3\left[x^2-(3y)^2\right]\\ &=3(x-3y)(x+3y)\end{aligned}$$

(b)

Group the first two terms and the last two terms:

$$\begin{aligned}ab-7a-3b^2+21b&=a(b-7)-3b(b-7)\\ &=(a-3b)(b-7)\end{aligned}$$

(a)

$$\begin{aligned}x^2-81y^2&=x^2-(9y)^2\\ &=(x-9y)(x+9y)\end{aligned}$$

(b)

Find two numbers with product $-14$ and sum $-5$: these are $-7$ and $2$.

$$\begin{aligned}x^2-5x-14&=(x+2)(x-7)\end{aligned}$$

(c)

Group $3xy-2y^2$ and $-6y+9x$; both pairs contain $(3x-2y)$:

$$\begin{aligned}3xy-2y^2-6y+9x&=y(3x-2y)+3(3x-2y)\\ &=(y+3)(3x-2y)\end{aligned}$$

(a)

$$\begin{aligned}2a^2-98b^2&=2(a^2-49b^2)\\ &=2(a+7b)(a-7b)\end{aligned}$$

(b)

We need $(5y+p)(3y+q)$ with $pq=-4$ and $5q+3p=-17$. Trying $p=1$, $q=-4$ gives $-20+3=-17$:

$$\begin{aligned}15y^2-17y-4&=(5y+1)(3y-4)\end{aligned}$$

Check: $(5y+1)(3y-4)=15y^2-20y+3y-4=15y^2-17y-4$.

(c)

$$\begin{aligned}10yz-15y+8xz-12x&=(10yz-15y)+(8xz-12x)\\ &=5y(2z-3)+4x(2z-3)\\ &=(2z-3)(5y+4x)\end{aligned}$$

(a)

$$\begin{aligned}3xy-6ay-4x+8a&=3y(x-2a)-4(x-2a)\\ &=(x-2a)(3y-4)\end{aligned}$$

(b)

$m^2-9n^2$ is a difference of two squares, so it factorises into the two given expressions:

$$\begin{aligned}m^2-9n^2&=m^2-(3n)^2\\ &=(m+3n)(m-3n)\\ &=(6)(-5)\\ &=-30\end{aligned}$$

(a)

$$\begin{aligned}x^2-121&=x^2-11^2\\ &=(x+11)(x-11)\end{aligned}$$

(b)

Choose $x$ so that $x^2-121=779$:

$$\begin{aligned}x^2&=779+121\\ x^2&=900\\ x&=30\quad(\text{taking the positive value})\end{aligned}$$

Substitute $x=30$ into the factorised form:

$$\begin{aligned}779&=(30+11)(30-11)\\ &=41\times 19\end{aligned}$$

The two factors are $19$ and $41$.

(a)

We need $(2x+p)(x+q)$ with $pq=-15$ and $2q+p=7$. Taking $p=-3$, $q=5$ gives $10-3=7$:

$$\begin{aligned}2x^2+7x-15&=(2x-3)(x+5)\end{aligned}$$

(b)

The expression is the one in (a) with $x$ replaced by $(2y+1)$. So let $x=2y+1$:

$$\begin{aligned}2(2y+1)^2+7(2y+1)-15&=\left[2(2y+1)-3\right]\left[(2y+1)+5\right]\\ &=(4y-1)(2y+6)\\ &=2(4y-1)(y+3)\end{aligned}$$

The bracket $(2y+6)$ has a common factor $2$, which is taken out to factorise completely.

(a)

$$\begin{aligned}(x+y)^2&=x^2+2xy+y^2\\ 50&=x^2+y^2+2(5)\\ x^2+y^2&=40\end{aligned}$$

(b)

$$\begin{aligned}(x-y)^2&=x^2-2xy+y^2\\ &=(x^2+y^2)-2xy\\ &=40-2(5)\\ &=30\end{aligned}$$

(a)

Multiply the last two brackets first: they form a difference of two squares.

$$\begin{aligned}\left(x+\tfrac{1}{4}\right)\left(x-\tfrac{1}{4}\right)&=x^2-\tfrac{1}{16}\end{aligned}$$

Now the product is another difference of two squares:

$$\begin{aligned}\left(x^2+\tfrac{1}{16}\right)\left(x^2-\tfrac{1}{16}\right)&=(x^2)^2-\left(\tfrac{1}{16}\right)^2\\ &=x^4-\frac{1}{256}\end{aligned}$$

(b)

Factorise the denominators: $3x-3=3(x-1)$ and $2-2x=-2(x-1)$. So both fractions share the factor $(x-1)$, and $x\neq 1$.

$$\begin{aligned}\frac{5x}{3(x-1)}-\frac{3}{2(x-1)}&=2\\ \frac{2(5x)-3(3)}{6(x-1)}&=2\\ \frac{10x-9}{6(x-1)}&=2\\ 10x-9&=12(x-1)\\ 10x-9&=12x-12\\ 2x&=3\\ x&=\frac{3}{2}\end{aligned}$$