Factorising expressionsIB MYP Maths Standard: Revision notes

Factorising is the reverse of expanding: it writes an expression as a product. First find the highest common factor (HCF) of every term, numbers and letters, and take it outside a bracket. $6x^2+9x=3x(2x+3)$ and $12a^2b-8ab^2=4ab(3a-2b)$. The bracket must contain no further common factor, otherwise the expression is not fully factorised. Always check by expanding: $3x(2x+3)=6x^2+9x$.

With four terms, group them in pairs and take a common factor from each pair. If the brackets match, they become a new common factor. $xy+3x+4y+12=x(y+3)+4(y+3)=(y+3)(x+4)$. If the brackets do not match, try pairing the terms in a different order or check the signs.

When one square is subtracted from another, $a^2-b^2=(a-b)(a+b)$. For example $x^2-49=(x-7)(x+7)$ and $9-y^2=(3-y)(3+y)$. There is no middle term because $+7x$ and $-7x$ cancel. Take out any common factor first: $3x^2-75=3(x^2-25)=3(x-5)(x+5)$. It works with numbers as well: $53^2-47^2=(53-47)(53+47)=6\times100=600$.

To factorise $x^2+bx+c$, find two numbers with product $c$ and sum $b$, then write $(x+p)(x+q)$. For $x^2+9x+20$ the numbers are $4$ and $5$, so $(x+4)(x+5)$. If $c$ is positive, both numbers have the same sign as $b$. If $c$ is negative, the numbers have different signs and the larger one has the sign of $b$: $x^2+3x-10=(x+5)(x-2)$ and $x^2-x-12=(x-4)(x+3)$. In this subtopic the number in front of $x^2$ is 1.

To simplify an algebraic fraction, factorise the top and bottom and cancel common factors: $\frac{x^2-9}{x+3}=\frac{(x-3)(x+3)}{x+3}=x-3$. To solve a quadratic equation, rearrange so that one side is zero, factorise, and set each bracket equal to zero, because if a product is zero then one factor must be zero. $x^2-5x+6=0$ gives $(x-2)(x-3)=0$, so $x=2$ or $x=3$. Also $x^2+7x=0$ gives $x(x+7)=0$, so $x=0$ or $x=-7$. In a real problem, reject any answer that does not make sense, such as a negative length.