式の展開と因数分解

Multiply the term outside the bracket by every term inside (the distributive law). For example $3x(2x-5)=6x^2-15x$. Watch the signs: when you multiply by a negative term the inner signs flip, as in $-2(x-4)=-2x+8$.

For $(x+p)(x+q)$ form the four products — First, Outer, Inner, Last — and add them. Example: $(3x+4)(x-2)=3x^2-6x+4x-8=3x^2-2x-8$. The most common slip is forgetting to combine the two middle ($x$) terms.

The perfect squares and the difference of two squares are worth memorising: $(a+b{)}^2=a^2+2ab+b^2$, $(a-b{)}^2=a^2-2ab+b^2$, and $a^2-b^2=(a+b)(a-b)$. The “difference of two squares” in particular lets you factorise instantly whenever you see two squares subtracted with no middle term, e.g. $9x^2-25=(3x+5)(3x-5)$.

The golden rule of factorising is to look for a common factor first. Pull out the largest number and letters shared by every term. Example: $6x^3-9x^2=3x^2(2x-3)$. Take both the numerical common factor (here 3) and the lowest power of the shared letter (here $x^2$).

When the leading coefficient is 1, find two numbers that multiply to $c$ and add to $b$. Calling them $p,q$ gives $(x+p)(x+q)$. Example: for $x^2+2x-15$, the pair multiplying to $-15$ and adding to $+2$ is $+5$ and $-3$, so $(x+5)(x-3)$. It speeds things up to notice the patterns: if $c$ is negative the two numbers have opposite signs; if $c$ is positive and $b$ negative, both numbers are negative.

When the leading coefficient is not 1, the reliable method is the “ac method” (splitting the middle term). Compute $a\times c$, find two numbers multiplying to it and adding to $b$, split the middle term, then factor in pairs. Example: for $6x^2+11x-10$, $a\times c=-60$; the pair multiplying to $-60$ and adding to $+11$ is $+15$ and $-4$, giving $6x^2+15x-4x-10=3x(2x+5)-2(2x+5)=(2x+5)(3x-2)$.

The 0580 Extended syllabus (E2.2) also asks you to expand products of more than two brackets, to factorise four terms by grouping, and to factorise expressions of the form $a{x}^3+b{x}^2+cx$. Expand three brackets two at a time. Example: for $(x-2)(x+3)(2x+1)$, first $(x-2)(x+3)=x^2+x-6$, then multiply by $(2x+1)$ to get $2x^3+3x^2-11x-6$. For grouping, take a common factor out of the first two terms and out of the last two; when the same bracket appears, factor it out: $2ax+2bx+5ay+5by=2x(a+b)+5y(a+b)=(a+b)(2x+5y)$. For an expression like $2x^3-x^2-15x$, take out the common factor $x$ first, then factorise the quadratic that is left: $x(2x+5)(x-3)$. The syllabus says “factorise” means factorise fully, so do not leave anything inside a bracket that could still be factorised.