Multiply the term outside by every term inside: $4x(x-5) = 4x^2-20x$. The main danger is a negative multiplier — in $-2(3x-7)$, both inner terms flip sign: $-2\times3x=-6x$ and $-2\times(-7)=+14$, giving $-6x+14$. A common mistake is only distributing the negative to the first term and forgetting it applies to the second too.

Every term in the first bracket multiplies every term in the second (FOIL, or a grid for a visual check), then like terms are collected: $(x-5)(x-4) = x^2-4x-5x+20 = x^2-9x+20$. This isn't just a standalone skill — expanding back out is the most reliable way to verify a factorisation is correct.

Find the HCF of the numerical coefficients and the lowest shared power of each variable, then write that HCF outside a bracket containing what's left.

Example: $15x^2y^3+9x^4y$. Numerical HCF of 15 and 9 is 3. Lowest power of $x$ (between $x^2,x^4$) is $x^2$. Lowest power of $y$ (between $y^3,y$) is $y$. Full HCF: $3x^2y$. Result: $3x^2y(5y^2+3x^2)$.

Find two numbers that multiply to give $c$ and add to give $b$, then write $(x+p)(x+q)$.

Example: $y^2-9y+20$ — needs two numbers multiplying to $+20$ and summing to $-9$: that's $-4$ and $-5$, giving $(y-4)(y-5)$.

The sign pattern is the usual sticking point: a positive $c$ with a negative $b$ (like this example) always means both factors are negative. By contrast, if $c$ is negative, the two factors must have opposite signs, and if $b$ is positive with $c$ positive, both factors are positive. Working through the sign of $b$ and $c$ before guessing at factor pairs saves a lot of trial and error. Always expand your brackets back out afterwards to confirm you recover the original expression exactly — this single habit catches almost every sign slip before it costs a mark.

Find two numbers that multiply to give $a\times c$ and add to give $b$, then use them to split the middle term into two parts before grouping. For $2x^2+7x+3$: $a\times c=6$, and the pair that multiplies to 6 and adds to 7 is $1$ and $6$. Rewrite as $2x^2+x+6x+3$, then group: $x(2x+1)+3(2x+1)=(2x+1)(x+3)$. This "splitting and grouping" approach extends the same logic used for $a=1$ quadratics, rather than requiring a completely new method.

If a quadratic has a common numerical factor across all terms, e.g. $4x^2-12x+8$, extract it first: $4(x^2-3x+2)$. This converts a harder-looking quadratic into the simple $a=1$ case, which then factorises normally as $4(x-1)(x-2)$.

Recognise this pattern whenever you see two perfect squares separated by a minus sign: $4x^2-25y^2 = (2x)^2-(5y)^2 = (2x-5y)(2x+5y)$.

Factorise $2x^2 + 8x - 10$ fully, then verify your answer by expansion.

1. Extract the common factor of 2: $2(x^2+4x-5)$.
2. Factorise the simple quadratic: find two numbers multiplying to $-5$, summing to $+4$ — that's $+5$ and $-1$. So $x^2+4x-5=(x+5)(x-1)$.
3. Full factorisation: $2(x+5)(x-1)$.
4. Verify: $2(x+5)(x-1) = 2(x^2-x+5x-5)=2(x^2+4x-5)=2x^2+8x-10$ ✓.