E2.2 - Factorising quadratic expressions

Monic quadratic factors

Can you find two brackets whose product is $x^2+5x+6$, without guessing endlessly? Expansion gives the clue: $(x+2)(x+3)=x^2+3x+2x+6$. The two bracket numbers multiply to $6$ and add to $5$. One pair does both jobs.

In general, expanding $(x+p)(x+q)$ gives $x^2+(p+q)x+pq$. To reverse this, look for numbers with the required sum and product. List factor pairs of the constant first, then test their sums. A positive product needs matching signs; a negative product needs opposite signs. Not every quadratic has integer bracket factors: no integer pair has product $1$ and sum $1$, so this search cannot factorise $x^2+x+1$ into integer brackets.

Change only the constant, from $12$ to $-18$. The bracket numbers must now have opposite signs, so their sum is a difference in size.

If both bracket numbers are negative, their product is positive but their sum is negative. For example, $x^2-7x+12=(x-3)(x-4)$.

For a negative constant, choose opposite signs and check which number must have the greater magnitude.

Non-monic quadratic factors

When the coefficient of $x^2$ is not $1$, the cross-products still determine the middle term. In $(2x+1)(x+3)$, the cross-products are $6x$ and $x$. Their coefficients multiply to $6$, which is also $2\times3$, the product of the leading coefficient and constant.

For $ax^2+bx+c$, find two numbers with product $ac$ and sum $b$. Split $bx$ using them, then factorise by grouping. This works because the split preserves the expression and makes a common bracket appear. First extract any common factor from the whole expression; “factorise” means factorise fully.

Change only the constant to $-3$. The splitting numbers now have negative product; keeping the old pair would change the question.

The negative sign belongs to the whole second group: $-2x-3=-(2x+3)$. Writing $-(2x-3)$ would change the constant.