6.03b - Solving factorised quadratics

Why a product can be zero

A rectangle has sides $x\,\mathrm{cm}$ and $(x+3)\,\mathrm{cm}$, and area $28\,\mathrm{cm^2}$. Could the equation for its sides give two values of $x$, even though only one describes the rectangle?

First consider zero products. $0\times7=0$, $5\times0=0$ and $0\times0=0$. But multiplying two nonzero real numbers cannot give zero: if $ab=0$ and $a\ne0$, division by $a$ gives $b=0$.

A root or solution of an equation is a value that makes it true. A factorised quadratic equal to zero lets you find roots by setting each factor equal to zero separately.

Change only the sign inside the second factor. Its root changes sign too.

The roots are alternatives. One value of $x$ does not have to make both factors zero. In contrast, $(x-4)^2=0$ has just one distinct root, $x=4$, appearing in both factors: a repeated root.

Solve each factor separately in the next product.

Factorising to find both roots

An expanded monic quadratic has the form $x^2+bx+c$. To solve its equation by factorisation, first make one side zero, then find two bracket constants with sum $b$ and product $c$.

Factorising alone does not finish a “Solve” question: the required answers are values of $x$, not just brackets. Conversely, the constants inside the brackets are not the roots until you solve each factor equal to zero.

Change only the position of the constant term: the next equation has the same relationship, but it must be rearranged before using the product rule.

If the constant is zero, factor out $x$: $x^2+5x=0$ becomes $x(x+5)=0$, giving $x=0$ or $x=-5$. Dividing by $x$ would lose zero. A repeated factor, such as $(x+3)^2=0$, gives only $x=-3$ as a distinct answer. Not every quadratic has integer factors; if no suitable pair exists, a different quadratic method may be needed.