MISSION · Algebra
Factoring Quadratics
Why two brackets beat one expanded expression
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The idea in one sentence
Factoring rewrites \(x^2 + bx + c\) as two brackets multiplied together, which is the form that hands you the solutions for free.
Why bother
Because of one fact about zero: if two things multiply to zero, at least one of them is zero. Nothing else in arithmetic behaves like this — a product of 12 could come from countless pairs, but a product of 0 pins one factor down completely.
So if \(x^2 - 5x + 6 = 0\) and you can write it as \((x - 2)(x - 3) = 0\), then either \(x - 2 = 0\) or \(x - 3 = 0\). The solutions are 2 and 3, read straight off, with no formula and no rearranging.
That’s what factoring is for. It converts a hard question — which \(x\) makes this expression vanish — into an easy one.
Finding the brackets
For \(x^2 + bx + c\), you need two numbers that multiply to \(c\) and add to \(b\). That’s the whole search.
\(x^2 + 7x + 12\): which pair multiplies to 12 and adds to 7? The factor pairs of 12 are 1 and 12, 2 and 6, 3 and 4. The last adds to 7, so it factors as \((x + 3)(x + 4)\).
The reason is just the expansion running backwards. \((x + p)(x + q)\) multiplies out to \(x^2 + (p + q)x + pq\) — the sum appears in the middle, the product at the end.
Signs follow from the same structure. A negative \(c\) means the two numbers have opposite signs. A positive \(c\) with a negative \(b\) means both are negative.
Two patterns worth recognising on sight
Difference of squares. \(x^2 - 9 = (x + 3)(x - 3)\). Any square minus any square splits like this, and the middle terms cancel when you check it. There is no equivalent for a sum of squares.
Common factor first. \(2x^2 + 10x + 12\) has a 2 in every term. Take it out — \(2(x^2 + 5x + 6)\) — and factor what’s left. Skipping this step makes every quadratic harder than it needs to be, and it’s the first thing to check, always.
When it doesn’t factor
Plenty of quadratics have no whole-number factorisation. \(x^2 + x + 1\) doesn’t, and neither does \(x^2 - 2\) over the integers, even though it has perfectly good roots at \(\pm\sqrt{2}\).
Factoring is a shortcut that works when the numbers are friendly. When they aren’t, the quadratic formula always works, and spending ten minutes hunting for factors that don’t exist is the most common way to waste time in an exam.
Common mix-ups
- Mixing up which pair multiplies and which adds. Multiply to \(c\), add to \(b\).
- Getting the signs backwards. \((x - 2)(x - 3)\) gives roots of \(+2\) and \(+3\), not \(-2\) and \(-3\).
- Forgetting the common factor before factoring the rest.
- Setting the product to something other than zero. The zero-product rule needs a zero: to solve \(x^2 - 5x + 6 = 2\), move the 2 across first.
Check yourself
Factor \(x^2 - 2x - 15\) and give its roots.
Two numbers multiplying to \(-15\) and adding to \(-2\) are \(-5\) and 3, so it is \((x - 5)(x + 3)\), with roots at \(x = 5\) and \(x = -3\).
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