Factoring Quadratics

Factoring reverses expanding

To factor a trinomial is to write it as a product of factors, the exact reverse of expanding a product into a trinomial. Since the two operations undo each other, the way to discover the factors is to look closely at what expanding produces. In the chapter on expanding you multiplied two simple factors of the form $x + p$ and $x + q$ and got

Read that equation from right to left and it stops being a multiplication rule and becomes a factoring recipe.

The picture behind the recipe is the same rectangle you drew for expanding, read the other way. Give a rectangle a width of $x + p$ and a height of $x + q$. The two splits carve it into four pieces whose areas are $x^2$, $px$, $qx$, and the small corner $pq$.

The two strips $px$ and $qx$ together make the middle term, since they combine to $(p + q)x$, and the corner $pq$ is the constant. So matching the middle term of a trinomial pins down $p + q$, and matching the constant pins down $pq$. Everything below is just carrying out that search.

Finding the two numbers

The recipe turns factoring into a small, orderly search. Run through the pairs of whole numbers that multiply to $c$, and pick the pair that also adds to $b$. The signs do much of the work of narrowing that search, so read them before you start listing.

Because $c = pq$ is the product of the two numbers, its sign tells you whether they share a sign. When $c$ is positive, $p$ and $q$ must have the same sign, because only two positives or two negatives multiply to a positive. In that case the pair’s sum $b$ tells you which sign they share: both positive when $b$ is positive, and both negative when $b$ is negative. When $c$ is negative, the two numbers must have opposite signs, because a positive times a negative is negative. When the signs are opposite, the number with the larger size carries the sign of $b$. Reading the signs this way cuts the list of candidate pairs roughly in half before you add anything.

The method does not care if $b$ happens to be $0$. To factor $x^2 - 9$, read it as $x^2 + 0x - 9$ and look for two numbers with sum $0$ and product $-9$. Opposite numbers of the same size add to $0$, and $3$ and $-3$ multiply to $-9$, so $x^2 - 9 = (x + 3)(x - 3)$. A missing middle term simply means the two numbers are opposites.

Solving a quadratic by factoring

Factoring is worth the effort because it feeds straight into the zero-product property from the last lesson. To solve $x^2 + bx + c = 0$, factor the left side into $(x + p)(x + q)$ and set the product to zero:

A product is zero only when one of its factors is zero, so $x + p = 0$ or $x + q = 0$, which gives $x = -p$ or $x = -q$. Watch the sign flip one more time: the number that sits inside a factor as $+p$ becomes the root $-p$. A factor and the root it produces always have opposite signs, and that is where a careless solve most often goes wrong.

Pulling out a common factor first

Two setups look harder than they are, and both begin with a factor you pull out before the sum-and-product search even starts.

The first is a trinomial whose three coefficients share a common numerical factor, such as $2x^2 + 6x + 4$. Its leading coefficient is not $1$, so the recipe does not apply directly. But every term is even, so factor the $2$ out front first:

Inside the parentheses the leading coefficient is back to $1$. The sum-and-product method now finishes the job: two numbers with sum $3$ and product $2$ are $1$ and $2$, so

Always pull out a common numerical factor first. It shrinks the numbers you have to work with and turns a stubborn-looking trinomial into a monic one you already know how to handle.

The second setup is a trinomial with no constant term, like $x^2 - 5x$. Here $c = 0$, and instead of hunting for a pair you simply pull out the shared $x$:

Every term contains an $x$, so $x$ is the common factor, and a single linear factor is left. Solving $x^2 - 5x = 0$ then gives $x = 0$ or $x = 5$. The root $x = 0$ is the one most often lost, because dividing both sides by $x$ quietly throws it away. Keep the factor rather than dividing, and both roots survive.

When a trinomial will not factor

Not every trinomial can be factored this way. The method depends on finding two whole numbers with the right sum and product, and sometimes no such pair exists. Take $x^2 + x + 1$. You would need two numbers that multiply to $1$ and add to $1$. The only integer pairs that multiply to $1$ are $1$ and $1$, which add to $2$, or $-1$ and $-1$, which add to $-2$. Neither adds to $1$, so the search fails, and $x^2 + x + 1$ is irreducible over the integers: it cannot be written as a product of factors with whole-number coefficients.

That does not mean the equation $x^2 + x + 1 = 0$ has no solutions. It means only that factoring over the integers is the wrong tool for it. Later lessons build methods, completing the square and the quadratic formula, that solve any quadratic whether it factors or not. For now, an honest run through the factor pairs of $c$ may turn up no pair that adds to $b$. When that happens, you can stop and report that the trinomial does not factor over the integers, rather than force an answer that is not there.

Common mistakes