Factoring polynomials

The teacher wrote three tasks on the board: 1) work out 47 · 53 + 47 · 47; 2) work out 57² − 43²; 3) solve x² = 5x. Murad reached for a calculator, while Aysel did all three in her head in a minute: 47 · (53 + 47) = 4700; (57 − 43)(57 + 43) = 14 · 100 = 1400; x(x − 5) = 0, so x = 0 or x = 5. Murad divided the equation by x, found only x = 5 and lost a root. Aysel’s secret is the same every time: she turned a sum into a product.

In the previous lessons, including “Special products (short multiplication formulas)”, we turned products into polynomials by expanding brackets. Now we go the other way. A product is valuable because a lot can be seen in it at once: if one factor is zero, the product is zero; if one factor is divisible by 7, so is the product; and in a fraction only factors can be cancelled. In the next lesson, “Algebraic fractions (rational expressions)”, you will need this skill at every step.

Taking out the common factor

This is the simplest method, and it is always tried first. The distributive law a(b + c) = ab + ac is read from right to left: the factor that repeats in every term is written once in front of the brackets, and inside the brackets stay the quotients of each term divided by that factor.

1. 1
   Coefficients

   Find the GCD of the coefficients (for 12 and 18 it is 6). If the first term is negative, a negative factor such as −6 is usually taken out.

2. 2
   Letters

   Take every letter that appears in all the terms, with its smallest exponent: for a³b, a²b² and a⁴b this is a²b.

3. 3
   Divide

   Divide every term by the common factor and write the quotient in the brackets. If a term equals the common factor itself, 1 stays in its place.

4. 4
   Check

   Expand the bracket in your head: you must get the original polynomial. The terms inside the bracket must have no common factor left.

The common factor can also be a whole bracket. Then we treat the bracket like a single letter. If two brackets are opposites of each other, such as 4 − n and n − 4, first take out the minus sign: 4 − n = −(n − 4).

Factoring by grouping

Sometimes the terms have no common factor all together, but when they are grouped in pairs, the same bracket appears in every group. In ab + 3a + 2b + 6, for example, no factor is shared by all four terms, but the first two terms are divisible by a and the last two by 2.

Factoring with the special product formulas

In the lesson “Special products (short multiplication formulas)” we used these identities from left to right, to expand brackets. To factor, we read them from right to left. The main job is to recognise the formula: count the terms and see which of them are perfect squares (1, 4, 9, 25x², a⁴, …) or perfect cubes (8, 27, 64, x³, 8a³, …).

Factoring a quadratic trinomial

Why does the method work? Expand: (x + m)(x + n) = x² + nx + mx + mn = x² + (m + n)x + mn. So the coefficient of x is the sum of m and n, and the constant term is their product. To factor, it is enough to find these two numbers.

When the leading coefficient is not 1 (a ≠ 1), split the middle term into two terms and use grouping. To do this, look for two numbers whose product is a · c and whose sum is b. Another way is completing the square: bring the expression to a difference of squares.

1. 1
   Compute a · c

   For 2x² + 7x + 3: a · c = 2 · 3 = 6.

2. 2
   Choose two numbers

   Numbers with product 6 and sum b = 7: 1 and 6.

3. 3
   Split the middle term

   Write 7x = x + 6x: 2x² + x + 6x + 3.

4. 4
   Group

   x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). The brackets must match.

Not every quadratic trinomial factors with whole numbers: for x² + x + 1 there are no integers with product 1 and sum 1. In the lesson “Quadratic equations” you will learn, using the discriminant, which trinomials can be factored at all, and the formula ax² + bx + c = a(x − x₁)(x − x₂).

Let us put all the methods into one plan. In harder tasks, two or three methods are usually used one after another.

1. 1
   Common factor

   Always start by taking out the common factor (with a minus sign if needed).

2. 2
   Count the terms

   Two terms: difference of squares, or sum (difference) of cubes. Three terms: perfect square or quadratic trinomial. Four or more: grouping.

3. 3
   Go all the way

   Look at every factor again: can it be factored further? For example, a⁴ − 1 = (a² − 1)(a² + 1) = (a − 1)(a + 1)(a² + 1).

4. 4
   Check

   Expand the brackets, or put the same number (say x = 1 or x = 2) into both expressions — the results must be equal.

What factoring is good for

The most important property of a product: a product of two numbers is zero only when at least one of them is zero. So we move all terms of an equation to one side and factor that side — a hard equation breaks into several easy ones.

Factoring also makes calculations easier. The difference of squares turns the difference of two big squares into one multiplication, and a common factor produces “round” numbers. To prove divisibility, we also bring the expression to the form of a product: if one factor is divisible by n, the whole product is divisible by n.