JentooSAT®

SAT® · Math: Advanced Math

Factoring Beyond the Basic Cases

Some polynomials will not factor until you reorganize them first: group four terms into pairs, treat x2 as a single variable, or spot a cubes pattern. Each trick turns a strange-looking expression into one you already know how to factor.

Basic factoring pulls out a common factor or splits a trinomial into two binomials. But plenty of polynomials do not hand over their factors in that form. A four-term polynomial like x3+3x2-4x-12 has no single term shared by all four pieces, but it does split into two pairs that each hide the same binomial. Group (x3+3x2) and (-4x-12): the first pair gives x2(x+3), the second gives -4(x+3). Now (x+3) is common to both, so the whole thing becomes (x+3)(x2-4), which factors again into (x+3)(x-2)(x+2).

A quartic like x4-13x2+36 looks intimidating until you notice it only has even powers of x. Let u = x2, and the expression becomes u2-13u+36, an ordinary trinomial that factors as (u-4)(u-9). Swap x2 back in: (x2-4)(x2-9). Both factors are differences of squares, so the complete factorization keeps going to (x-2)(x+2)(x-3)(x+3). The lesson repeats everywhere in this unit: never stop factoring just because you found two pieces, check whether either piece can split again.

Cubes follow their own fixed patterns. A difference of cubes a3-b3 factors as (a-b)(a2+ab+b2), and a sum a3+b3 factors as (a+b)(a2-ab+b2). For 8x3-27, treat it as (2x)3-33, so a=2x and b=3: the factorization is (2x-3)(4x2+6x+9). The quadratic piece never factors further over real numbers, so the only real solution comes from 2x-3=0, giving x=3/2. These patterns connect straight into solving equations and simplifying rational expressions, where spotting the hidden structure is most of the work.

Key terms

Factoring by grouping
Splitting a four-term polynomial into two pairs, factoring each pair, then pulling out their shared binomial factor.
Difference of cubes
A pattern a3-b3 that always factors as (a-b)(a2+ab+b2).
Sum of cubes
A pattern a3+b3 that always factors as (a+b)(a2-ab+b2).
Difference of squares
A pattern a2-b2 that factors as (a-b)(a+b), and can be reapplied if a new difference of squares appears.
Substitution variable
A stand-in like u=x2 used to rewrite a quartic as a simpler trinomial before factoring.

Watch it explained

  1. How it worksHow does factoring by grouping crack a four-term polynomial like x3+3x2-4x-12?
  2. Worked exampleFind the complete factorization of x4-13x2+36 by treating x2 as one single variable.
  3. Worked exampleShow me how to factor 8x3-27 as a difference of cubes and read off its only real solution.
  4. Going furtherWhat changes when a polynomial hides a perfect-square trinomial inside it, like x4+6x2+9?

Check yourself

Your turnmedium1 of 3
Find the complete factorization of 7x3 + 7x2 + 7x + 7.

End of topic.