The expression 2(x)^(3) − 8x conceals which values of x make it zero. Written as 2x(x − 2)(x + 2), it immediately shows three possibilities: 0, 2, − 2. The expression’s value has stayed the same; its new form makes a particular property easier to see.

This is factoring: replacing a polynomial with an identically equal product. It reverses polynomial multiplication and uses special product identities.

Start with what the terms share

Both terms of 2(x)^(3) − 8x contain the factor 2x. Read the distributive law backwards:

2(x)^(3) − 8x = 2x((x)^(2) − 4).

The parentheses contain what must multiply 2x to recover each original term. This is valid at x = 0 too: we are rewriting a polynomial by a multiplication law, not dividing an equation by an unknown x.

Now (x)^(2) − 4 is a difference of squares, giving 2x(x − 2)(x + 2). We often need to combine methods, rather than choose just one.

2x³ − 8x=2x(x² − 4)=2 · x · (x − 2) · (x + 2)x = 0x = 2x = −2
All three rows are the same polynomial. The lower arrows connect each variable factor’s zero to x; the constant 2 is never zero.

For 6(a)^(2)b + 9a(b)^(2), a useful common factor is 3ab: the coefficients share the divisor 3, and each term contains at least one a and one b. The result is 3ab(2a + 3b). Multiplying back recovers the original sum, checking the result independently of our guess.

Grouping makes a shared factor visible

In xy + 3x + 2y + 6, no variable factor belongs to all four terms. But two pairs create matching parentheses:

xy + 3x + 2y + 6 = x(y + 3) + 2(y + 3).

The shared factor is now the entire expression y + 3, so the result is (x + 2)(y + 3). Grouping helps when the first step produces a repeated bracket; arbitrary groups do not guarantee that.

Watch the sign too: a(u − v) + b(v − u) = a(u − v) − b(u − v) = (a − b)(u − v). The expressions u − v and v − u are opposites, not identical. Methods and checks in OpenStax §1.5.

Why a product reveals zeros

A zero of an expression is a variable value making the expression equal zero. A product of real numbers is zero if and only if at least one factor is zero. If every factor is nonzero, their product is nonzero too.

In 2x(x − 2)(x + 2), the constant 2 cannot become zero. That leaves x = 0, x − 2 = 0 or x + 2 = 0: respectively x = 0,2, − 2. Substitution into the original expression confirms all three. At x = 1, both forms give −6, but one check does not establish an identity for every x; the chain of valid transformations does.

A quadratic after its roots are known

For a(x)^(2) + bx + c with a ≠ 0, the quadratic formula and Vieta give a general method. If real roots x_(1),x_(2) exist (repeat a double root), then a(x)^(2) + bx + c = a(x − x_(1))(x − x_(2)). Keep the leading coefficient a.

For example, 2(x)^(2) − 14x + 20 has roots 2 and 5, giving 2(x − 2)(x − 5). Their sum 7 produces coefficient −14; their product 10, together with the factor 2, produces constant 20. Without real roots there is no factorization into real linear factors. This complements grouping when a suitable pair is hard to recognize.

Branch · A limit of the methodClarifying what ‘cannot be factored’ means.

Why real linear factors do not always exist

For every real x, (x)^(2) + 1 ≥ 1. Thus no real number makes this polynomial zero. A product of two real linear factors of positive degree would necessarily have such a zero, so that factorization is impossible. Of course 1 ⋅ ((x)^(2) + 1) is possible; here we mean a nontrivial product of lower-degree polynomials. The conclusion depends on the coefficients we allow.

Choose a form for a purpose. Sums are convenient for addition; products help with zeros and later with cancelling fractions. Look for a common factor, recognize an identity or useful grouping, repeat as needed, and check by expanding.