Proving and Applying the Factor Theorem

From a Polynomial Value to a Factor

A polynomial can be represented in several ways. In expanded form, it is straightforward to evaluate at a particular number. In factored form, its linear factors are visible. The Factor Theorem connects these two descriptions: evaluating a polynomial at \(a\) tells us whether \(x-a\) is one of its factors.

This connection is useful when a particular number is suspected to be a zero, or when a proposed linear expression is suspected to divide a polynomial exactly. The test is simple, but its sign convention matters: the factor \(x-a\) is tested by evaluating at \(a\), while \(x+a\) is tested by evaluating at \(-a\).

Zeros and Factors

Throughout, let \(p(x)\) be a polynomial with real coefficients and let \(a\) be a real number. A polynomial \(p\) has a zero at \(a\) if \(p(a)=0\). Equivalently, \(a\) is a zero of \(p\) when the equation \(p(a)=0\) holds. This terminology describes a particular input and its output; it does not by itself give a factorization.

The expression \(x-a\) is a factor of \(p(x)\) if there is a polynomial \(q(x)\) such that \(p(x)=(x-a)q(x)\). In that case, \(p\) is divisible by \(x-a\) with no remainder. The polynomial \(q(x)\) is the other factor in this product.

The Factor Theorem

The Remainder Theorem says that when a polynomial \(p(x)\) is divided by \(x-a\), the remainder is \(p(a)\). Polynomial division applies because \(x-a\) is a nonzero polynomial. Since the divisor has degree one, the remainder is a constant (possibly zero). The following argument shows why this remainder result gives an exact test for a factor.

Proof. By polynomial division, there are polynomials \(q(x)\) and a constant \(r\) such that

Substitute \(x=a\). The term containing \(x-a\) becomes zero, so

Thus the remainder when \(p(x)\) is divided by \(x-a\) is exactly \(p(a)\). If \(p(a)=0\), then \(r=0\), and the division equation becomes \(p(x)=(x-a)q(x)\). By the definition of factor, \(x-a\) is a factor of \(p(x)\). Conversely, if \(x-a\) is a factor, then \(p(x)=(x-a)q(x)\) for some polynomial \(q(x)\). Substituting \(x=a\) gives \(p(a)=(a-a)q(a)=0\). Both directions hold, which proves the theorem.

The proof establishes an “if and only if,” so it gives two useful directions. A zero at \(a\) guarantees the factor \(x-a\); and a factor \(x-a\) guarantees a zero at \(a\). It does not say that every number is a zero, or that every factor of a polynomial is linear.

Applying the Theorem to a Proposed Factor

To test a proposed factor, first express it as \(x-a\). Then evaluate the polynomial at \(a\). A result of zero proves that the expression is a factor; a nonzero result proves that it is not. When the factor is written as \(x+a\), rewrite it as \(x-(-a)\) before choosing the input.

Using Zeros to Recognize Factors

The theorem can also be applied when a number is supplied as a possible zero rather than when a factor is supplied. If evaluation confirms that \(a\) is a zero, the corresponding factor is \(x-a\). If several values are confirmed as zeros, each gives a corresponding linear factor. Any resulting full factorization should still be checked, since the Factor Theorem alone does not specify the remaining factor.

A Common Sign Error

A frequent mistake is to test \(x+a\) by evaluating at \(a\). The theorem does not match a factor with a value by copying the sign: it matches \(x-a\) with \(a\). The safe step is to rewrite the proposed factor in that form before substituting.

For example, consider \(p(x)=x^2-5x+6\) and the proposed factor \(x+2\). A mistaken test might use \(p(2)\):

This calculation is correct, but it tests the factor \(x-2\), not \(x+2\). For \(x+2=x-(-2)\), the correct test is \(p(-2)\):

Since \(p(-2)\ne 0\), \(x+2\) is not a factor. The zero at \(2\) instead confirms that \(x-2\) is a factor. Indeed, \(p(x)=(x-2)(x-3)\), as multiplication gives \(x^2-3x-2x+6=x^2-5x+6\).

Why the Test Matters

The Factor Theorem turns a question about divisibility into a substitution. Instead of performing polynomial division every time a particular linear factor is proposed, one evaluation decides whether the remainder is zero. It also gives a direct way to translate between two pieces of information: a known zero \(a\) gives the factor \(x-a\), and a known factor \(x-a\) gives the zero \(a\).

The conclusion is exact for polynomials and linear factors of the stated form. If \(p(a)\ne 0\), the theorem rules out \(x-a\) as a factor. If \(p(a)=0\), it confirms that factor, but it does not, by itself, show what all the other factors are. To verify a proposed complete factorization, multiply its factors and compare the result with the original polynomial.