Use the Remainder Theorem to find remainders and decide whether x − a is a factor

Learning goal: Understand and apply the Remainder Theorem: for a polynomial p(x) and a number a, the remainder on division by x − a is p(a); and in particular p(a) = 0 if and only if (x − a) is a factor of p(x). Use both directions of the theorem to find remainders without dividing and to factor polynomials.

Division with remainder, for integers and for polynomials. When 23 is divided by 5, the quotient is 4 and the remainder is 3, and the check is 23 = 5 × 4 + 3, with the remainder smaller than the divisor. Polynomials divide the same way. Dividing p(x) = x³ − 4x² + x + 6 by x − 1 by long division gives the quotient x² − 3x − 2 and a remainder of 4, and the check is p(x) = (x − 1)(x² − 3x − 2) + 4. Multiply it out: (x − 1)(x² − 3x − 2) = x³ − 3x² − 2x − x² + 3x + 2 = x³ − 4x² + x + 2, and adding 4 gives x³ − 4x² + x + 6, which is p(x). The rule for "smaller" is that the remainder has lower degree than the divisor. The divisor x − 1 has degree 1, so the remainder has degree 0: it is a constant.

The Remainder Theorem and why it is true. Because the divisor x − a has degree 1, the division of any polynomial p(x) by x − a can always be written as p(x) = (x − a)·q(x) + r, where q(x) is the quotient polynomial and r is a constant. This equation is an identity: it holds for every value of x. So substitute x = a. The factor (x − a) becomes (a − a) = 0, the whole product (x − a)·q(x) becomes 0, and the equation collapses to p(a) = r. That is the Remainder Theorem: the remainder on division by x − a is p(a). The proof is three lines long, and the only fact it uses is that a constant remainder does not depend on x. In the example above, p(1) = 1 − 4 + 1 + 6 = 4, which matches the remainder 4 from long division without doing any division at all.

The factor half of the theorem. A polynomial x − a is a factor of p(x) exactly when the division leaves no remainder, that is, when r = 0. Since r = p(a), we get: p(a) = 0 if and only if (x − a) is a factor of p(x). Both directions are needed. If p(a) = 0, then p(x) = (x − a)·q(x) + 0, so x − a is a factor. Conversely, if x − a is a factor, then p(x) = (x − a)·q(x), and substituting x = a gives p(a) = 0·q(a) = 0. A number a with p(a) = 0 is called a zero of p, so the theorem says the zeros of p are exactly the numbers a for which x − a is a factor.

A fully worked model. Take p(x) = x³ − 4x² + x + 6 and test a = 2: p(2) = 8 − 16 + 2 + 6 = 0. By the theorem, x − 2 is a factor. Divide to find the other factor. Synthetic division with 2 uses the coefficients 1, −4, 1, 6: bring down 1; 1 × 2 = 2, and −4 + 2 = −2; −2 × 2 = −4, and 1 + (−4) = −3; −3 × 2 = −6, and 6 + (−6) = 0. The bottom row 1, −2, −3, 0 means the quotient is x² − 2x − 3 and the remainder is 0, as predicted. Now x² − 2x − 3 = (x − 3)(x + 1), so p(x) = (x − 2)(x − 3)(x + 1). Confirm with the theorem: p(3) = 27 − 36 + 3 + 6 = 0 and p(−1) = −1 − 4 − 1 + 6 = 0, so x − 3 and x + 1, which is x − (−1), are both factors. The table in the diagram records these four evaluations.

Using the theorem to find a remainder without dividing. What is the remainder when 2x⁴ − 3x² + 5x − 1 is divided by x + 1? Write the divisor as x − (−1), so a = −1. Then p(−1) = 2·1 − 3·1 + 5·(−1) − 1 = 2 − 3 − 5 − 1 = −7. The remainder is −7. This is much faster than long division on a degree-4 polynomial, and it is the intended use of the theorem: evaluation and division-by-a-linear are the same computation.

Using the theorem to find an unknown coefficient. For which k is x − 3 a factor of x³ + kx² − 5x + 6? The factor condition is p(3) = 0: 27 + 9k − 15 + 6 = 0, so 18 + 9k = 0 and k = −2. Check: with k = −2, p(3) = 27 − 18 − 15 + 6 = 0.

The misconception to remove. Students often reverse the sign: to test the factor x + 2, they compute p(2). The divisor x + 2 equals x − (−2), so the number to substitute is −2. A safe habit is to write the divisor in the form x − a first and read off a. The other common error is to treat a nonzero p(a) as meaningless; it is not meaningless, it is the remainder, and it tells you exactly how far x − a is from being a factor.

What the theorem does not say. It applies only to division by a linear divisor x − a with leading coefficient 1. For a divisor such as 2x − 1, write it as 2(x − ½): the remainder on division by x − ½ is p(½), and division by 2x − 1 gives the same remainder with the quotient halved. For a divisor of degree 2 or more, the remainder need not be a constant (its degree is only guaranteed to be less than the divisor's), and the theorem does not apply directly.

Worked remainder by evaluation: Find the remainder when 2x³ − x + 5 is divided by x + 2 and by x − 1. For x + 2, a = −2: p(−2) = 2(−8) − (−2) + 5 = −16 + 2 + 5 = −9, so the remainder is −9 and x + 2 is not a factor. For x − 1, a = 1: p(1) = 2 − 1 + 5 = 6, remainder 6. Neither divisor is a factor, and the theorem gave both remainders with no division.

Worked factor test and complete factoring: Is x − 2 a factor of x⁴ − 16? Since p(2) = 16 − 16 = 0, yes. Synthetic division by 2 on the coefficients 1, 0, 0, 0, −16: bring down 1; 2, so 0 + 2 = 2; 4, so 0 + 4 = 4; 8, so 0 + 8 = 8; 16, so −16 + 16 = 0. The quotient is x³ + 2x² + 4x + 8. Test a = −2 on the original: p(−2) = 16 − 16 = 0, so x + 2 is also a factor; indeed x³ + 2x² + 4x + 8 = x²(x + 2) + 4(x + 2) = (x + 2)(x² + 4). So x⁴ − 16 = (x − 2)(x + 2)(x² + 4). The factor x² + 4 has no real zero, since x² + 4 ≥ 4 for every real x, so no further real linear factor x − a exists; over the complex numbers x² + 4 = (x − 2i)(x + 2i), so the complete factorization there is (x − 2)(x + 2)(x − 2i)(x + 2i).

Worked identification of all linear factors: Let p(x) = x³ + 2x² − 5x − 6. Try small integers. p(1) = 1 + 2 − 5 − 6 = −8, not a zero. p(−1) = −1 + 2 + 5 − 6 = 0, so x + 1 is a factor. p(2) = 8 + 8 − 10 − 6 = 0, so x − 2 is a factor. Dividing by (x + 1)(x − 2) = x² − x − 2 gives x + 3, since (x² − x − 2)(x + 3) = x³ + 3x² − x² − 3x − 2x − 6 = x³ + 2x² − 5x − 6. Therefore p(x) = (x + 1)(x − 2)(x + 3), and p(−3) = −27 + 18 + 15 − 6 = 0 confirms the third factor.

Worked proof sketch: Explain why the remainder on dividing p(x) by x − a equals p(a). Division gives p(x) = (x − a)q(x) + r with r constant, because the remainder must have degree less than 1. This holds for every x, so put x = a: p(a) = (a − a)q(a) + r = 0 + r = r. The remainder is p(a). If p(a) = 0 the remainder is 0, so x − a is a factor; if x − a is a factor the remainder is 0, so p(a) = 0.

Representation task: For p(x) = x³ − 4x² + x + 6, make a four-column table with the headings a, p(a), remainder on division by x − a, and "factor?". Fill it for a = −2, −1, 0, 1, 2, 3, computing p(a) by substitution and the remainder by synthetic division for at least two of the rows so that you see the two columns agree. Then write the complete factorization of p(x) and, underneath, one sentence for each direction of the "if and only if" in the Remainder Theorem, in your own words.

Independent performance: Build a cubic polynomial with three integer zeros of your choosing by multiplying out (x − a)(x − b)(x − c); keep the factored form hidden. Exchange the expanded cubic with a partner. For the cubic you receive: find the remainder on division by x + 1 and by x − 5 using the Remainder Theorem, state whether either is a factor, then find all three linear factors by testing values and dividing. Success criteria: every remainder is computed by substitution and at least one is confirmed by synthetic division; the sign of a is correct for divisors of the form x + c; the final factorization multiplies back to the given cubic; and the write-up states explicitly why p(a) = 0 guarantees the factor.