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Polynomial Rules






The Theorems Behind the Toolkit

Working with polynomials involves a great deal of computation — dividing, testing values, searching for roots. But underneath that work sits a collection of theorems that explain why certain shortcuts are valid, why certain searches are guaranteed to terminate, and why the structure of a polynomial's coefficients reveals information about its solutions before a single calculation is performed.

These results connect seemingly separate tasks. Evaluating a polynomial at a point turns out to be the same as finding a division remainder. Knowing one root immediately produces a factor. The coefficients of a polynomial place strict limits on which rational numbers could possibly be roots. Each theorem removes guesswork and replaces it with structure.



The Remainder Theorem

When a polynomial P(x)P(x) is divided by a linear binomial (xc)(x - c), the remainder is a constant — and that constant equals P(c)P(c), the value of the polynomial evaluated at x=cx = c.

This means the remainder can be found without performing any division at all. To find the remainder when P(x)=x34x+2P(x) = x^3 - 4x + 2 is divided by (x3)(x - 3), simply compute P(3)=2712+2=17P(3) = 27 - 12 + 2 = 17. The remainder is 1717.

The theorem applies to any polynomial and any value of cc. If P(x)=2x4x2+5P(x) = 2x^4 - x^2 + 5 and the divisor is (x+1)(x + 1), read (x+1)(x + 1) as (x(1))(x - (-1)) and evaluate P(1)=2(1)1+5=6P(-1) = 2(1) - 1 + 5 = 6. The remainder is 66, confirmed instantly without setting up long division or synthetic division.

Why the Remainder Theorem Works

The proof follows directly from the division algorithm. Dividing P(x)P(x) by (xc)(x - c) produces a quotient Q(x)Q(x) and a remainder RR:

P(x)=(xc)Q(x)+RP(x) = (x - c) \cdot Q(x) + R


Because the divisor (xc)(x - c) has degree 11, the remainder RR must have degree less than 11 — meaning RR is a constant, independent of xx.

Now substitute x=cx = c into both sides:

P(c)=(cc)Q(c)+R=0Q(c)+R=RP(c) = (c - c) \cdot Q(c) + R = 0 \cdot Q(c) + R = R


The factor (cc)(c - c) annihilates the quotient term, leaving P(c)=RP(c) = R. Evaluation and division are not separate operations — evaluating PP at cc is equivalent to extracting the remainder of division by (xc)(x - c).

The Factor Theorem

The Factor Theorem is a direct consequence of the Remainder Theorem: (xc)(x - c) is a factor of P(x)P(x) if and only if P(c)=0P(c) = 0.

The logic is immediate. If the remainder when dividing by (xc)(x - c) is zero, then P(x)=(xc)Q(x)P(x) = (x - c) \cdot Q(x) with no leftover term — so (xc)(x - c) divides P(x)P(x) exactly. Conversely, if (xc)(x - c) is a factor, then P(c)=(cc)Q(c)=0P(c) = (c - c) \cdot Q(c) = 0.

This creates a bridge between roots and factors. The value cc is a root of P(x)P(x) precisely when (xc)(x - c) is a factor of P(x)P(x). Finding one gives the other automatically.

To test whether (x2)(x - 2) is a factor of x36x2+11x6x^3 - 6x^2 + 11x - 6, evaluate P(2)=824+226=0P(2) = 8 - 24 + 22 - 6 = 0. Since the result is zero, (x2)(x - 2) is indeed a factor, and 22 is a root.

Using the Factor Theorem

The Factor Theorem works in both directions — it can verify suspected factors and it can build polynomials from known roots.

To test potential roots, substitute each candidate into P(x)P(x) and check for zero. For P(x)=x37x+6P(x) = x^3 - 7x + 6, testing x=1x = 1 gives P(1)=17+6=0P(1) = 1 - 7 + 6 = 0, confirming that (x1)(x - 1) is a factor. Dividing out this factor via synthetic division yields the quotient x2+x6x^2 + x - 6, which factors further as (x+3)(x2)(x + 3)(x - 2). The complete factorization is (x1)(x+3)(x2)(x - 1)(x + 3)(x - 2).

To construct a polynomial from its roots, reverse the process. If a cubic polynomial has roots 11, 2-2, and 44, the Factor Theorem guarantees that (x1)(x - 1), (x+2)(x + 2), and (x4)(x - 4) are all factors. Their product gives:

(x1)(x+2)(x4)=x33x26x+8(x - 1)(x + 2)(x - 4) = x^3 - 3x^2 - 6x + 8


Any nonzero scalar multiple a(x1)(x+2)(x4)a(x - 1)(x + 2)(x - 4) shares the same roots, so the polynomial is determined up to a constant factor unless additional information — such as a specific function value — pins it down.

The Rational Root Theorem

For polynomials with integer coefficients, the Rational Root Theorem restricts which rational numbers can possibly be roots. If P(x)=anxn+an1xn1++a1x+a0P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 has integer coefficients and pq\frac{p}{q} is a rational root expressed in lowest terms, then pp divides the constant term a0a_0 and qq divides the leading coefficient ana_n.

This transforms an infinite search into a finite one. Instead of testing every rational number, only the fractions formed by divisors of a0a_0 over divisors of ana_n need to be checked.

For P(x)=2x3+x213x+6P(x) = 2x^3 + x^2 - 13x + 6, the constant term is 66 with divisors ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6, and the leading coefficient is 22 with divisors ±1,±2\pm 1, \pm 2. The candidate list is:

±1,±2,±3,±6,±12,±32\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}


Twelve candidates replace infinitely many possibilities. Each can be tested by substitution or synthetic division until all rational roots are found or the list is exhausted.

Applying the Rational Root Theorem

The theorem produces a candidate list, but the actual work lies in testing each candidate efficiently. A systematic approach avoids redundant effort.

Start by listing all divisors of the constant term a0a_0 — these are the possible values of pp. Then list all divisors of the leading coefficient ana_n — these are the possible values of qq. Form every fraction ±pq\pm \frac{p}{q} and reduce to lowest terms to eliminate duplicates.

For 2x3+x213x+62x^3 + x^2 - 13x + 6, begin testing with the simplest candidates. P(1)=2+113+6=4P(1) = 2 + 1 - 13 + 6 = -4, so 11 is not a root. P(2)=16+426+6=0P(2) = 16 + 4 - 26 + 6 = 0, so 22 is a root. By the Factor Theorem, (x2)(x - 2) is a factor. Synthetic division gives a quotient of 2x2+5x32x^2 + 5x - 3, which factors as (2x1)(x+3)(2x - 1)(x + 3).

The complete factorization is (x2)(2x1)(x+3)(x - 2)(2x - 1)(x + 3), giving roots 22, 12\frac{1}{2}, and 3-3 — all members of the candidate list.

Testing via synthetic division is often more efficient than direct substitution, since each test simultaneously produces the quotient polynomial if the candidate turns out to be a root.

Limitations of the Rational Root Theorem

The Rational Root Theorem identifies only rational roots. Irrational roots and complex roots lie entirely outside its reach.

The polynomial x22x^2 - 2 illustrates this clearly. Its constant term is 2-2 and its leading coefficient is 11, producing candidates ±1,±2\pm 1, \pm 2. Testing each: P(1)=1P(1) = -1, P(1)=1P(-1) = -1, P(2)=2P(2) = 2, P(2)=2P(-2) = 2. None equals zero. The actual roots are ±2\pm\sqrt{2}, both irrational — invisible to the theorem.

A polynomial may have no rational roots at all. The polynomial x4+x3+x2+x+1x^4 + x^3 + x^2 + x + 1 has integer coefficients and candidates ±1\pm 1, but neither works. All four of its roots are complex, and no amount of rational testing will find them.

The theorem is best understood as a first filter, not a complete solution. When it finds roots, synthetic division reduces the polynomial's degree, potentially making the remaining roots accessible through the quadratic formula or other methods. When it finds none, it confirms that the roots are irrational or complex and signals the need for different tools.

The Integer Root Theorem

When the leading coefficient of a polynomial is 11 — a monic polynomial — the Rational Root Theorem simplifies considerably. Every rational root pq\frac{p}{q} has qq dividing 11, which forces q=±1q = \pm 1, which means pq\frac{p}{q} is an integer. So all rational roots of a monic polynomial are integers, and they must divide the constant term.

For x36x2+11x6x^3 - 6x^2 + 11x - 6, the only candidates are the divisors of 6-6: ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6. No fractions need to be considered.

Testing: P(1)=16+116=0P(1) = 1 - 6 + 11 - 6 = 0, so 11 is a root. Dividing out (x1)(x - 1) leaves x25x+6=(x2)(x3)x^2 - 5x + 6 = (x - 2)(x - 3). All three roots — 11, 22, 33 — are integers dividing 66, exactly as the theorem predicts.

This simplification makes monic polynomials particularly approachable. The candidate list is shorter, the candidates are easier to test, and the search is faster. Recognizing that a polynomial is monic before beginning the root search saves unnecessary work with fractions.

Descartes' Rule of Signs

Descartes' Rule of Signs estimates the number of positive and negative real roots of a polynomial by counting sign changes in its coefficients.

To count positive real roots, write P(x)P(x) in standard form and count how many times the sign of the coefficients changes from one term to the next. For P(x)=x32x2x+2P(x) = x^3 - 2x^2 - x + 2, the coefficient signs are +,,,++, -, -, +. The sign changes twice: from ++ to - and from - to ++. The polynomial has either 22 or 00 positive real roots — the actual count equals the number of sign changes or less by an even number.

To count negative real roots, apply the same procedure to P(x)P(-x). Substituting x-x gives P(x)=x32x2+x+2P(-x) = -x^3 - 2x^2 + x + 2, with signs ,,+,+-, -, +, +. One sign change occurs, so the polynomial has exactly 11 negative real root.

The rule does not identify the roots or specify which of the possible counts is correct. For P(x)=x32x2x+2P(x) = x^3 - 2x^2 - x + 2, the actual roots are 11, 22, and 1-1 — two positive and one negative, consistent with the prediction. But the rule alone cannot distinguish between 22 positive roots and 00 positive roots without further analysis.

Upper and Lower Bounds

Synthetic division provides a way to establish bounds on the real roots of a polynomial, narrowing the interval where roots can exist.

The upper bound test works as follows: divide P(x)P(x) by (xc)(x - c) where c>0c > 0 using synthetic division. If every entry in the bottom row is non-negative (positive or zero), then cc is an upper bound — no real root of P(x)P(x) exceeds cc.

The lower bound test is similar but uses a negative value of cc. Divide P(x)P(x) by (xc)(x - c) where c<0c < 0. If the entries in the bottom row alternate in sign (with zeros counting as either sign), then cc is a lower bound — no real root is less than cc.

For P(x)=x34x2+x+6P(x) = x^3 - 4x^2 + x + 6, synthetic division by (x5)(x - 5) produces the row 1,1,6,361, 1, 6, 36 — all non-negative, so 55 is an upper bound. Synthetic division by (x+2)(x + 2), meaning c=2c = -2, produces 1,6,13,201, -6, 13, -20 — the signs alternate, so 2-2 is a lower bound. All real roots of this polynomial lie in the interval [2,5][-2, 5].

These bounds are especially useful when combined with the Rational Root Theorem, eliminating candidates that fall outside the established range before any testing begins.

The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra makes a sweeping guarantee: every polynomial of degree n1n \geq 1 with complex coefficients has exactly nn roots in the complex numbers, counted with multiplicity.

A quadratic has 22 roots. A cubic has 33. A degree-1010 polynomial has 1010. No polynomial of positive degree can avoid having roots — the theorem eliminates the possibility of a polynomial that simply never equals zero.

Some roots may repeat. The polynomial x33x2+3x1=(x1)3x^3 - 3x^2 + 3x - 1 = (x - 1)^3 has degree 33 but only one distinct root, x=1x = 1, appearing with multiplicity 33. The count of nn roots includes these repetitions.

For polynomials with real coefficients, an additional constraint applies: complex roots come in conjugate pairs. If a+bia + bi is a root, then abia - bi is also a root. This means a real polynomial of odd degree must have at least one real root — an odd number of roots cannot be paired off entirely into conjugate pairs.

The theorem guarantees existence but offers no method for finding roots. For degree 55 and above, no general formula in radicals exists — a result proven by Abel and Galois. Numerical methods, the Rational Root Theorem, and other tools from this page fill the gap between knowing roots exist and actually computing them.

Vieta's Formulas

Vieta's formulas express symmetric relationships between the roots of a polynomial and its coefficients, without requiring the roots to be known individually.

For a monic quadratic x2+bx+cx^2 + bx + c with roots r1r_1 and r2r_2:

r1+r2=br1r2=cr_1 + r_2 = -b \qquad r_1 \cdot r_2 = c


The sum of the roots equals the negative of the linear coefficient, and the product equals the constant term.

For a monic cubic x3+bx2+cx+dx^3 + bx^2 + cx + d with roots r1r_1, r2r_2, r3r_3:

r1+r2+r3=br_1 + r_2 + r_3 = -b


r1r2+r1r3+r2r3=cr_1r_2 + r_1r_3 + r_2r_3 = c


r1r2r3=dr_1 \cdot r_2 \cdot r_3 = -d


The pattern extends to any degree: the kk-th elementary symmetric polynomial of the roots equals (1)k(-1)^k times the coefficient of xnkx^{n-k} in a monic degree-nn polynomial.

These relationships are useful for verifying factorizations. After finding roots 11, 22, and 33 for x36x2+11x6x^3 - 6x^2 + 11x - 6, Vieta's formulas confirm: 1+2+3=6=(6)1 + 2 + 3 = 6 = -(-6), 12+13+23=111 \cdot 2 + 1 \cdot 3 + 2 \cdot 3 = 11, and 123=6=(6)1 \cdot 2 \cdot 3 = 6 = -(-6). All three match the coefficients. They also allow solving problems where individual roots are not needed — for instance, finding the sum of the squares of the roots using (r1+r2+r3)22(r1r2+r1r3+r2r3)=3622=14(r_1 + r_2 + r_3)^2 - 2(r_1r_2 + r_1r_3 + r_2r_3) = 36 - 22 = 14.

Common Mistakes

The theorems on this page are closely related, and confusing their roles or misapplying their details produces errors that are easy to make and sometimes hard to spot.

The Remainder Theorem and the Factor Theorem are frequently conflated. The Remainder Theorem says dividing P(x)P(x) by (xc)(x - c) leaves a remainder of P(c)P(c). The Factor Theorem adds one specific condition: if that remainder is zero, then (xc)(x - c) is a factor. Applying the Factor Theorem when P(c)0P(c) \neq 0 — concluding that (xc)(x - c) is somehow a partial factor — reflects a misunderstanding of what the theorem actually states.

With the Rational Root Theorem, two errors dominate. The first is forgetting negative candidates — the divisors of a0a_0 and ana_n include both positive and negative values. The second is omitting fractional candidates when the leading coefficient is not 11. For 2x3+x62x^3 + x - 6, candidates include ±12\pm\frac{1}{2} and ±32\pm\frac{3}{2}, not just integers.

In Descartes' Rule of Signs, the most common error occurs when computing P(x)P(-x). Every odd-powered term changes sign while even-powered terms stay the same. Writing P(x)P(-x) incorrectly — changing the wrong terms or missing one — corrupts the sign-change count.

With Vieta's formulas, sign errors are persistent. The sum of roots for x2+bx+cx^2 + bx + c is b-b, not bb. The product of roots for a cubic x3+bx2+cx+dx^3 + bx^2 + cx + d is d-d, not dd. The alternating sign pattern (1)k(-1)^k is easy to lose track of, especially at higher degrees.