Polynomials: identities, factoring and synthetic division

Polynomials: a reasoned guide

From basic manipulations to the algebraic structure you will meet at university. The goal is not to memorize a bag of tricks, but to recognize which idea applies and why it works.

1. Essential language

A polynomial in one variable with coefficients in a set $K$ has the form

P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

where the exponents are non-negative integers and the coefficients belong to $K$ , for example $ℚ$ , $ℝ$ or $ℂ$ .

Degree

If $a_{n} \neq 0$ , its degree is $n$ . We write $\deg (P) = n$ .

Special terms

$a_{n}$ is the leading coefficient; $a_{0}$ is the constant term. A polynomial whose leading coefficient is 1 is monic.

Zero polynomial

All its coefficients are zero. Its degree is not treated like that of other polynomials; a common convention is $\deg (0) = - \infty$ .

Equality

Two polynomials are equal when the coefficients of corresponding powers are equal. Agreeing at one or a few values of $x$ is not enough.

A polynomial and an expression are not the same thing.

\frac{1}{x}

\sqrt{x}

and

x^{- 2}

are not polynomials in

x

, because they respectively contain a variable in the denominator, a fractional exponent and a negative exponent.

2. Operations and degrees

Operation	Rule	Remark
Sum	Add the coefficients of terms with the same power.	$\deg (P + Q) \leq \max {\deg (P), \deg (Q)}$ ; the degree may decrease through cancellation.
Product	Use the distributive property and add the exponents of powers being multiplied.	Over ℚ, ℝ or ℂ, if $P, Q \neq 0$ , then $\deg (P Q) = \deg (P) + \deg (Q)$ .
Power	Multiply the polynomial by itself repeatedly.	$\deg (P^{k}) = k \cdot \deg (P)$ for $P \neq 0$ and $k > 0$ .
Composition	Substitute $Q (x)$ for $x$ in $P$ .	$\deg (P \circ Q) = \deg (P) \cdot \deg (Q)$ if both have positive degree.

(2 x^{2} - 3 x + 1) + (x^{2} + 5 x - 4) = 3 x^{2} + 2 x - 3

3. Algebraic identities

These are identities: they hold for every choice of the expressions $A$ and $B$ . You should be able to use them in both directions, to expand and to factor.

Name	Identity	Pattern to recognize
Square of a sum	${(A + B)}^{2} = A^{2} + 2 A B + B^{2}$	Two squares and a positive twice-product.
Square of a difference	${(A - B)}^{2} = A^{2} - 2 A B + B^{2}$	Two squares and a negative twice-product.
Sum times difference	$(A + B) (A - B) = A^{2} - B^{2}$	A difference of squares.
Cube of a sum	${(A + B)}^{3} = A^{3} + 3 A^{2} B + 3 A B^{2} + B^{3}$	Coefficients 1, 3, 3, 1 and complementary powers.
Cube of a difference	${(A - B)}^{3} = A^{3} - 3 A^{2} B + 3 A B^{2} - B^{3}$	Coefficients 1, 3, 3, 1 with alternating signs.
Sum of cubes	$A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})$	The binomial keeps the sign; in the trinomial the mixed term has the opposite sign.
Difference of cubes	$A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$	The binomial keeps the sign; every sign in the trinomial is positive.

The general formula: the binomial theorem

{(A + B)}^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) A^{n - k} B^{k}

The binomial coefficients $(\binom{n}{k})$ give the rows of Pascal’s triangle. This formula reappears in combinatorics, probability, calculus and linear algebra.

4. Factoring techniques

Factoring means writing a polynomial as a product of lower-degree polynomials. The factorization depends on the allowed coefficient set.

Common-factor extraction

Extract the greatest common factor: $6 x^{3} - 9 x^{2} = 3 x^{2} (2 x - 3)$ .

Factoring by grouping

Group terms that share a common factor: $x^{3} + x^{2} + 2 x + 2 = (x + 1) (x^{2} + 2)$ .

Using identities in reverse

$9 x^{2} - 25 = (3 x - 5) (3 x + 5)$ .

Substitution

Recognize a simpler structure: letting $t = x^{2}$ , $x^{4} - 5 x^{2} + 4$ becomes $t^{2} - 5 t + 4$ .

Finding a root

If you find $α$ such that $P (α) = 0$ , then $(x - α)$ is a factor. Divide and continue with the quotient.

Completing the square

$x^{2} + 6 x + 5 = {(x + 3)}^{2} - 4$ , then use the difference of squares.

First practical rule: always look for a common factor, including a numerical or negative one. It is the cheapest check and makes every later structure easier to see.

5. Quadratic trinomials

For the monic trinomial $x^{2} + s x + p$ , look for two numbers $u$ and $v$ whose sum is $s$ and whose product is $p$ :

x^{2} + (u + v) x + u v = (x + u) (x + v)

In the general case $a x^{2} + b x + c$ , calculate the discriminant $Δ = b^{2} - 4 a c$ . If the roots are $x_{1}$ and $x_{2}$ , then

a x^{2} + b x + c = a (x - x_{1}) (x - x_{2})

If $Δ > 0$ , there are two distinct real roots.
If $Δ = 0$ , there is one repeated real root and the trinomial is a perfect square up to the factor $a$ .
If $Δ < 0$ , it has no factorization into real linear factors, but it does factor over ℂ.

Viète’s formulas give $x_{1} + x_{2} = - \frac{b}{a}$ and $x_{1} x_{2} = \frac{c}{a}$ : they are useful far beyond factoring.

6. Euclidean division of polynomials

If $A (x)$ and $B (x) \neq 0$ have coefficients in a field, there are unique polynomials $Q$ and $R$ such that

A (x) = B (x) Q (x) + R (x)

Long-division algorithm

Order dividend and divisor by descending powers, inserting zero coefficients for missing terms.
Divide the leading term of the current dividend by the leading term of the divisor to get the next quotient term.
Multiply the divisor by that term and subtract.
Repeat until the remainder has lower degree than the divisor.

(2 x^{3} + 3 x^{2} - 5 x + 6) : (x + 2)

Check: $(x + 2) (2 x^{2} - x - 3) + 12$ reconstructs the dividend.

7. Remainder theorem, factor theorem and roots

Remainder theorem

The remainder when dividing $P (x)$ by $x - α$ is the number $P (α)$ .

Factor theorem

$P (α) = 0$ if and only if $x - α$ divides $P (x)$ . In Italy this is often called Ruffini’s theorem.

Multiplicity

$α$ has multiplicity $m$ if ${(x - α)}^{m}$ divides $P$ , but the next power does not.

The remainder theorem follows directly from division:

P (x) = (x - α) Q (x) + r

A nonzero polynomial of degree

n

has at most

n

distinct roots. Counting multiplicity, a complex polynomial of degree

n

has exactly

n

roots: this is the fundamental theorem of algebra.

8. Synthetic division

Ruffini’s rule is synthetic division: it shortens division by a monic linear divisor $x - α$ . It does not apply directly to divisors such as $x^{2} + 1$ or $2 x - 3$ .

Example: divide $2 x^{3} - 3 x^{2} - 11 x + 6$ by $x - 3$ .

$α = 3$	2	−3	−11	6
products		6	9	−6
sums	2	3	−2	0

Write every coefficient, including zeros for missing terms.
Bring down the first coefficient.
Multiply it by $α$ , write the result under the next coefficient and add.
Repeat. The last number is the remainder; the others are the quotient coefficients.

Here the quotient is $2 x^{2} + 3 x - 2$ and the remainder is zero. Therefore

2 x^{3} - 3 x^{2} - 11 x + 6 = (x - 3) (2 x^{2} + 3 x - 2)

9. Rational root theorem

For a polynomial with integer coefficients

P (x) = a_{n} x^{n} + \dots + a_{0}

If a reduced fraction $\frac{p}{q}$ is a rational root, then $p$ divides the constant term $a_{0}$ and $q$ divides the leading coefficient $a_{n}$ .

Monic polynomial

If the leading coefficient is 1, every possible rational root is an integer dividing the constant term.

The theorem narrows the search; it does not guarantee a root

Candidates must be checked by evaluating $P (\frac{p}{q})$ . A divisor of the constant term is not automatically a root.

Zero constant term

If $a_{0} = 0$ , factor out $x$ : zero is already a root.

For $2 x^{3} - 3 x^{2} - 11 x + 6$ , the only rational candidates are $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$ . The search is now finite.

10. Factorization depends on the field

Polynomial	Over ℚ	Over ℝ	Over ℂ
$x^{2} - 2$	It has no factorization into rational linear factors.	$(x - \sqrt{2}) (x + \sqrt{2})$	The same real factorization.
$x^{2} + 1$	Irreducible.	Irreducible.	$(x - i) (x + i)$
$x^{4} + 4$	$(x^{2} - 2 x + 2) (x^{2} + 2 x + 2)$		Each quadratic factor splits further into linear factors.

Irreducible means that it cannot be factored into nonconstant polynomials over the chosen field.
Over ℂ every nonconstant polynomial factors into linear factors.
Over ℝ the only irreducible factors have degree 1, or degree 2 with negative discriminant.
Over ℚ irreducible factors may have any degree.

Gauss’s lemma, which you will meet in algebra, provides a controlled passage between primitive polynomials in

ℤ [x]

and factorizations in

ℚ [x]

. In practice, after extracting the greatest common divisor of the coefficients, looking for rational factors is equivalent to looking for suitable integer factors.

11. Polynomial GCD and the Euclidean algorithm

As with integers, the greatest common divisor of two polynomials is found through repeated division:

A = B Q_{1} + R_{1}

The last nonzero remainder, made monic, is the GCD. This method is more fundamental than full factorization: it works even when the factors are not apparent.

Coprime polynomials

$P$ and $Q$ are coprime if their GCD is 1, up to nonzero constants.

Bézout’s identity

There are polynomials $U$ and $V$ such that $U P + V Q = GCD (P, Q)$ .

Repeated roots

In characteristic zero, $P$ has repeated roots if and only if $GCD (P, P^{'}) \neq 1$ .

12. Rational expressions

A rational expression is a quotient $\frac{P (x)}{Q (x)}$ with $Q (x) \neq 0$ . Factoring numerator and denominator allows simplification, but does not erase the original domain restrictions.

\frac{x^{2} - 1}{x^{2} - x} = \frac{(x - 1) (x + 1)}{x (x - 1)} = \frac{x + 1}{x}

Cancel factors, not terms in a sum. In

\frac{x + 1}{x}

you cannot “cancel the

x

”.

13. Working strategy

Normalize. Order by degree, combine like terms and explicitly include any zero coefficients.
Extract common factors. Extract the greatest numerical and algebraic common factor.
Look at the form. Count the terms and look for identities, grouping or a substitution such as $t = x^{2}$ .
Use the degree. For degree 2, use the discriminant and roots; for higher degrees, look for possible rational roots.
Divide. Once you find a root, use synthetic or long division and repeat on the quotient.
Decide where to stop. A factor may be irreducible over ℚ but not over ℝ or ℂ: always state the field.
Check. Multiply the factors back out, compare the degree and leading coefficient, then check the constant term.

14. Common mistakes

The “distributed” square

${(A + B)}^{2} \neq A^{2} + B^{2}$ : the term $2 A B$ is missing.

Sign before parentheses

$- (A - B) = - A + B$ . The sign changes for every term.

Missing terms in synthetic division

In $x^{3} - 4 x + 1$ the coefficients are $1, 0, - 4, 1$ , not three numbers.

Sign of the divisor

To divide by $x + 3$ with synthetic division, use $α = - 3$ .

Mistaking a candidate for a root

The rational root theorem only produces a list of candidates; each one must be checked.

Stopping division too early

The remainder is valid only when its degree is strictly less than the divisor’s degree.

Lost domain restrictions

Canceling a denominator factor does not restore the values that made it zero to the domain.

Factorization without a field

Saying that a polynomial “does not factor” is incomplete: specify whether you are working over ℚ, ℝ or ℂ.

15. Graded exercises

Try to justify every step and reveal the solution only at the end.

1. Common factor and difference of squares

Factor $3 x^{3} - 12 x$ .

Show solution

First factor out $3 x$ , then recognize a difference of squares:

$3 x (x^{2} - 4) = 3 x (x - 2) (x + 2)$ .

2. Non-monic trinomial

Factor $6 x^{2} + x - 2$ .

Show solution

The roots are $\frac{1}{2}$ and $- \frac{2}{3}$ . Therefore

$6 x^{2} + x - 2 = (2 x - 1) (3 x + 2)$ .

3. Substitution

Factor over ℝ $x^{4} - 5 x^{2} + 4$ .

Show solution

With $t = x^{2}$ you get $t^{2} - 5 t + 4 = (t - 1) (t - 4)$ .

Therefore $(x - 1) (x + 1) (x - 2) (x + 2)$ .

4. Remainder without division

Find the remainder when dividing $x^{5} - 2 x^{2} + 7$ by $x - 2$ .

Show solution

By the remainder theorem, it is enough to calculate $P (2) = 32 - 8 + 7 = 31$ .

5. Synthetic division and complete factorization

Factor over ℚ $x^{3} - 4 x^{2} + x + 6$ .

Show solution

The integer candidates are $\pm 1, \pm 2, \pm 3, \pm 6$ . Since $P (2) = 0$ , synthetic division gives the quotient $x^{2} - 2 x - 3$ .

Result: $(x - 2) (x - 3) (x + 1)$ .

6. Multiplicity

Find the roots and their multiplicities for $x^{4} - 2 x^{3} + x^{2}$ .

Show solution

$x^{2} (x^{2} - 2 x + 1) = x^{2} {(x - 1)}^{2}$ .

The roots 0 and 1 both have multiplicity 2.

7. Field of factorization

Factor $x^{4} - 2$ over ℚ and over ℝ.

Show solution

It is irreducible over ℚ. Over ℝ, letting $a = \sqrt[4]{2}$ , we have $x^{4} - 2 = (x - a) (x + a) (x^{2} + a^{2})$ .

The last factor, $x^{2} + \sqrt{2}$ , is irreducible over ℝ.

8. Rational expression

Simplify $\frac{x^{2} - 4}{x^{2} - 3 x + 2}$ and state the restrictions.

Show solution

$\frac{(x - 2) (x + 2)}{(x - 1) (x - 2)} = \frac{x + 2}{x - 1}$ , with $x \neq 1, 2$ .

What you should know before university

Move confidently between expanded and factored form.
Use algebraic identities and the distributive property without losing signs or terms.
Perform polynomial division and check it with $A = B Q + R$ .
Connect roots, factors, remainders, signs and multiplicities.
Factor low-degree polynomials using common factors, substitutions, the discriminant and synthetic division.
Simplify rational expressions while preserving domain restrictions.
Distinguish examples, conjectures and proofs, and justify why each technique applies.
Remember that “irreducible” and “fully factored” depend on the coefficient field.

In algebra courses these objects become the ring

K [x]

; in calculus you will use factorizations and multiplicities for limits, signs and expansions; in linear algebra polynomials enter through eigenvalues, characteristic polynomials and diagonalization.