Polynomials: Degree, Zeroes, Division and Factorising

What makes an expression a polynomial

A polynomial is an algebraic expression where every variable is raised only to a non-negative whole-number power. That single rule is what disqualifies some expressions that otherwise look similar:

• $4x^2 - 15x + 9$ is a polynomial (powers 2, 1, 0, all non-negative whole numbers).
• $5x + 2y^{-1} - 120$ is not a polynomial: $y$ has a negative power.
• $y^{6/7} + 2z - \tfrac{1}{2}$ is not a polynomial: $6/7$ isn't a whole number.
• $\sqrt{x} + 1$ and $\dfrac{1}{x} + x$ are not polynomials either: $\sqrt{x} = x^{1/2}$ and $\dfrac{1}{x} = x^{-1}$.

The coefficients can be any numbers, including fractions and surds: $\sqrt{2}x^2 + \tfrac{1}{3}$ is a polynomial. Only the powers of the variable are restricted.

Monomial, binomial, trinomial

The parts of a polynomial separated by $+$ or $-$ are its terms. The number of terms names the expression:

• Monomial: one term, e.g. $5z$ or $4x^2$.
• Binomial: two terms, e.g. $3p - 2q$.
• Trinomial: three terms, e.g. $4x^2 + 6x + 7$.

Finding the degree

In one variable, the degree is just the highest power present. With multiple variables, add the exponents within each term, then take the highest total across all terms.

Worked example: find the degree of $5x^2y^3 + 2x^2y^2 - 8xy$.

Add exponents within each term:

The highest total is 5, so the polynomial has degree 5.

Polynomials named by degree

A non-zero constant like 7 has degree 0, because $7 = 7x^0$. The zero polynomial (just 0) is the exception: its degree is left undefined, since $0 = 0x^5 = 0x^{100}$ fits any power.

Worked example: the degree of a product

Find the degree of $(x^2 + 1)(x^3 - 2x)$ without fully expanding it.

Solution: The highest-power term of the product comes from multiplying the highest-power terms of each factor, $x^2 \cdot x^3 = x^5$, and nothing else can cancel it. So the degree is $2 + 3 =$ 5. Expanding confirms it: $x^5 - 2x^3 + x^3 - 2x = x^5 - x^3 - 2x$.

Worked example: the value of a polynomial

Find $p(2)$ and $p(-1)$ for $p(x) = x^3 - 2x + 1$.

Solution: Substitute each value for $x$:

A value of $x$ that makes $p(x) = 0$ is called a zero of the polynomial. Zeroes are where the next sections pick up.

Zeroes and coefficients of a quadratic

For a quadratic polynomial $ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:

Worked example. Find the zeroes of $6x^2 - 7x - 3$ and verify the relationship. With $a=6$, $b=-7$, $c=-3$, the quadratic formula gives $x = \dfrac{7 \pm \sqrt{49 + 72}}{12} = \dfrac{7 \pm 11}{12}$, so the zeroes are $\dfrac{3}{2}$ and $-\dfrac{1}{3}$.

Building a quadratic from its zeroes. A quadratic whose zeroes have sum $S$ and product $P$ is $x^2 - Sx + P$. For sum $\sqrt{2}$ and product $\tfrac{1}{3}$ that is $x^2 - \sqrt{2}x + \tfrac{1}{3}$, or, multiplying by 3 to clear the fraction, $3x^2 - 3\sqrt{2}x + 1$. Any non-zero multiple of a valid answer is also valid.

The division algorithm

Dividing $p(x)$ by a nonzero $g(x)$ always gives a quotient $q(x)$ and remainder $r(x)$ with:

Worked example: long division

Divide $p(x) = x^3 - 3x^2 + 5x - 3$ by $g(x) = x^2 - 2$.

Solution: $x^3 \div x^2 = x$; subtracting $x(x^2-2) = x^3-2x$ leaves $-3x^2+7x-3$. Then $-3x^2 \div x^2 = -3$; subtracting $-3(x^2-2) = -3x^2+6$ leaves $7x-9$, whose degree is below the divisor's. So the quotient is $x - 3$ and the remainder $7x - 9$.

Check: $(x^2-2)(x-3) + 7x - 9 = x^3 - 3x^2 - 2x + 6 + 7x - 9 = x^3 - 3x^2 + 5x - 3$.

The remainder and factor theorems

Dividing by a linear $x - a$, the remainder is simply $p(a)$. For the same $p(x)$: $p(2) = 8 - 12 + 10 - 3 = 3$, so dividing by $x - 2$ leaves remainder 3 (the quotient is $x^2 - x + 3$). And $p(1) = 1 - 3 + 5 - 3 = 0$, so $x - 1$ is a factor: $p(x) = (x-1)(x^2-2x+3)$.

Worked example. $ax^3+3x^2-3$ and $2x^3-5x+a$ leave the same remainder when divided by $x-4$. Find $a$. Each remainder is the polynomial's value at $x = 4$: $64a + 48 - 3 = 64a + 45$ and $128 - 20 + a = 108 + a$. Setting them equal, $63a = 63$, so $a = 1$.

Worked example: finding remaining zeroes

Two zeroes of $3x^4+6x^3-2x^2-10x-5$ are $\pm\sqrt{5/3}$. Find the others.

Solution: $\left(x-\sqrt{5/3}\right)\left(x+\sqrt{5/3}\right) = x^2-\dfrac{5}{3}$ is a factor. Dividing by it gives quotient $3x^2+6x+3$ and remainder 0, and $3x^2+6x+3 = 3(x+1)^2$. So the other two zeroes are both $-1$, a repeated zero.

Check: $\left(x^2-\dfrac{5}{3}\right)(3x^2+6x+3) = 3x^4+6x^3+3x^2-5x^2-10x-5 = 3x^4+6x^3-2x^2-10x-5$.

Factorising with identities

Recognising a cube pattern turns a hard-looking factorisation into one line:

Worked example. Factorise $64a^3 - 27b^3 - 144a^2b + 108ab^2$. With $x = 4a$ and $y = 3b$, $x^3 = 64a^3$, $y^3 = 27b^3$ and $3xy(x-y) = 3(4a)(3b)(4a-3b) = 144a^2b - 108ab^2$. So the expression is $(x-y)^3 = (4a-3b)^3$.

Worked example. A rectangle's area is $25a^2-35a+12$. Find possible expressions for its sides. Put $x = 5a$: the area is $x^2-7x+12 = (x-4)(x-3)$, so $25a^2-35a+12 = (5a-4)(5a-3)$, and the sides could be $5a - 4$ and $5a - 3$.

Common mistakes

• Taking the single largest exponent in several variables. In $5x^2y^3$ the degree is $2 + 3 = 5$, not 3.
• Calling $\sqrt{x}$ or $\dfrac{1}{x}$ terms polynomial. They're powers $\tfrac{1}{2}$ and $-1$, which aren't allowed. A surd coefficient like $\sqrt{2}x$ is fine.
• Saying a constant has no degree. A non-zero constant has degree 0. Only the zero polynomial's degree is undefined.
• Reading the degree before simplifying. $(x^2 + x) - x^2$ looks quadratic but simplifies to $x$, which has degree 1.
• Skipping missing powers. Divide $x^3 + 1$ as $x^3 + 0x^2 + 0x + 1$, or the columns misalign.
• Stopping too early or too late. Stop exactly when the remainder's degree drops below the divisor's.
• Sign errors when subtracting. Subtract the whole product line, every term.
• Forgetting to check. $g(x)q(x) + r(x)$ must rebuild $p(x)$ exactly.
• Dropping the minus sign in the sum of zeroes. It is $-\dfrac{b}{a}$: for $6x^2 - 7x - 3$ the zeroes add to $+\dfrac{7}{6}$.

Where this comes up in exams

The 2025-26 CBSE Class IX syllabus lists the definition of a polynomial in one variable, its degree, zeroes, the Remainder Theorem, the Factor Theorem and factorising cubic polynomials, and the Class X syllabus lists the relationship between zeroes and coefficients of quadratic polynomials. Neither names the division algorithm itself, which is the method behind both theorems.

Checked against CBSE Mathematics curriculum, Classes IX-X (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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