What a polynomial function is

A polynomial function can be written as a finite sum of non-negative integer powers of its input variable, with each power multiplied by a constant coefficient. The usual form is

A coefficient is a constant number multiplying a power of the input variable in an algebraic term. The degree is the greatest exponent of $x$ whose coefficient is non-zero. It doesn’t depend on the number of terms present. What matters is the highest surviving power.

Unless a domain is stated, this course takes the domain of a polynomial function to be $\mathbb{R}$. Its graph is continuous and smooth, with no gaps, corners or vertical asymptotes. This makes polynomials useful for modelling in sciences: they can provide a simple symbolic model of a relationship that changes steadily.

Equations, zeros and roots

A zero of a function is an input value that gives an output of $0$. A root of a polynomial equation is a solution found by setting the polynomial equal to $0$. Therefore, if

On a graph, real roots appear as the $x$-intercepts. Algebraically, we write $p(c)=0$; graphically, the curve meets the $x$-axis at $x=c$. Both representations carry the same mathematical information. Symbolic, graphical and numerical views of functions are equivalent here, although each one makes certain features easier to spot.

A factor is a polynomial expression that divides another polynomial with zero remainder. When $x-c$ is a factor of $p(x)$, the graph has a zero at $x=c$. The converse also holds: if $c$ is a zero, then $x-c$ is a factor. This is the factor theorem, which is used properly in the next section.

A multiplicity is a positive integer showing how many times the same factor occurs. For instance, if $(x-c)^2$ is a factor, then the root $c$ has multiplicity $2$. On a sketch, a root with odd multiplicity usually crosses the $x$-axis, while one with even multiplicity usually touches the axis and turns. This is useful for checking a sketch, but it shouldn’t replace the algebra.

Working with polynomial expressions

Two polynomial expressions are identical if and only if their corresponding coefficients are equal. This simple fact sits behind many coefficient-comparison questions. If two cubics are equal for every value of $x$, their $x^3$, $x^2$, $x$ and constant coefficients must match.

Let $q(x)$ be another polynomial function, with $q$ naming the second polynomial function. To form $p(x)+q(x)$, add like powers of $x$. For multiplication, expand in the usual way and collect like terms. Provided neither polynomial is the zero polynomial, the degree of $p(x)q(x)$ equals the degree of $p(x)$ plus the degree of $q(x)$.

Polynomial division follows the same structure as integer division. If $d(x)$ is the divisor polynomial, $Q(x)$ the quotient polynomial and $R(x)$ the remainder polynomial, then

The degree of $R(x)$ must be less than the degree of $d(x)$. When dividing by a linear factor, the remainder is simply a constant.

Real and complex roots

A polynomial equation of degree $n$ has $n$ complex roots when multiplicity is counted. Over the real numbers, fewer roots may be visible because non-real complex roots don’t appear as $x$-intercepts. For a polynomial with real coefficients, non-real roots come in conjugate pairs. If $z$ is a non-real complex root, where $z$ is a complex number, then its conjugate $\overline{z}$ is also a root. This connects directly with complex roots of polynomial equations from AHL number and algebra.

An odd-degree polynomial with real coefficients must therefore have at least one real root. Its ends extend in opposite vertical directions, so a continuous curve cannot pass from one side to the other without crossing the $x$-axis. This gives a graphical interpretation of the algebraic fact—not a weaker version, but another representation.

The division identity behind both theorems

Divide a polynomial by the linear expression $x-c$. The division identity is

Substitute $x=c$:

$$p(c)=(c-c)Q(c)+R=R$$

That single line explains why the theorem works. There’s no need for long division to find the remainder; just evaluate the polynomial at the value that makes the divisor zero.

Remainder theorem

The remainder theorem states that when a polynomial $p(x)$ is divided by $x-c$, its remainder is $p(c)$. Watch the sign. For division by $x-3$, use $p(3)$. For division by $x+3$, use $p(-3)$.

This result is especially useful when a polynomial has unknown coefficients. Suppose the remainder after division by $x-2$ is $5$. You can write $p(2)=5$ straight away. A second division with a different remainder gives another equation. That’s usually quicker and cleaner than carrying out two polynomial divisions.

Factor theorem

The factor theorem states that $x-c$ is a factor of $p(x)$ if and only if $p(c)=0$. It is simply the zero-remainder case of the remainder theorem.

A practical method is:

• choose a possible value of $c$;
• calculate $p(c)$;
• if $p(c)=0$, write down the factor $x-c$;
• divide or use another method to reduce the polynomial and continue.

If a polynomial has integer coefficients and leading coefficient $1$, every integer root must divide the constant term $a_0$. That doesn’t mean every divisor is a root, but it does produce a short list of sensible values to test. For example, when $a_0=12$, the possible integer roots must be among the positive and negative divisors of $12$. Test those candidates rather than guessing random numbers.

Factorising over different number systems

Over the complex numbers, a degree $n$ polynomial splits into $n$ linear factors, counting multiplicity. With real coefficients over the real numbers, it may instead factor into linear factors together with irreducible quadratic factors. An irreducible quadratic factor is a quadratic polynomial with real coefficients and no real roots, so it cannot be split any further into real linear factors.

Pay attention when a question asks for factorisation over $\mathbb{R}$. Stop once the remaining quadratic factors are irreducible. If the question works over $\mathbb{C}$, those quadratics can be split using complex roots.

Using technology appropriately

Some polynomial equations have no pleasant analytic method in an examination setting. If technology is allowed or required, use it to find roots or intersections. You still need to interpret the result mathematically: identify which roots are real, which are repeated, and which factors follow from them. Technology supplies numerical evidence; the factor and remainder theorems account for the algebra behind it.

The IB formulae you need

Consider the polynomial equation

$$\sum_{r=0}^{n}a_rx^r=0$$

Just three coefficients determine the sum and product of all $n$ roots, counted with multiplicity: $a_n$, $a_{n-1}$ and $a_0$.

Let the roots be $\alpha_1,\alpha_2,\ldots,\alpha_n$. Here, $\alpha_i$ is the $i$th root, and $i$ counts the roots from $1$ to $n$. Then

$$\sum_{i=1}^{n}\alpha_i=-\frac{a_{n-1}}{a_n}, \qquad \prod_{i=1}^{n}\alpha_i=\frac{(-1)^n a_0}{a_n}$$

where $\sum$ tells you to add the listed roots and $\prod$ tells you to multiply them. These are the syllabus formulae. Be careful with the sign in the product: an odd degree gives a negative sign in front of $a_0/a_n$, while an even degree gives a positive sign.

Why the formulae are true

If a degree $n$ polynomial has roots $\alpha_1,\alpha_2,\ldots,\alpha_n$, it can be written over the complex numbers as

$$p(x)=a_n(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$$

Expanding this product gives the following coefficient of $x^{n-1}$:

$$-a_n(\alpha_1+\alpha_2+\cdots+\alpha_n)$$

Match this with $a_{n-1}x^{n-1}$ to get the sum formula. The constant term is

$$a_n(-\alpha_1)(-\alpha_2)\cdots(-\alpha_n)=a_n(-1)^n\alpha_1\alpha_2\cdots\alpha_n$$

and matching it with $a_0$ gives the product formula.

This is the part of Viete's theorem required in IB Mathematics: analysis and approaches. The full theorem also includes relationships involving pairwise products, triple products and so on. Here, the required content is the sum of all roots and their product.

How to use the formulae

The signs are particularly straightforward for a monic cubic. If

$$x^3+Ax^2+Bx+C=0$$

where $A$ is the coefficient of $x^2$, $B$ is the coefficient of $x$ and $C$ is the constant term, the roots have sum $-A$ and product $-C$. The two syllabus formulae don’t directly give the middle coefficient $B$. You may instead find it by substituting a known root into the equation or by expanding factors.

For a quadratic equation $ax^2+bx+c=0$, where $a$ is the quadratic coefficient, $b$ is the linear coefficient and $c$ is the constant term, the same rules give sum $-b/a$ and product $c/a$. The AHL formula isn’t a new trick. It extends the quadratic result to degree $n$.

For example, consider

$$2x^3-5x^2+7x-4=0$$

The sum of its three roots is

$$-\frac{-5}{2}=\frac{5}{2}$$

and their product is

$$\frac{(-1)^3(-4)}{2}=2$$

Neither result requires you to find the individual roots.

Roots as facts, representations as interpretations

A polynomial may appear as an expanded equation, a factorised expression, a table of values or a graph. Its roots remain the same mathematical facts, but each representation presents them differently: as solutions, factors, zeros in a table, or intercepts on a graph. Moving between these forms isn’t just decoration; it is often the shortest route to the solution.