IB Analysis and Approaches HL · Algebra review and mathematical induction · Section 2 of 6

Polynomials: remainder and factor theorems

3 worked examples · By Álvaro López

Key idea

A polynomial can be divided just like an integer, and the remainder when dividing by $x - c$ can be found without dividing: just calculate $P (c)$ . With this idea you can find parameters, zeros and factorisations, a very common type of question in Paper 1 and Paper 2 of the IB.

Polynomials, degree and zeros

For example, $P (x) = 4 x^{3} - x + 7$ has degree $3$ , leading coefficient $4$ and constant term $7$ (the coefficient of $x^{2}$ is $0$ ). When a question asks for “a polynomial with real coefficients”, all the $a_{i}$ are real numbers.

For $P (x) = x^{2} - 5 x + 6$ we have $P (2) = 4 - 10 + 6 = 0$ and $P (3) = 9 - 15 + 6 = 0$ : the zeros are $2$ and $3$ . Zeros of $P$ , roots of $P (x) = 0$ , and points where the graph of $y = P (x)$ crosses the $x$ -axis are three ways of saying the same thing. In Paper 2 the graphic display calculator finds them directly, but in Paper 1 you have to get them by hand.

Dividing polynomials

Dividing $P (x)$ by $D (x)$ works just like dividing integers: $17 = 5 \cdot 3 + 2$ means dividend = divisor $\cdot$ quotient + remainder, with a remainder smaller than the divisor.

This last condition says when to stop: you keep dividing while the remainder has a degree equal to or greater than that of the divisor. If the divisor has degree 1, the remainder is a constant; if it has degree 2, the remainder has degree at most 1, that is, it has the form $p x + q$ .

For example, to divide $x^{4} - 3 x^{2} + 2 x - 5$ by $x^{2} - x + 1$ :

$x^{4} \div x^{2} = x^{2}$ . Subtract $x^{2} (x^{2} - x + 1) = x^{4} - x^{3} + x^{2}$ , leaving $x^{3} - 4 x^{2} + 2 x - 5$ .
$x^{3} \div x^{2} = x$ . Subtract $x (x^{2} - x + 1) = x^{3} - x^{2} + x$ , leaving $- 3 x^{2} + x - 5$ .
$- 3 x^{2} \div x^{2} = - 3$ . Subtract $- 3 (x^{2} - x + 1) = - 3 x^{2} + 3 x - 3$ , leaving $- 2 x - 2$ .

As $- 2 x - 2$ has degree $1 < 2$ , we stop: quotient $x^{2} + x - 3$ and remainder $- 2 x - 2$ . Check: $(x^{2} - x + 1) (x^{2} + x - 3) = x^{4} - 3 x^{2} + 4 x - 3$ , and adding the remainder gives back $x^{4} - 3 x^{2} + 2 x - 5$ .

The remainder theorem

The reason is one line. By the division algorithm, $P (x) = (x - c) Q (x) + R$ with $R$ constant, because the divisor has degree 1. If we substitute $x = c$ , the term $(x - c) Q (x)$ vanishes and $P (c) = R$ remains. Dividing by $x - c$ is like “evaluating at $c$ ”, which is much quicker.

Be careful with the sign: to divide by $x + 3$ write it as $x - (- 3)$ , and the remainder is $P (- 3)$ . For a divisor of the form $a x - b$ the remainder is $P (\frac{b}{a})$ , the value that makes the divisor zero.

The factor theorem

It is the remainder theorem in the case $R = 0$ : if the remainder when dividing by $x - c$ is zero, the division is exact and $P (x) = (x - c) Q (x)$ ; and conversely. Three statements are therefore equivalent: $c$ is a zero of $P$ , $x = c$ is a root of $P (x) = 0$ , and $x - c$ divides $P (x)$ .

So $P (x) = x^{2} - 5 x + 6$ has $x - 2$ and $x - 3$ as factors, and factorises as $(x - 2) (x - 3)$ .

If the divisor has degree 2 and two known roots, for example $x^{2} - 4 = (x - 2) (x + 2)$ , the remainder $p x + q$ satisfies $R (2) = P (2)$ and $R (- 2) = P (- 2)$ : because $P (x) = (x^{2} - 4) Q (x) + R (x)$ and the first term vanishes at $2$ and at $- 2$ . This gives two equations for $p$ and $q$ .

The fundamental theorem of algebra and factorisation

This tells us two useful things. A polynomial of degree $n$ has at most $n$ distinct zeros. And if a zero $c$ is found, the factor theorem lets us divide and reduce the problem to a polynomial of degree one less. Complex numbers appear in another topic of the course; for now, it is enough to know that a factor such as $x^{2} + x + 1$ does not factorise with real numbers, but does with complex numbers.

Worked examples

Example 1 Basic

Dividing by x - c and checking with the remainder theorem

Let $P (x) = 2 x^{3} - 5 x^{2} + x + 7$ .

a) Divide $P (x)$ by $x - 3$ and write down the quotient and the remainder.

b) Without dividing, find the remainder when $P (x)$ is divided by $x + 1$ .

Solution

a) By synthetic division with $c = 3$ , using the coefficients $2, - 5, 1, 7$ :

Bring down the $2$ .
$- 5 + 3 \cdot 2 = 1$ .
$1 + 3 \cdot 1 = 4$ .
$7 + 3 \cdot 4 = 19$ .

The last number is the remainder and the others are the coefficients of the quotient, of degree $2$ : the quotient is $2 x^{2} + x + 4$ and the remainder is $19$ . That is,

$2 x^{3} - 5 x^{2} + x + 7 = (x - 3) (2 x^{2} + x + 4) + 19$

Check with the remainder theorem: $P (3) = 2 \cdot 27 - 5 \cdot 9 + 3 + 7 = 54 - 45 + 10 = 19$ , which matches the remainder.

b) To divide by $x + 1 = x - (- 1)$ the remainder is $P (- 1)$ :

$P (- 1) = 2 (- 1)^{3} - 5 (- 1)^{2} + (- 1) + 7 = - 2 - 5 - 1 + 7 = - 1$

The remainder is $- 1$ .

Example 2 Medium

Parameter, factorisation and roots

Consider the polynomial $P (x) = x^{3} + k x^{2} - 5 x - 6$ , where $k \in R$ , and it is known that $x - 2$ is a factor of $P (x)$ .

a) Find the value of $k$ .

b) Fully factorise $P (x)$ with that value of $k$ .

c) Find all the solutions of the equation $P (x) = 0$ .

Solution

a) Factor theorem. If $x - 2$ is a factor, then $P (2) = 0$ :

$P (2) = 8 + 4 k - 10 - 6 = 4 k - 8 = 0 ⟺ k = 2$

b) With $k = 2$ , $P (x) = x^{3} + 2 x^{2} - 5 x - 6$ .

Divide by $x - 2$ . By synthetic division with $c = 2$ and coefficients $1, 2, - 5, - 6$ : bring down $1$ ; $2 + 2 \cdot 1 = 4$ ; $- 5 + 2 \cdot 4 = 3$ ; $- 6 + 2 \cdot 3 = 0$ . The remainder is $0$ (as expected) and the quotient is $x^{2} + 4 x + 3$ .
Factorise the quotient. We look for two numbers that add to $4$ and multiply to $3$ : they are $1$ and $3$ . So $x^{2} + 4 x + 3 = (x + 1) (x + 3)$ .
Result.

$P (x) = (x - 2) (x + 1) (x + 3)$

c) $P (x) = 0$ if and only if one of the factors equals $0$ : $x = 2$ , $x = - 1$ or $x = - 3$ .

Example 3 Hard

Two conditions, factorisation and a remainder of degree 1

The polynomial $P (x) = x^{3} + a x^{2} + b x + 6$ , with $a, b \in R$ , is divisible by $x - 1$ and leaves a remainder of $8$ when divided by $x + 1$ .

a) Find $a$ and $b$ .

b) Factorise $P (x)$ with these values.

c) Find the remainder when $P (x)$ is divided by $x^{2} - 1$ .

Solution

a) Translate the data. Divisible by $x - 1$ means $P (1) = 0$ . Remainder $8$ when dividing by $x + 1 = x - (- 1)$ means $P (- 1) = 8$ .

$P (1) P (- 1) = 1 + a + b + 6 = 0 = - 1 + a - b + 6 = 8 \Rightarrow a + b = - 7 \Rightarrow a - b = 3$

Adding the two equations, $2 a = - 4$ and $a = - 2$ . From the first, $b = - 7 - a = - 5$ . So,

$P (x) = x^{3} - 2 x^{2} - 5 x + 6$

Check: $P (1) = 1 - 2 - 5 + 6 = 0$ and $P (- 1) = - 1 - 2 + 5 + 6 = 8$ .

b) $x - 1$ is a factor. We divide with $c = 1$ and coefficients $1, - 2, - 5, 6$ : bring down $1$ ; $- 2 + 1 = - 1$ ; $- 5 + (- 1) = - 6$ ; $6 + (- 6) = 0$ . The quotient is $x^{2} - x - 6$ .

To factorise it, two numbers that add to $- 1$ and multiply to $- 6$ are $- 3$ and $2$ : $x^{2} - x - 6 = (x - 3) (x + 2)$ . Therefore,

$P (x) = (x - 1) (x - 3) (x + 2)$

c) The divisor $x^{2} - 1 = (x - 1) (x + 1)$ has degree $2$ , so the remainder has the form $R (x) = p x + q$ and

$P (x) = (x^{2} - 1) Q (x) + p x + q$

We substitute the roots of the divisor, $x = 1$ and $x = - 1$ , where the first term vanishes:

$P (1) = 0 P (- 1) = 8 = p + q = - p + q$

Adding, $2 q = 8$ , so $q = 4$ and $p = - 4$ . The remainder is $- 4 x + 4$ .

Check by division: $P (x) = (x^{2} - 1) (x - 2) + (- 4 x + 4)$ , since $(x^{2} - 1) (x - 2) = x^{3} - 2 x^{2} - x + 2$ and adding $- 4 x + 4$ gives $x^{3} - 2 x^{2} - 5 x + 6$ .

In the exam: always write the set-up ( $P (1) = 0$ , $P (- 1) = 8$ ) before solving the system: it is what earns the method marks. In part c), a remainder of degree 1 is set up with two unknowns, not one.

Common mistakes

Summary

A polynomial of degree $n$ is $a_{n} x^{n} + \dots + a_{0}$ with $a_{n} \neq = 0$ ; a zero $c$ satisfies $P (c) = 0$ and is a root of $P (x) = 0$ .
Division algorithm: $P (x) = D (x) Q (x) + R (x)$ , with $R = 0$ or the degree of $R$ less than that of $D$ . Dividing by a divisor of degree 1 leaves a constant remainder.
Remainder theorem: the remainder when $P (x)$ is divided by $x - c$ is $P (c)$ . For $x + 3$ use $P (- 3)$ .
Factor theorem: $x - c$ is a factor of $P (x)$ if and only if $P (c) = 0$ .
To find parameters: translate each piece of data into an equation of the form $P (c) = value$ and solve the system.
With a divisor of degree 2 with roots $r_{1}$ and $r_{2}$ , the remainder $p x + q$ satisfies $R (r_{1}) = P (r_{1})$ and $R (r_{2}) = P (r_{2})$ .
Fundamental theorem of algebra: a polynomial of degree $n$ factorises into $n$ complex linear factors; it has at most $n$ distinct zeros. (HL: the non-real zeros of a real polynomial come in conjugate pairs.)
To factorise a cubic: find a zero, divide, factorise the quotient of degree 2.

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