Polynomial

An expression like $2x^3 - 5x^2 + 4x - 1$ is built from one letter, x, raised to the powers 3, 2, 1 and 0, each multiplied by a number, and added up. An algebraic expression in which every power of the variable is a whole number (0, 1, 2, 3, …) is called a polynomial.

So $x^2 - 5x + 6$, 7x and $4x^4 - 3$ are polynomials, but $x^{-1} + 2$ (that is 1/x + 2) and $\sqrt{x} + 1$ (that is $x^{1/2} + 1$) are not: their powers are −1 and ½.

A polynomial of degree n, its highest power, can be written as a function:

$$f(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \dots + a_2 x^2 + a_1 x + a_0$$

where the coefficients $a_n, a_{n-1}, \dots, a_1, a_0$ are real numbers, $a_n \neq 0$, and n is a whole number. Two of them have names: $a_n$, in front of the highest power, is the leading coefficient, and a₀, with no x, is the constant term. Written this way, highest power first, the polynomial is in descending order (standard form).

Degree	Name	Example
0	constant	7
1	linear	3x − 2
2	quadratic	$x^2 - 5x + 6$
3	cubic	$x^3 - x^2 - 10x - 8$
4	quartic (biquadratic)	$x^4 - 3x^2 + 5x - 4$

Polynomials are also named by how many terms they have: a monomial has one ($7x^3$), a binomial two (x − 2), a trinomial three ($x^2 - 5x + 6$).

A polynomial is a function, so it has a value for each x: put the number in place of x everywhere. For $f(x) = x^3 - 2x + 5$, f(2) = 8 − 4 + 5 = 9 and f(−1) = −1 + 2 + 5 = 6. That is why polynomials turn up wherever one quantity depends smoothly on another: a box’s volume on its cut, the height of a thrown ball on time, a factory’s cost on how much it makes.

When 17 is divided by 5, the quotient is 3 and the remainder 2, and the four numbers are tied together: 17 = 5 × 3 + 2. Polynomials divide the same way.

Whenever a polynomial f(x) is divided by a divisor d(x), there are a quotient q(x) and a remainder R such that

$$f(x) = d(x) \times q(x) + R \qquad \text{or} \qquad \frac{f(x)}{d(x)} = q(x) + \frac{R}{d(x)}$$

This is the division algorithm. The remainder always has a lower degree than the divisor; dividing by a linear divisor such as x − 2, the remainder is a constant. If R = 0, then f(x)/d(x) = q(x) exactly: d(x) divides f(x).

The activity is no coincidence. It is a theorem:

Remainder theorem: when a polynomial f(x) of degree n is divided by x − a, the remainder is f(a), and the quotient has degree n − 1.

The same idea gives the remainder for any linear divisor. Make the divisor zero, and put that value of x into f(x):

Divisor	It is zero when	Remainder
x − a	x = a	f(a)
x + a = x − (−a)	x = −a	f(−a)
$ax - b = a\left(x - \frac{b}{a}\right)$	$x = \frac{b}{a}$	$f\left(\frac{b}{a}\right)$
$ax + b = a\left\{x - \left(-\frac{b}{a}\right)\right\}$	$x = -\frac{b}{a}$	$f\left(-\frac{b}{a}\right)$

For ax − b, the proof is the same: f(x) = (ax − b) q(x) + R, and putting x = b/a makes ax − b = 0.

Synthetic division: the division and f(a) in one

You met synthetic division in Grade 9: a short way to divide by x − a using only the coefficients. Write a on the left and the coefficients (every power, 0 for a missing one) in a row. Bring down the first; then, again and again, multiply the last number of the bottom row by a, write it under the next coefficient, and add. For $x^4 - 3x^2 + 5x - 4$ divided by x − 2:

$$\begin{array}{r|rrrrr} 2 & 1 & 0 & -3 & 5 & -4 \\ & & 2 & 4 & 2 & 14 \\ \hline & 1 & 2 & 1 & 7 & \boxed{10} \end{array}$$

The bottom row is the quotient’s coefficients, $x^3 + 2x^2 + x + 7$, and the last number, boxed, is the remainder: Activity 1 again, in four lines.

Why is that last number f(2)? Follow the bottom row: 1, then 1 × 2 + 0 = 2, then 2 × 2 − 3 = 1, then 1 × 2 + 5 = 7, then 7 × 2 − 4 = 10. That is

$$f(2) = \big(((1 \times 2 + 0) \times 2 - 3) \times 2 + 5\big) \times 2 - 4$$

and multiplying out the brackets gives $2^4 - 3 \times 2^2 + 5 \times 2 - 4$: exactly f(2). Synthetic division is working out f(a), one multiply and one add at a time. (Computers work out a polynomial’s value this way too. It is called Horner’s method, and no general method uses fewer multiplications.)

Worked examples

Question 1

Question 2

Question 3

Use the remainder theorem to find the remainder.

Question 4

Question 5

When f(x) is divided by d(x) with remainder 0, then f(x) = d(x) × q(x): d(x) is a factor of f(x). Joined to the remainder theorem, this gives a test for factors that needs no division at all.

Factor theorem: if f(a) = 0, then x − a is a factor of the polynomial f(x).

Converse of the factor theorem: if x − a is a factor of f(x), then f(a) = 0.

Together: x − a is a factor of f(x) exactly when f(a) = 0. A number a with f(a) = 0 is a root (or zero) of the polynomial, and on the graph of y = f(x) it is where the curve meets the X-axis. So three things are one: a root a, a factor x − a, and a crossing of the X-axis at x = a.

Worked examples

Question 1

Questions 2 and 3

Question 4

Use the factor theorem to prove each.

Question 5

Question 6

The activity shows the plan: find one root, and the factor theorem and synthetic division do the rest. But where do you look for that first root? Not among all the numbers: there is a short list.

Rational root theorem: if a polynomial equation with whole-number coefficients
$$a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0$$
has a rational root p/q in its lowest terms, then p is a factor of the constant term a₀ and q is a factor of the leading coefficient $a_n$. So the only possible rational roots are

$$\pm \frac{\text{factors of } a_0}{\text{factors of } a_n}$$

each reduced to its lowest terms. They are only candidates: some of them are roots and most are not. The values of x that satisfy f(x) = 0 are called the roots or zeros of the equation.

Why is it true? Take a cubic $ax^3 + bx^2 + cx + d = 0$ with the root p/q in lowest terms. Putting it in and multiplying by $q^3$:
$$ap^3 + bp^2q + cpq^2 + dq^3 = 0$$
The first three terms each have a factor p, so $dq^3$ must too. But p and q have no common factor, so p divides d. In the same way the last three terms have a factor q, so q divides $ap^3$, and so q divides a.

Steps for factorizing a polynomial of degree 3

• a.Arrange the terms of f(x) in descending order of the powers of x.
• b.List the possible roots ±p/q, p a factor of the constant term a₀ and q of the leading coefficient $a_n$, in lowest terms. Try them in f(x), ±1 and ±2 first, then ±3, $\pm \frac{1}{2}$, $\pm \frac{2}{3}$ and the rest, until one makes f(x) = 0.
• c.When f(a) = 0, divide by x − a by synthetic division to find the quotient q(x).
• d.Factorize the quotient, a quadratic.
• e.Write f(x) as the product of all three factors.

Worked examples

Back to the box

So the opening question has two answers: cut 2 cm squares for a shallow box 24 cm × 16 cm × 2 cm, or 6 cm squares for a deep one 16 cm × 8 cm × 6 cm. The equation also had a third root, 16, which the algebra finds but the card can’t make: a root of the equation is an answer to the real problem only if it makes sense there.

Questions 1 and 2

Question 3

Factorize.

Question 4

Solve.

In the SEE’s Optional Mathematics paper (75 marks, 3 hours), Polynomial is one of the units of Algebra (Units 1 to 6 of this book). CDC’s specification grid gives Algebra 17 marks, in three kinds of question:

Group	Kind of question	Algebra’s marks	Levels
A	multiple choice, 1 mark each	2	knowledge 1, application 1
B	within the content area: questions on Algebra alone, each of 2 to 4 parts	10	knowledge 2, understanding 2, application 4, higher ability 2
C	across content areas: one question joining Algebra with another area	5	understanding 2, application 2, higher ability 1

In CDC’s model paper, polynomials have a group B question of their own, worth 2 marks: given f(x) = x − 1 and the cubic $p(x) = x^3 - 6x^2 + 11x - 6$ (Exercise 2.3, question 1), write the condition for f(x) to be a factor of p(x) (1 mark, knowledge: p(1) = 0), and use the rational root theorem for the possible roots of p(x) (1 mark, application: ±1, ±2, ±3, ±6). The same paper’s Limits and Continuity question uses the fact that every polynomial is continuous.

What each kind asks, and where its marks are:

• Statements and conditions: the remainder or factor theorem in words, or “the condition for x − a to be a factor”: f(a) = 0. One mark each; a proof is two or three lines.
• Remainders and missing constants: what makes the divisor zero, the value put in, the equation in k solved. A mark for setting up f(a) = R, a mark for the answer.
• Possible roots: the list ± factors of the constant over factors of the leading coefficient, written out in full.
• Factorize or solve a cubic: a root found (shown by f(a) = 0), synthetic division, the quadratic factorized, the answer. Each is usually worth a mark.

Here is a question of the longer kind, written for this book, with how it is marked.