Mathematics 1 Week 4 Notes: Polynomial Functions, Division Algorithm and Zeros

01

Polynomial functions

Polynomial Functions

The layman's view

A polynomial is a mathematical expression that consists of numerous mathematical terms added together.

The mathematician's view

A polynomial is an algebraic expression in which the only arithmetic is addition, subtraction, multiplication and “natural” exponents of the variables.

The word polynomial comes from polynomen: poly means many and nomen means name. In this chapter we deal only with polynomials in a single variable.

Definition: polynomial functionA polynomial function

f(x)

of degree $n$ is

f(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_0 x^0, \qquad a_n \neq 0,\ n \in \mathbb{N}

Figure 1. The general form of a polynomial function: terms with decreasing powers of x, the first coefficient not zero

Reading the diagram

Each box is one term: a number (the coefficient) times a power of $x$ . The highest power, $n$ , is the degree, and its coefficient $a_n$ must not be zero, otherwise that term would vanish and the degree would be smaller. The last term has power 0, because $x^0 = 1$ .

Domain: all real numbers

A polynomial function can be treated as a function from $\mathbb{R}$ to $\mathbb{R}$ , so its domain is $\mathbb{R}$ .

Range: depends on f(x)

The range is not the same for every polynomial; it depends on the particular function $f(x)$ .

1.1Reading the definition: what is and is not a polynomial

Only addition, subtraction, multiplication and natural-number exponents are allowed. This decides what is and is not a polynomial.

Expression	Polynomial?	Why
$5x^3 + x^2 + x$	Yes	Natural exponents 3, 2, 1 combined by addition
$2x + 4$	Yes	Exponents 1 and 0
$7$	Yes	A constant: $7x^0$
$1/x$	No	Division by the variable means the exponent $-1$ , which is not natural
$\sqrt{x}$	No	It is $x^{1/2}$ , and $1/2$ is not a natural exponent

The slides have a heading “Example: Identification of a polynomial” but the example itself is blank in the PDF. The table above only applies the definition given on the slides.

1.2Going deeper

Coefficients can be any real numbers (including fractions and negatives); only the exponents must be natural numbers (extra remark).
A polynomial function gives one output per input. Put in $x = 2$ in $5x^3 + x^2 + x$ : $40 + 4 + 2 = 46$ .
Why the first coefficient cannot be 0. If $a_n = 0$ the term vanishes and the polynomial has a smaller degree, as with $a = 0$ in the quadratic of Week 3.

Exam tips

Check each exponent: whole number, non-negative, and no variable in a denominator or under a root.

Common mistakes

Calling $1/x$ or $\sqrt{x}$ a polynomial.
Forgetting that a constant such as 7 is a polynomial.

Next · Polynomial functionsClassification of Polynomials

02

Polynomial functions

Classification of Polynomials

Polynomials can be sorted in three ways: by the number of variables, by the degree, and by the number of terms.

2.1Based on the number of variables

Type of polynomial	Example
Polynomial in one variable	$5x^3 + x^2 + x$
Polynomial in two variables	$5x^3y + x^2 + xy$
Polynomial in more than two variables	$5x^3yz + x^2zy + xy$

2.2Based on the degree

Degree of a variable in one term

The exponent on the variable in a term. In $x^5y^4$ the degree of $x$ is 5.

Degree of a term

Add the degrees of all the variables in the term. The degree of $x^5y^4$ is $5 + 4 = 9$ .

Degree of the polynomial

The highest degree of any term with a non-zero coefficient. For $f(x) = x^3 + x^{10}$ the degree is 10.

The zero polynomial

The degree of the zero polynomial is undefined.

Example 1: degrees in x³ + x⁵y⁴

For the polynomial in two variables $x^3 + x^5y^4$ , find the degree of each term and of the polynomial.

Term $x^3$ : the only variable is $x$ , with degree 3, so the term has degree 3.
Term $x^5y^4$ : the degrees are 5 and 4, so the term has degree $5 + 4 = 9$ .
Polynomial: the highest term degree is 9, so the polynomial has degree 9.

The lecture gives the degrees of the term $x^5y^4$ ; the degree of the whole expression follows from the rule stated above. Polynomials are named by their degree:

Degree	Name	Examples
0	Constant polynomial	$5, 6, e, \pi$
1	Linear polynomial	$2x + 4,\ -3x$
2	Quadratic polynomial	$2x^2 + 4,\ -3x^2$
3	Cubic polynomial	$2x^3 + 4,\ -3x^3$
4	Quartic polynomial	$2x^4 + 4,\ -3x^4$

Link to earlier weeksLinear polynomials are the straight lines of Week 2 and quadratic polynomials are the parabolas of Week 3. Week 4 studies polynomials of any degree.

2.3Based on the number of terms

Name	Explanation	Examples
Monomial	Polynomial with one term	$x^{10},\ 6x^9,\ e,\ 5 + 4$
Binomial	Polynomial with two terms	$x^{10} + 10,\ 6x^9 - x^2,\ x^2 + 3x + x$
Trinomial	Polynomial with three terms	$x^{10} + x^2 + 10,\ 6x^9 - x^2 + 2,\ x^2 + x + e$

Like terms are combined before counting: $5 + 4$ is the single term 9 (a monomial) and $x^2 + 3x + x$ is $x^2 + 4x$ (a binomial). These are the lecture's own examples; the remark explains why they sit in those rows.

2.4Going deeper

For one variable, the degree is just the largest exponent with a non-zero coefficient: $3x^4 - 2x + 7$ has degree 4.
Constants other than 0 have degree 0 ( $7 = 7x^0$ ); only the zero polynomial has no degree.
Names combine. $x^3 + x$ is a cubic binomial in one variable.

Exam tips

Combine like terms first, then count terms and find the degree.

Common mistakes

Taking the degree of $x^5y^4$ as 5 instead of 9.
Counting $x^2 + 3x + x$ as three terms.

Next · Algebra of polynomialsAddition and Subtraction

03

Algebra of polynomials

Addition and Subtraction

3.1Addition

When two polynomials are added, the like terms are combined: terms with the same variable and the same exponent.

Addition algorithmLet

p(x) = \sum_{k=0}^{n} a_k x^k

and

q(x) = \sum_{j=0}^{m} b_j x^j

. Then

p(x) + q(x) = \sum_{k=0}^{m \vee n} (a_k + b_k)\, x^k

Here $m \vee n$ denotes whichever of $m$ and $n$ is the maximum. If one polynomial is shorter, its missing coefficients are taken as 0.

Example 2: sum of two general cubics

Both have degree 3, so $m \vee n = 3$ . Expand the sum over $k = 0, 1, 2, 3$ :

(a_3 + b_3)x^3 + (a_2 + b_2)x^2 + (a_1 + b_1)x + (a_0 + b_0)

Example 3: sum of two numerical cubics

Find the sum of $p(x) = x^3 + 3x^2 + 5x - 10$ and $q(x) = 3x^3 + 5x^2 - 6x - 20$ .

List the coefficients by power $a_3, a_2, a_1, a_0 = 1, 3, 5, -10$ and $b_3, b_2, b_1, b_0 = 3, 5, -6, -20$ .
Add the coefficients of each power.

	x³	x²	x	x⁰
p(x)	1	3	5	−10
q(x)	3	5	−6	−20
p + q	1 + 3 = 4	3 + 5 = 8	5 + (−6) = −1	−10 + (−20) = −30

Answer

$4x^3 + 8x^2 - x - 30$

Check by substitutionPut

x = 1

into everything:

p(1) = 1 + 3 + 5 - 10 = -1

,

q(1) = 3 + 5 - 6 - 20 = -18

, and the sum at

x = 1

is

4 + 8 - 1 - 30 = -19 = -1 + (-18)

. A sum of polynomials always has the sum of their values. This check is added; the slides only show the addition algorithm.

3.2Subtraction

Subtraction algorithmWith

p

and

q

as above,

p(x) - q(x) = \sum_{k=0}^{m \vee n} (a_k - b_k)\, x^k

. It works exactly like addition: subtract the coefficients of like terms.

Example 4: subtract −3x² + 2x − 2 from x³ + 2x² + 8x

Subtract $q(x) = -3x^2 + 2x - 2$ from $p(x) = x^3 + 2x^2 + 8x$ .

Fill in the missing coefficients with zero so that both polynomials have the same powers: $p$ has no constant term, $q$ has no $x^3$ term.

	x³	x²	x	x⁰
p(x)	1	2	8	0
q(x)	0	−3	2	−2

Subtract power by power.

	x³	x²	x	x⁰
p − q	1 − 0 = 1	2 − (−3) = 5	8 − 2 = 6	0 − (−2) = 2

Answer

$p(x) - q(x) = x^3 + 5x^2 + 6x + 2$

Common slipSubtracting a negative coefficient adds it:

2 - (-3) = 5

, and

0 - (-2) = +2

. Write the missing coefficients as 0 first so that nothing is left out.

3.3Going deeper

The degree of a sum can drop. $(x^3 + x) + (-x^3 + 2)$ has degree 1, because the $x^3$ terms cancel (extra remark).
The degree of a sum is at most the larger degree, so polynomials stay polynomials under addition and subtraction.
Vertical layout. Line up the powers in columns, as in the tables above, and work column by column.

Exam tips

Write the powers in descending order and fill every gap with 0.
Check the answer at $x = 1$ : the coefficients add to the values.

Common mistakes

Forgetting to change every sign of the second polynomial when subtracting.
Adding unlike terms such as $x^2$ and $x$ .

Next · Algebra of polynomialsMultiplication

04

Algebra of polynomials

Multiplication

Multiplication algorithmLet

p(x) = \sum_{k=0}^{n} a_k x^k

and

q(x) = \sum_{j=0}^{m} b_j x^j

. Then

p(x)\, q(x) = \sum_{k=0}^{m+n} \left( \sum_{j=0}^{k} a_j\, b_{k-j} \right) x^k

The coefficient of $x^k$ is the sum of all products $a_j b_{k-j}$ whose subscripts add up to $k$ . Coefficients beyond the degree ( $a_j = 0$ for $j > n$ and $b_j = 0$ for $j > m$ ) count as 0.

A discrepancy in the slideThe slides print the inner term as

(a_j - b_{k-j})

with a minus sign, but their own table of calculations uses the products

a_j b_{k-j}

. The formula above follows the table.

Why the subscripts add to k

A term $a_j x^j$ times a term $b_i x^i$ gives $a_j b_i x^{j+i}$ . So the product has an $x^k$ term exactly when $j + i = k$ , that is $i = k - j$ . This reason is added to explain the formula; the slides state only the formula.

Example 5: multiply x² + x + 1 by x² + 2x + 1

List the coefficients (the degree is 2 for both, so $k$ runs from 0 to $2 + 2 = 4$ , and all coefficients beyond $k = 2$ are 0):

k	0	1	2
a_k	1	1	1
b_k	1	2	1

For each k add all products with subscripts adding to $k$ :

k	Coefficient	Calculation
0	a₀b₀	1
1	a₁b₀ + a₀b₁	1 + 2 = 3
2	a₀b₂ + a₁b₁ + a₂b₀	1 + 2 + 1 = 4
3	a₀b₃ + a₁b₂ + a₂b₁ + a₃b₀	0 + 1 + 2 + 0 = 3
4	a₀b₄ + a₁b₃ + a₂b₂ + a₃b₁ + a₄b₀	0 + 0 + 1 + 0 + 0 = 1

Write the answer. The coefficients 1, 3, 4, 3, 1 are those of $x^0, x^1, x^2, x^3, x^4$ :

Answer

$(x^2 + x + 1)(x^2 + 2x + 1) = x^4 + 3x^3 + 4x^2 + 3x + 1$

Check by substitutionAt

x = 1

:

p(1) = 3

and

q(1) = 4

, so the product is 12; the coefficients

1 + 3 + 4 + 3 + 1

also add to 12. The check at

x = 1

is added; the slides do not include it.

4.1Going deeper

The degree of a product is $m + n$ . Here $2 + 2 = 4$ . The leading coefficient is $a_n b_m$ (extra remark).
It is just distributing. The formula is the rule “multiply every term of one polynomial by every term of the other, then combine like terms” written with subscripts.
Check by counting. A product of a 3-term and a 3-term polynomial has $3 \times 3 = 9$ products before combining.

Exam tips

Make a table of $a_k$ and $b_k$ and work down $k = 0, 1, 2, \ldots, m + n$ .
Check with $x = 1$ : the product of the values equals the sum of the coefficients.

Common mistakes

Multiplying coefficients only on the same power (that is addition's rule).
Missing a cross term such as $a_1 b_0$ .

Next · Algebra of polynomialsDivision and the Division Algorithm

05

Algebra of polynomials

Division and the Division Algorithm

Definition: polynomial divisionThe division of a polynomial by a monomial, binomial or another polynomial, using various methods, is known as polynomial division. One method is the division algorithm (long division) below.

5.1Terminology

Figure 2. The four parts of a division, shown for p(x) = x⁴ + 2x² + 3x + 2 divided by q(x) = x² + x + 1

Reading the diagram

Dividing $p(x)$ (the dividend) by $q(x)$ (the divisor) gives a quotient $h(x)$ and a remainder $r(x)$ . Written as a fraction, $\frac{p(x)}{q(x)} = x^2 - x + 2 + \frac{2x}{q(x)}$ . Multiplying by $q(x)$ gives $p(x) = q(x)h(x) + r(x)$ .

5.2The division algorithm

Arrange the terms in descending order of the degree and add the missing exponents with 0 as coefficient.
Divide the first term of the dividend by the first term of the divisor to get a monomial.
Multiply the monomial with the divisor and subtract the result from the dividend.
Check whether the resulting polynomial has degree less than the divisor. If true, write it as the remainder; otherwise go back to Step 2.

Why Step 1 mattersMissing powers are written with coefficient 0 so that every column lines up. For example

x^4 + 2x^2 + 3x + 2

is arranged as

x^4 + 0x^3 + 2x^2 + 3x + 2

.

5.3Worked example: x⁴ + 2x² + 3x + 2 divided by x² + x + 1

Example 6: long division, step by step

Figure 3. The same division laid out as a long division: the quotient on top, each product being subtracted under the line above it, and the remainder at the bottom

Arrange. $p(x) = x^4 + 0x^3 + 2x^2 + 3x + 2$ , $q(x) = x^2 + x + 1$ .
Round 1. $x^4 \div x^2 = x^2$ . Multiply: $x^2(x^2 + x + 1) = x^4 + x^3 + x^2$ . Subtract: $-x^3 + x^2 + 3x + 2$ .
Round 2. $-x^3 \div x^2 = -x$ . Multiply: $-x(x^2 + x + 1) = -x^3 - x^2 - x$ . Subtract: $2x^2 + 4x + 2$ .
Round 3. $2x^2 \div x^2 = 2$ . Multiply: $2(x^2 + x + 1) = 2x^2 + 2x + 2$ . Subtract: $2x$ .
Stop. The remainder $2x$ has degree 1, which is less than the degree 2 of the divisor.

x^4 + 2x^2 + 3x + 2 = (x^2 + x + 1)(x^2 - x + 2) + 2x

Answer

Quotient $x^2 - x + 2$ , remainder $2x$ .

CheckMultiply back:

(x^2 + x + 1)(x^2 - x + 2) = x^4 + 0x^3 + x^2 + x + 2

, and adding the remainder

2x

gives

x^4 + 2x^2 + 3x + 2

, the dividend. The multiplying-back check is added; the slides show the long division and the quotient and remainder only.

5.4Lecture examples

Example 7: quotient of x⁵ − 4x⁴ + 3x³ + 3x² − 4 by (x − 2)²

If $p(x) = x^5 - 4x^4 + 3x^3 + 3x^2 - 4$ and $q(x) = (x - 2)^2$ , then $\frac{p(x)}{q(x)}$ is

(a) $x^3 + x + 1$
(b) $x^4 - x^2 - 1$
(c) $x^3 - x - 1$
(d) $x^4 - x^2 + 1$

Expand the divisor: $q(x) = (x - 2)^2 = x^2 - 4x + 4$ . Arrange the dividend with the missing $x$ term: $x^5 - 4x^4 + 3x^3 + 3x^2 + 0x - 4$ .
Round 1. $x^5 \div x^2 = x^3$ ; $x^3(x^2 - 4x + 4) = x^5 - 4x^4 + 4x^3$ . Subtract: $-x^3 + 3x^2 + 0x - 4$ .
Round 2. $-x^3 \div x^2 = -x$ ; $-x(x^2 - 4x + 4) = -x^3 + 4x^2 - 4x$ . Subtract: $-x^2 + 4x - 4$ .
Round 3. $-x^2 \div x^2 = -1$ ; $-1(x^2 - 4x + 4) = -x^2 + 4x - 4$ . Subtract: $0$ .

Figure 4. Long division of the degree 5 polynomial: the remainder is 0

Answer

The remainder is 0, so the division is exact and the quotient is $x^3 - x - 1$ : option (c).

Why not the othersThe degree of the quotient is the degree of

p

minus the degree of

q

,

5 - 2 = 3

, so (b) and (d), which have degree 4, cannot be right. Between (a) and (c): round 1 leaves

-x^3 + 3x^2 - 4

, so the next quotient term is

-x^3 \div x^2 = -x

, negative; option (a) has

+x

, so it is ruled out.

Example 8: what should be subtracted from 6x⁴ + 5x³ + 4x − 4 to make it divisible by 2x² + x − 1?

(a) $4x$
(b) $4x - 3$
(c) $6x - 3$
(d) $2x - 3$

Idea

Divide $P(x)$ by the divisor. If the remainder is subtracted from $P(x)$ , what is left has no remainder, so it is divisible.

Arrange. $6x^4 + 5x^3 + 0x^2 + 4x - 4$ divided by $2x^2 + x - 1$ .
Round 1. $6x^4 \div 2x^2 = 3x^2$ ; $3x^2(2x^2 + x - 1) = 6x^4 + 3x^3 - 3x^2$ . Subtract: $2x^3 + 3x^2 + 4x - 4$ .
Round 2. $2x^3 \div 2x^2 = x$ ; $x(2x^2 + x - 1) = 2x^3 + x^2 - x$ . Subtract: $2x^2 + 5x - 4$ .
Round 3. $2x^2 \div 2x^2 = 1$ ; $1(2x^2 + x - 1) = 2x^2 + x - 1$ . Subtract: $4x - 3$ .

Figure 5. Long division of the degree 4 polynomial: quotient 3x² + x + 1 and remainder 4x − 3

Answer

The remainder is $4x - 3$ (degree 1, less than 2). Hence $4x - 3$ should be subtracted: option (b).

Check: $(6x^4 + 5x^3 + 4x - 4) - (4x - 3) = 6x^4 + 5x^3 - 1 = (2x^2 + x - 1)(3x^2 + x + 1)$ . The last line is added as a check of the answer; the slides stop at the remainder.

Example 9: degrees of the quotient and the remainder

Dividing $p(x)$ by $q(x)$ gives quotient $h(x)$ and remainder $r(x)$ . (a) What is the maximum degree of $r(x)$ ? (b) If $\deg p(x) < \deg q(x)$ , which statements are correct: $h(x) = 0$ ; $\deg h(x) = \deg q(x)$ ; $\deg r(x) = \deg q(x)$ ; $\deg r(x) = \deg p(x)$ ?

(a) The division stops only when the remainder has degree strictly less than the divisor. So the maximum degree of $r(x)$ is $\deg q(x) - 1$ .
(b) If $p$ has smaller degree than $q$ , no division round can even begin, so the quotient is the zero polynomial, $h(x) = 0$ , and the remainder is $p(x)$ itself, so $\deg r(x) = \deg p(x)$ . The correct statements are $h(x) = 0$ and $\deg r(x) = \deg p(x)$ .

A note on the slideThe slides write “

\deg h(x) = 0

” for the zero quotient, although elsewhere they say the degree of the zero polynomial is undefined. The meaning is simply that the quotient is the zero polynomial.

Remainder

$\deg r(x) < \deg q(x)$ , so at most $\deg q(x) - 1$ .

Small dividend

$\deg p < \deg q$ : quotient $h(x) = 0$ and remainder $r(x) = p(x)$ .

Example 10: find a so that two polynomials leave the same remainder

The polynomials $x^3 + ax^2 + 5x + 7$ and $x^3 + 2x^2 + 3x + 2a$ leave the same remainder when divided by $x - 2$ . Find $a$ .

Idea

The remainder on division by $x - 2$ is the value of the polynomial at $x = 2$ . So both polynomials must have the same value at $x = 2$ .

First polynomial at $x = 2$ : $8 + 4a + 10 + 7 = 4a + 25$ .
Second polynomial at $x = 2$ : $8 + 8 + 6 + 2a = 2a + 22$ .
Equate them:

4a + 25 = 2a + 22 \;\Rightarrow\; 2a = -3 \;\Rightarrow\; a = -\tfrac{3}{2}

Check. With $a = -\frac{3}{2}$ both values equal $4a + 25 = 19$ and $2a + 22 = 19$ . The slides use this substitution idea directly (without naming it) and simply state that both polynomials must have the same value at $x = 2$ .

5.5Going deeper

Same as number long division. $17 = 5 \times 3 + 2$ has the same shape as $p = q\,h + r$ , with the remainder smaller than the divisor.
Dividing by $x - a$ . The remainder is the constant $p(a)$ : that is the idea behind Example 10 (the remainder theorem, an extra name not used on the slides).
Zero remainder means divisible. In Example 7 the remainder is 0, so $q$ is a factor of $p$ .
Keep columns aligned. Write each product under the matching powers, as in the long-division figures.

Exam tips

After each subtraction, drop the leading term (it is zero) and bring down the rest.
Check with $p = q\,h + r$ .

Common mistakes

Forgetting the 0 coefficient for a missing power.
Not changing every sign when subtracting the product.
Stopping too early, or too late: the remainder's degree must be below the divisor's.

Next · Graphs and zerosCharacteristics of Polynomial Graphs

06

Graphs and zeros

Characteristics of Polynomial Graphs

Smooth: no sharp corners

Polynomial functions of second degree or higher have graphs that do not have sharp corners; they are smooth curves.

Continuous: no breaks

Polynomial functions have graphs that have no breaks. Curves with no breaks are called continuous.

6.1Which graphs are polynomial functions?

Example 11: spot the polynomial graphs

Which of the graphs below represent polynomial functions?

Figure 6. The four graphs of the question, redrawn: (a) a smooth curve, (b) a straight line, (c) a V with a sharp corner at x = 0, (d) a curve with a break

(a) A smooth curve with no break: it represents a polynomial function.
(b) A straight line, which is smooth and continuous: it represents a polynomial function.
(c) It has a sharp corner at $x = 0$ , so it is not a polynomial function.
(d) It is not continuous (the curve is broken), so it is not a polynomial function.

Answer

The graphs in options (a) and (b) represent polynomial functions.

A note on the figureThe four graphs are redrawn from the lecture's picture, so the exact shapes are approximate; what matters is the smoothness and continuity of each.

Option	Smooth?	Continuous?	Polynomial?
(a)	Yes	Yes	Yes
(b)	Yes	Yes	Yes
(c)	No: sharp corner at x = 0	Yes	No
(d)	Yes	No: broken curve	No

6.2Going deeper

A pencil test. If you can trace the whole graph without lifting the pencil (continuous) and without a sharp turn (smooth), it could be a polynomial.
These are necessary, not sufficient. Every polynomial graph is smooth and continuous, but a smooth continuous graph such as $y = 2^x$ is not a polynomial (extra remark).
The V is $|x|$ . The absolute value has a sharp corner at 0, so it is not a polynomial function (extra remark).

Exam tips

Look for corners and breaks first.

Common mistakes

Calling a straight line not a polynomial (it is a degree 1 polynomial).

Next · Graphs and zerosZeros of a Polynomial Function

07

Graphs and zeros

Zeros of a Polynomial Function

Definition: zerosIf

f

is a polynomial function, the values of

x

for which

f(x) = 0

are called the zeros of

f

.

Factor method: set each factor to zero

If the equation of the polynomial function can be factored, set each factor equal to zero and solve for the zeros.

A zero gives a factor: x = a gives (x − a)

Any value $x = a$ that is a zero of a polynomial function yields a factor of the polynomial of the form $(x - a)$ .

x-intercepts

At an x-intercept the output value is zero. So to find the x-intercepts of the graph, find the input values whose output is zero: the zeros of $f$ .

7.1Illustrations

Example 12: zeros and factors of two polynomials

Find the zeros of $f(x) = x^2 + 2x - 24 = (x - 4)(x + 6)$ and of $g(x) = (x + 2)(x - 1)(x - 3)$ , and show them on the graph.

f: set each factor to zero: $x - 4 = 0$ gives $x = 4$ , and $x + 6 = 0$ gives $x = -6$ .
g: $x + 2 = 0$ , $x - 1 = 0$ , $x - 3 = 0$ give $x = -2, 1$ and $3$ .

Figure 7. The zeros are the x-intercepts: where each graph meets the x-axis. Each zero a matches a factor (x − a). Left: f(x) = x² + 2x − 24. Right: g(x) = (x + 2)(x − 1)(x − 3)

Polynomial	Zeros	Factors
$x^2 + 2x - 24$	4, −6	$(x - 4)(x + 6)$
$x^3 - 2x^2 - 5x + 6$	−2, 1, 3	$(x + 2)(x - 1)(x - 3)$

Check

g(x) = x^3 - 2x^2 - 5x + 6

since

(x + 2)(x - 1) = x^2 + x - 2

and

(x^2 + x - 2)(x - 3) = x^3 - 2x^2 - 5x + 6

. And

g(-2) = g(1) = g(3) = 0

. These two examples are illustrations of the facts above; the slides state the facts without an example. The first polynomial is the one of Week 3, whose roots were found by factoring.

7.2Finding zeros in general

Quadratic: degree 2

Use the quadratic formula (Week 3).

Cubic and quartic: degrees 3 and 4

Formulas exist, but they are not simple enough to remember.

Higher degrees: general case

Formulas do not exist for general higher-degree polynomials.

For general polynomials, finding the zeros can be a challenging problem. Factoring is the main tool available when the polynomial factors.

7.3Going deeper

The number of zeros. A polynomial of degree $n$ has at most $n$ zeros: $f$ (degree 2) has 2, $g$ (degree 3) has 3 (extra remark).
Zero to factor and back. From the zero $x = -6$ you write the factor $(x + 6)$ : the sign flips.
The graph and the zeros. A graph can also miss the x-axis completely, as $x^2 + 2x + 5$ does: no real zeros (Week 3, discriminant $-16$ ).

Exam tips

Check each zero by substituting it into $f$ .

Common mistakes

Writing the factor $(x - 6)$ for the zero $x = -6$ .

Next · All topicsWeek 4 Summary Sheet

08

All topics

Week 4 Summary Sheet

Everything in Week 4 on one sheet, for a last-minute revision.

Key definitions

Polynomial function

f(x) = a_n x^n + \cdots + a_0 x^0,\quad a_n \neq 0,\ n \in \mathbb{N}

Degree

Highest degree of any term with non-zero coefficient. Zero polynomial: undefined.

Sum and difference

\sum_{k=0}^{m \vee n} (a_k \pm b_k)\, x^k

Product

\sum_{k=0}^{m+n} \left( \sum_{j=0}^{k} a_j b_{k-j} \right) x^k

Names at a glance

By degree	By terms	By variables
0 constant, 1 linear, 2 quadratic	monomial: 1 term	one variable
3 cubic, 4 quartic	binomial: 2 terms	two variables
	trinomial: 3 terms	more than two variables

Division and zeros

Idea	Statement
Division	$p = q \cdot h + r$ , $\deg r < \deg q$
Small dividend	$\deg p < \deg q$ gives $h = 0$ and $r = p$
Zero	$f(a) = 0$ means $(x - a)$ is a factor and $a$ is an x-intercept
Graph	smooth and continuous: no sharp corners, no breaks

Values computed in the notes

Operation	Result
$(x^3 + 3x^2 + 5x - 10) + (3x^3 + 5x^2 - 6x - 20)$	$4x^3 + 8x^2 - x - 30$
$(x^3 + 2x^2 + 8x) - (-3x^2 + 2x - 2)$	$x^3 + 5x^2 + 6x + 2$
$(x^2 + x + 1)(x^2 + 2x + 1)$	$x^4 + 3x^3 + 4x^2 + 3x + 1$
$(x^4 + 2x^2 + 3x + 2) \div (x^2 + x + 1)$	quotient $x^2 - x + 2$ , remainder $2x$
$(x^5 - 4x^4 + 3x^3 + 3x^2 - 4) \div (x - 2)^2$	quotient $x^3 - x - 1$ , remainder 0
$(6x^4 + 5x^3 + 4x - 4) \div (2x^2 + x - 1)$	quotient $3x^2 + x + 1$ , remainder $4x - 3$
Equal remainders on division by $(x - 2)$	$a = -\frac{3}{2}$

Revision overview

Definitionf(x) = aₙxⁿ + … + a₀x⁰
aₙ ≠ 0, n a natural number
Domain R, graph smooth and continuous

Classificationby variables, degree, number of terms
degree 0 to 4: constant to quartic

Add and subtractcombine like terms:
(aₖ ± bₖ) xᵏ

Polynomial functionsMathematics for Data Science 1

Multiplycoefficient of xᵏ is
a₀bₖ + a₁bₖ₋₁ + … + aₖb₀

Dividedivision algorithm, 4 steps
remainder degree < divisor degree

Zerosf(x) = 0, the x-intercepts
zero a gives factor (x − a)
no simple formula beyond quadratics

Revision: the whole of Week 4 on one map

Back to top

Mathematics forData Science 1

Chapters

Polynomial Functions

1.1Reading the definition: what is and is not a polynomial

1.2Going deeper

Classification of Polynomials

2.1Based on the number of variables

2.2Based on the degree

Example 1: degrees in x³ + x⁵y⁴

2.3Based on the number of terms

2.4Going deeper

Addition and Subtraction

3.1Addition

Example 2: sum of two general cubics

Example 3: sum of two numerical cubics

3.2Subtraction

Example 4: subtract −3x² + 2x − 2 from x³ + 2x² + 8x

3.3Going deeper

Multiplication

Example 5: multiply x² + x + 1 by x² + 2x + 1

4.1Going deeper

Division and the Division Algorithm

5.1Terminology

5.2The division algorithm

5.3Worked example: x⁴ + 2x² + 3x + 2 divided by x² + x + 1

Example 6: long division, step by step

5.4Lecture examples

Example 7: quotient of x⁵ − 4x⁴ + 3x³ + 3x² − 4 by (x − 2)²

Example 8: what should be subtracted from 6x⁴ + 5x³ + 4x − 4 to make it divisible by 2x² + x − 1?

Example 9: degrees of the quotient and the remainder

Example 10: find a so that two polynomials leave the same remainder

5.5Going deeper

Characteristics of Polynomial Graphs

6.1Which graphs are polynomial functions?

Example 11: spot the polynomial graphs

6.2Going deeper

Zeros of a Polynomial Function

7.1Illustrations

Example 12: zeros and factors of two polynomials

7.2Finding zeros in general

7.3Going deeper

Week 4 Summary Sheet

Key definitions

Names at a glance

Division and zeros

Values computed in the notes

Mathematics for
Data Science 1