HL · Further functions 1 · Polynomial functions: graphs, zeros, factor & remainder theorems

Module 17 of 46Lesson 1 of 9 · HL · Further functions 1 · Polynomial functions: graphs, zeros, factor & remainder theorems

Polynomials, rational functions, symmetry, inequalities & modulus · Essential understanding 1

Polynomial functions: graphs, zeros, factor & remainder theorems

IB AA AHL 2.12with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: A polynomial is a sum of powers of

x

with whole-number powers, such as

2x^3 - 5x^2 + x + 7

. Its zeros (the

x

-values where the graph meets the

x

-axis) tell you its factors, and its factors tell you its zeros. The remainder theorem and the factor theorem let you test a factor in seconds, with no long division, and they are the key to every cubic or quartic on Paper 1.

By the end of this lesson you can:

sketch a polynomial from its factorised form, showing zeros, the $y$ -intercept and the end behaviour
explain what a repeated zero does to the graph (touch or flat crossing)
use the remainder theorem to find the remainder on dividing by $(x - a)$ or $(px - q)$ without dividing
use the factor theorem to find a factor, then factorise a cubic completely and solve $f(x) = 0$
find unknown coefficients from a factor or a remainder, and build a polynomial from its graph

🧑🏾 Imagine this…

Onalenna is cutting a wooden board with a saw that always cuts at a whole number of centimetres. If a board is exactly

6

cm wide, the cuts at

1

2

3

and

6

divide it with nothing left over. A cut at

4

leaves a piece of

2

cm over. Polynomials work the same way:

x - 3

divides

f(x)

with nothing left over exactly when

f(3) = 0

. The “left over” is the remainder, and the remainder theorem says you can find it just by putting

x = 3

into

f

Polynomial

a function

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

with real coefficients and whole-number powers. Examples:

x^3 - 4x

5 - x^2

. Not polynomials:

\dfrac{1}{x}

\sqrt{x}

2^x

Degree

the highest power of

x

. A cubic has degree 3 and a quartic degree 4.

a_n

is the leading coefficient and

a_0

is the constant term (the

y

-intercept).

Zero (root)

a number

a

with

f(a) = 0

. The graph meets the

x

-axis at

(a, 0)

. “Zero” is used for the function, “root” for the equation

f(x) = 0

Factor

(x - a)

is a factor of

f(x)

when

f(x) = (x - a)\,Q(x)

for some polynomial

Q(x)

, with no remainder.

Multiplicity

(x - a)^m

is the highest power of

(x - a)

dividing

f(x)

, then

a

is a zero of multiplicity

m

(x-2)^2

gives a double zero.

Remainder

the constant left over when

f(x)

is divided by

(x - a)

f(x) = (x - a)\,Q(x) + R

★ Shape of a polynomial graph

A polynomial of degree $n$ has at most $n$ real zeros and at most $n - 1$ turning points. It is smooth and has no breaks, gaps or asymptotes. Its domain is all real $x$ .
Far left and far right (end behaviour): odd degree, the ends go in opposite directions ( $a_n > 0$ : down on the left, up on the right). Even degree, both ends go the same way ( $a_n > 0$ : both up).
Zero of multiplicity 1: the graph crosses the axis. Multiplicity 2: the graph touches the axis and turns back (a turning point on the axis). Multiplicity 3: it crosses with a flat “S” shape.
Factorised form $f(x) = k(x - a)(x - b)(x - c)$ : zeros at $a, b, c$ ; the $y$ -intercept is $f(0)$ .

★ Remainder theorem and factor theorem

Remainder theorem: when $f(x)$ is divided by $(x - a)$ , the remainder is $f(a)$ .
For a divisor $(px - q)$ , the remainder is $f\!\left(\dfrac{q}{p}\right)$ , because $px - q = 0$ when $x = \dfrac{q}{p}$ .
Factor theorem: $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$ . (It is the remainder theorem with remainder 0.)
Hunting for a zero: if $f$ has integer coefficients and a whole-number zero, that zero divides the constant term $a_0$ . To factorise $x^3 + \dots - 12$ , try $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ .
After one factor is found: divide (or match coefficients) to get a quadratic, then factorise the quadratic.

⚠️ Watch out

The remainder on dividing by $(x - 3)$ is $f(3)$ , not $f(-3)$ . The sign flips: divisor $(x + 2)$ means $a = -2$ , so use $f(-2)$ .
For a divisor like $(2x - 1)$ , do not use $f(1)$ . Use $f\!\left(\dfrac{1}{2}\right)$ .
On a sketch, a double zero must touch the axis, not cross it. Marks are lost for crossing, and for forgetting the $y$ -intercept.
$f(a) = 0$ tells you $(x - a)$ is a factor. It does not tell you that there are no other zeros.

🧪 Try it yourselfZeros, multiplicity and end behaviour

This is

y = k(x - p)(x - q)(x - r)

. Start with

k = 1

p = -2

q = 1

r = 3

: three crossings. Slide

q

towards

p

until they are equal: the graph touches at the double zero. Make

k

negative and watch the ends swap. Set

k = 2

to match Example 5.

Worked examples

Example 1.1Sketching from the factorised form

Sketch

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

, showing the zeros, the

y

-intercept and the behaviour for large

|x|

🧠 Think it through

Read the zeros from the brackets: $x = -1$ (power 1, so it crosses) and $x = 2$ (power 2, so it touches).
Degree: $1 + 2 = 3$ , odd. The leading coefficient is $-1$ (negative), so the graph goes up on the far left and down on the far right.
$y$ -intercept: put $x = 0$ .

✍️ By hand

Zeros:

x = -1

(crossing) and

x = 2

(double, touching). (A1)

f(0) = -(1)(-2)^2 = -4

, so the

y

-intercept is

(0, -4)

. (A1)

Leading term

-x^3

: as

x \to -\infty

f \to +\infty

; as

x \to +\infty

f \to -\infty

. (A1)

Shape: comes down from the top left, crosses at

-1

, dips to a minimum, rises to touch the axis at

2

, then falls away. (A1)

🔢 On your TI-Nspire CX II

Graph -(x+1)*(x-2)^2 to check. The touch at x = 2 is clear.

✅ Answer Zeros −1 (crossing) and 2 (touching); y-intercept (0, −4); up on the far left, down on the far right.

The turning point between −1 and 2 is not asked for on Paper 1 unless calculus is used. Just show a sensible dip.

Example 1.2The remainder theorem

Let

f(x) = 2x^3 - 5x^2 + x + 7

. Find the remainder when

f(x)

is divided by (a)

(x - 3)

, (b)

(2x + 1)

, (c)

(x + 2)

🧠 Think it through

(a) $a = 3$ . The remainder is $f(3)$ .
(b) Solve $2x + 1 = 0$ : $x = -\dfrac{1}{2}$ . The remainder is $f\!\left(-\dfrac{1}{2}\right)$ .
(c) $x + 2 = x - (-2)$ , so $a = -2$ .

✍️ By hand
(a) 
.  (M1)(A1)
(b) 
.  (M1)(A1)
(c) 
.  (A1)

🔢 On your TI-Nspire CX II

Define the function once with Define f(x)=2x^3-5x^2+x+7, then evaluate. Type fractions with the fraction template or as f(-1/2).

✅ Answer (a) 19. (b) 5. (c) −31.

Check (a) by division: 2x³ − 5x² + x + 7 = (x − 3)(2x² + x + 4) + 19.

Example 1.3The factor theorem: factorise a cubic

Show that

(x - 3)

is a factor of

f(x) = 2x^3 - 3x^2 - 11x + 6

. Hence factorise

f(x)

completely and solve

f(x) = 0

🧠 Think it through

Factor theorem: $(x - 3)$ is a factor if $f(3) = 0$ .
Write $f(x) = (x - 3)(2x^2 + bx + c)$ and match coefficients. The $x^3$ term gives $2$ ; the constant term gives $-3c = 6$ .
Match the $x^2$ coefficient to find $b$ , then factorise the quadratic.

✍️ By hand
, so  is a factor.  (M1)(AG)
Let . Constant: .
 terms: .  (M1)(A1)
Check the  term:  ✓.
.
.  (A1)
Roots: .  (A1)

🔢 On your TI-Nspire CX II

Check the roots by typing polyRoots(2x^3-3x^2-11x+6,x) on a Calculator page.
📍 Where to find it on the TI-Nspire
polyRoots( is in menu💻 click menu on keypad▸3: Algebra▸3: Polynomial Tools▸2: Real Roots of Polynomial

✅ Answer f(3) = 0; f(x) = (x − 3)(2x − 1)(x + 2); roots 3, ½ and −2.

Always check one more coefficient (here the x term) after matching. It catches arithmetic slips, and it costs ten seconds.

Example 1.4Finding unknown coefficients

The polynomial

f(x) = x^3 + px^2 + qx - 6

has

(x - 2)

as a factor. When

f(x)

is divided by

(x + 1)

the remainder is

12

. Find

p

and

q

, and write down

f(x)

🧠 Think it through

Factor $(x - 2)$ means $f(2) = 0$ : one equation in $p$ and $q$ .
Remainder 12 on dividing by $(x + 1)$ means $f(-1) = 12$ : a second equation.
Solve the pair of linear equations.

✍️ By hand
.  (M1)(A1)
.  (M1)(A1)
Add the two equations: , so .  Then .  (A1)
Check:  ✓.
.  (A1)

🔢 On your TI-Nspire CX II

Check the pair of equations with linSolve({2p+q=-1,p-q=19},{p,q}), then test your polynomial.
📍 Where to find it on the TI-Nspire
linSolve( is in menu💻 click menu on keypad▸3: Algebra▸2: Solve System of Linear Equations…

✅ Answer p = 6, q = −13; f(x) = x³ + 6x² − 13x − 6.

Command terms: “find” needs the values; “write down f(x)” needs the polynomial with the numbers put back in.

Example 1.5Building a polynomial from its graph

A cubic graph crosses the

x

-axis at

x = -2

x = 1

and

x = 3

, and passes through

(0, 12)

. Find the equation of the cubic in the form

y = ax^3 + bx^2 + cx + d

🧠 Think it through

Zeros $-2, 1, 3$ mean factors $(x + 2)$ , $(x - 1)$ , $(x - 3)$ . There is also an unknown multiplier $k$ .
Use the point $(0, 12)$ to find $k$ .
Then expand. Multiply two brackets first, then the third.

✍️ By hand
.  (M1)
At : 
.  (A1)
.  (M1)
.  (A1)

🔢 On your TI-Nspire CX II

Graph 2x^3-4x^2-10x+12 and check the three zeros and the y-intercept.

✅ Answer y = 2(x + 2)(x − 1)(x − 3) = 2x³ − 4x² − 10x + 12.

Forgetting the multiplier k is the most common loss of marks here. Any zeros-only polynomial is a whole family of graphs; the extra point picks one.

Challenge questions

⭐ Challenge 1.6Unknown coefficients and exact rootsPaper 1 style

Let

g(x) = x^3 + ax^2 + bx - 4

. The remainder when

g(x)

is divided by

(x - 1)

3

, and

(x + 2)

is a factor of

g(x)

(a) Find $a$ and $b$ .
(b) Factorise $g(x)$ as $(x + 2)$ times a quadratic, and hence find all the zeros of $g$ exactly.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Remainder 3 on dividing by $(x - 1)$ : $g(1) = 3$ . Factor $(x + 2)$ : $g(-2) = 0$ .
For (b), match coefficients with $(x + 2)(x^2 + px + q)$ , then use the quadratic formula.

✍️ By hand
(a) .  (M1)(A1)
.  (M1)(A1)
Add: , so  and .  (A1)
(b) 
.  (M1)(A1)
Check: constant  ✓.
.  (M1)
Zeros: .  (A1)

✅ Answer (a) a = 4, b = 2. (b) g(x) = (x + 2)(x² + 2x − 2); zeros −2, −1 + √3, −1 − √3.

⭐ Challenge 1.7An open boxPaper 2 style

Kabo makes an open box from a rectangular sheet of card,

20

cm by

30

cm, by cutting a square of side

x

cm from each corner and folding up the sides.

(a) Show that the volume is $V = 4x^3 - 100x^2 + 600x$ cm $^3$ , and state the possible values of $x$ .
(b) Verify that $x = 5$ gives $V = 1000$ , and use the factor theorem to find the other value of $x$ in the possible range that gives $V = 1000$ .
(c) Use your GDC to find the largest possible volume.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

The box has length $30 - 2x$ , width $20 - 2x$ and height $x$ .
Both $30 - 2x$ and $20 - 2x$ must be positive, so the smaller one gives the limit on $x$ .
For (b) form the cubic $4x^3 - 100x^2 + 600x - 1000 = 0$ and divide by 4 first.

✍️ By hand
(a) 
.  (M1)(AG)
We need  and : .  (A1)
(b) : .
At :  ✓, so  is a factor.  (M1)
.  (M1)
, that is  or .  (A1)
Only  lies in .  (R1)

🔢 On your TI-Nspire CX II

Graph 4x^3-100x^2+600x with window x: 0 to 10, y: 0 to 1200. Menu ▸ Analyze Graph ▸ Maximum.

✅ Answer (a) V = 4x³ − 100x² + 600x, 0 < x < 10. (b) x = 5 and x = 10 − 5√2 ≈ 2.93. (c) maximum volume ≈ 1056 cm³ at x ≈ 3.92 cm.

The root 17.07 satisfies the equation but is outside the context. Always compare each solution with the domain.

Next: Lesson 2 · Sum and product of roots

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · Further functions 1 · Polynomial functions: graphs, zeros, factor & remainder theorems

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Polynomial functions: graphs, zeros, factor & remainder theorems

Worked examples

Challenge questions