Module 18 of 46Lesson 1 of 5 · HL · AA 2.12 — Polynomial functions; factor and remainder theorems; sum and product of roots

Course map · Topic 2: Functions

2.12 Polynomial functions; factor and remainder theorems; sum and product of roots

HL only12 worked questions

A polynomial is built from whole-number powers of $x$ . HL asks you to find remainders without dividing, factorise cubics and quartics using the factor theorem, build polynomials from their roots, and use the sum and product of roots without finding the roots at all.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

Let

p(x)=2x^{3}-3x^{2}+x-5

. Find the remainder when

p(x)

is divided by

(a) $x-2$ [2]
(b) $2x+1$ [2]

Show the full step-by-step solution

Step 1🧠 ThinkNo long division needed

The remainder theorem says: put in the

x

-value that makes the divisor zero. For

x-2

that is

x=2

; for

2x+1

it is

x=-\dfrac12

Part (a)

Step 1✍️ By handp(2)

p(2)=16-12+2-5

=1

Part (b)

Step 1✍️ By handp(−½)

p\!\left(-\dfrac12\right)=2\left(-\dfrac18\right)-3\left(\dfrac14\right)-\dfrac12-5

=-\dfrac14-\dfrac34-\dfrac12-5=-\dfrac{13}{2}

Final answer (a)

1

(b)

-\dfrac{13}{2}

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[5 marks]

Let

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

(a) Show that $(x+3)$ is a factor of $f(x)$ .[2]
(b) Hence factorise $f(x)$ fully and solve $f(x)=0$ .[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkFactor theorem

(x+3)

is a factor exactly when

f(-3)=0

Step 2✍️ By handEvaluate f(−3)

f(-3)=-27+18+15-6=0

M1A1

(x+3)

is a factorAG

Part (b)

Step 1✍️ By handCompare coefficients: (x + 3)(x² + bx − 2)

The

x^{2}

term must be

x^{2}

(to give

x^{3}

), and the constant must be

-2

(because

3\times(-2)=-6

x^{2}

terms:

b+3=2\Rightarrow b=-1

f(x)=(x+3)(x^{2}-x-2)

M1A1

=(x+3)(x-2)(x+1)\Rightarrow x=-3,\ -1,\ 2

✅ Check it Sum of roots

=-3-1+2=-2=-\dfrac{2}{1}

✔ (matches

-\dfrac{a_2}{a_3}

Final answer (a)

f(-3)=0

(b)

(x+3)(x-2)(x+1)

;

x=-3,-1,2

Question 3●○○ Warm-up✍️ By hand · Paper 1 · no calculator[3 marks]

Write down the sum and the product of the roots of

3x^{4}-5x^{3}+2x-7=0

Show the full step-by-step solution

Step 1🧠 ThinkRead the coefficients

n=4

a_4=3

a_3=-5

a_0=-7

. The missing

x^{2}

term does not matter here.

Step 2✍️ By handUse the formulas

Sum

=-\dfrac{a_3}{a_4}=-\dfrac{-5}{3}=\dfrac53

Product

=\dfrac{(-1)^{4}a_0}{a_4}=\dfrac{-7}{3}=-\dfrac73

M1A1

Final answer Sum

\dfrac53

; product

-\dfrac73

Question 4●●○ Standard✍️ By hand · Paper 1 · no calculator[7 marks]

Let

p(x)=x^{3}+ax^{2}+bx-6

. The polynomial has a factor

(x-1)

, and leaves a remainder of 8 when divided by

(x-2)

(a) Find $a$ and $b$ .[4]
(b) Show that $p(x)=0$ has only one real root.[3]

Show the full step-by-step solution

Step 1🧠 ThinkTwo facts → two equations

“Factor

(x-1)

” means

p(1)=0

. “Remainder 8 on division by

(x-2)

” means

p(2)=8

. Solve the two equations together.

Part (a)

Step 1✍️ By handForm and solve

p(1)=1+a+b-6=0\Rightarrow a+b=5

M1A1

p(2)=8+4a+2b-6=8\Rightarrow 2a+b=3

Subtract:

a=-2,\ b=7

Part (b)

Step 1✍️ By handFactorise and test the quadratic

p(x)=(x-1)(x^{2}-x+6)

\Delta=(-1)^{2}-4(1)(6)=-23\lt 0

so the quadratic has no real roots; the only real root is

x=1

Final answer (a)

a=-2,\ b=7

(b)

p(x)=(x-1)(x^{2}-x+6)

and

\Delta=-23\lt 0

, so only

x=1

Question 5●●○ Standard✍️ By hand · Paper 1 · no calculator[4 marks]

A cubic polynomial

f(x)

has roots

1,\ -2

and

3

, and its graph meets the

y

-axis at

(0,12)

. Find

f(x)

in the form

ax^{3}+bx^{2}+cx+d

Show the full step-by-step solution

Step 1🧠 ThinkBuild from the roots, then fix the height

Each root

r

gives a factor

(x-r)

. A number

a

in front stretches the graph without moving the roots — the

y

-intercept tells us

a

Step 2✍️ By handFind a

f(x)=a(x-1)(x+2)(x-3)

f(0)=a(-1)(2)(-3)=6a=12\Rightarrow a=2

Step 3✍️ By handExpand

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

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

f(x)=2x^{3}-4x^{2}-10x+12

Final answer

f(x)=2x^{3}-4x^{2}-10x+12

Question 6●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

The equation

2x^{2}-6x+3=0

has roots

\alpha

and

\beta

. Without solving the equation, find

(a) $\alpha+\beta$ and $\alpha\beta$ [2]
(b) $\alpha^{2}+\beta^{2}$ [2]
(c) $\dfrac1\alpha+\dfrac1\beta$ [1]

Show the full step-by-step solution

Step 1🧠 ThinkRewrite everything using the sum and the product

We know the sum and the product straight from the coefficients. Every symmetric expression can be rewritten in terms of those two.

Part (a)

Step 1✍️ By handSum and product

\alpha+\beta=-\dfrac{-6}{2}=3,\quad\alpha\beta=\dfrac32

A1A1

Part (b)

Step 1✍️ By handUse (α + β)² = α² + 2αβ + β²

\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta=9-3

=6

Part (c)

Step 1✍️ By handCommon denominator

\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{3}{\frac32}=2

Final answer (a)

3

and

\dfrac32

(b)

6

(c)

2

Question 7●●○ Standard🖩 GDC · Paper 2[4 marks]

Use your GDC to factorise

2x^{3}+3x^{2}-11x-6

fully.

Show the full step-by-step solution

Step 1🧠 ThinkRoots give factors

If the roots are

r_1,r_2,r_3

and the leading coefficient is 2, then the polynomial is

2(x-r_1)(x-r_2)(x-r_3)

. The GDC finds the roots; we write the factors.

Step 2🖩 On the GDCpolyRoots

menu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›1: Find Roots of Polynomial…›polyRoots(2x^3+3x^2−11x−6,x)›enter💻 Enter

📍 Where to find it on the TI-Nspire

polyRoots(is inmenu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›2: Real Roots of Polynomial

Exact rational roots

Step 3✍️ By handWrite the factors

2(x+3)\left(x+\dfrac12\right)(x-2)

M1A1

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

A1A1

💡 Remember Put the leading 2 into the bracket with the fraction:

2\left(x+\dfrac12\right)=2x+1

Final answer

(x+3)(2x+1)(x-2)

Question 8●●○ Standard🖩 GDC · Paper 2[6 marks]

Let

p(x)=x^{4}-2x^{3}-7x^{2}+8x+5

(a) Find the roots of $p(x)=0$ .[2]
(b) Find the coordinates of the turning points of the graph of $y=p(x)$ .[2]
(c) Find the set of values of $k$ for which $p(x)=k$ has exactly four real solutions.[2]

Show the full step-by-step solution

Part (a)

Step 1🖩 On the GDCpolyRoots

polyRoots(x^4−2x^3−7x^2+8x+5,x)›enter💻 Enter

📍 Where to find it on the TI-Nspire

polyRoots(is inmenu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›2: Real Roots of Polynomial

Step 2✍️ By handAnswers

x=-2.25,\ -0.466,\ 1.47,\ 3.25

Part (b)

Step 1🖩 On the GDCGraph p, then Minimum (twice) and Maximum

menu💻 click menu on keypad›6: Analyze Graph›2: Minimum

Step 2✍️ By handTurning points

Minima

(-1.56,\,-11)

and

(2.56,\,-11)

; maximum

(0.5,\,7.06)

A1A1

Part (c)

Step 1🧠 ThinkSlide a horizontal line y = k

The line

y=k

cuts the W-shaped curve four times only when it lies strictly between the two minima (height −11) and the maximum (height 7.0625).

Step 2✍️ By handAnswer

-11\lt k\lt 7.0625\ \left(=\dfrac{113}{16}\right)

A1A1

Final answer (a)

-2.25,\ -0.466,\ 1.47,\ 3.25

(b) min

(-1.56,-11)

(2.56,-11)

; max

(0.5,\ 7.06)

(c)

-11\lt k\lt 7.0625

Question 9●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

When the polynomial

p(x)

is divided by

(x-1)

the remainder is 4, and when it is divided by

(x+2)

the remainder is

-5

. Find the remainder when

p(x)

is divided by

(x-1)(x+2)

Show the full step-by-step solution

Step 1🧠 ThinkDividing by a quadratic leaves a linear remainder

When you divide by something of degree 2, the remainder has degree at most 1, so it looks like

ax+b

. Write

p(x)=(x-1)(x+2)q(x)+ax+b

and substitute the two easy values

x=1

and

x=-2

(they make the first part zero).

Step 2✍️ By handSet up

p(x)=(x-1)(x+2)q(x)+ax+b

p(1)=a+b=4

p(-2)=-2a+b=-5

Step 3✍️ By handSolve

Subtract:

3a=9\Rightarrow a=3,\ b=1

M1A1

Remainder

3x+1

Final answer

3x+1

Question 10●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

The roots of

x^{3}-6x^{2}+11x+k=0

form an arithmetic sequence.

(a) Show that one root is 2.[2]
(b) Find $k$ and the other two roots.[4]

Show the full step-by-step solution

Step 1🧠 ThinkArithmetic roots: a − d, a, a + d

Three numbers in an arithmetic sequence can be written

a-d,\ a,\ a+d

. Their sum is simply

3a

— and we know the sum from the coefficients.

Part (a)

Step 1✍️ By handSum of roots

(a-d)+a+(a+d)=3a=-\dfrac{-6}{1}=6

a=2

Part (b)

Step 1✍️ By hand2 is a root, so p(2) = 0

8-24+22+k=0\Rightarrow k=-6

M1A1

x^{3}-6x^{2}+11x-6=(x-2)(x^{2}-4x+3)=(x-2)(x-1)(x-3)

Roots

1,\ 2,\ 3

✅ Check it Product

=1\times2\times3=6=-\dfrac{k}{1}

✔

Final answer (a) sum

=3a=6\Rightarrow a=2

(b)

k=-6

; roots

1

and

3

Question 11●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

A box with no lid is made from a 30 cm by 20 cm sheet of card by cutting a square of side

x

cm from each corner and folding up the sides.

(a) Show that the volume is $V=4x^{3}-100x^{2}+600x$ cm³ and state the possible values of $x$ .[3]
(b) Find the values of $x$ for which the volume is 1000 cm³.[2]
(c) Find the maximum possible volume.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handLength × width × height

V=x(30-2x)(20-2x)

=x(600-100x+4x^{2})=4x^{3}-100x^{2}+600x

Need

20-2x\gt 0

0\lt x\lt 10

Part (b)

Step 1🖩 On the GDCSolve V = 1000 with polyRoots

polyRoots(4x^3−100x^2+600x−1000,x)›enter💻 Enter

📍 Where to find it on the TI-Nspire

polyRoots(is inmenu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›2: Real Roots of Polynomial

Step 2✍️ By handReject the impossible root

x=2.93

x=5

(

17.1\gt 10

is rejected)A1A1

Part (c)

Step 1🖩 On the GDCMaximum on the graph for 0 < x < 10

menu💻 click menu on keypad›6: Analyze Graph›3: Maximum

Step 2✍️ By handAnswer

V_{\max}\approx1060

cm³ (1056.31) when

x=3.92

A1A1

Final answer (a) shown,

0\lt x\lt 10

(b)

x=2.93

x=5

(c)

1060

cm³ (3 s.f.)

Question 12●●● Exam-level✍️ + 🖩 · Paper 3[14 marks]

This question investigates the family of cubic equations

x^{3}-3x+k=0

k\in\mathbb{R}

, with roots

\alpha,\ \beta,\ \gamma

(real or complex).

(a) Write down $\alpha+\beta+\gamma$ and $\alpha\beta\gamma$ in terms of $k$ , and find $\alpha\beta+\beta\gamma+\gamma\alpha$ .[3]
(b) Hence show that $\alpha^{2}+\beta^{2}+\gamma^{2}=6$ for every value of $k$ .[2]
(c) Use your GDC to find the roots when $k=1$ , and verify the result of (b).[3]
(d) The graph of $y=x^{3}-3x$ has turning points at $(-1,2)$ and $(1,-2)$ . Find the values of $k$ for which the equation has three distinct real roots.[3]
(e) Find all three roots when $k=3$ , and explain why (b) still holds even though $\alpha^{2}+\beta^{2}+\gamma^{2}$ involves complex numbers.[3]

Show the full step-by-step solution

Step 1🧠 ThinkThe plan

We use the coefficients to learn about the roots without finding them, then test our discovery on the calculator for particular values of

k

Part (a)

Step 1✍️ By handRead the coefficients (a₃ = 1, a₂ = 0, a₁ = −3, a₀ = k)

\alpha+\beta+\gamma=-\dfrac{0}{1}=0

\alpha\beta\gamma=-\dfrac{k}{1}=-k

\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{a_1}{a_3}=-3

Part (b)

Step 1✍️ By handSquare the sum

\sum\alpha^{2}=\left(\sum\alpha\right)^{2}-2\sum\alpha\beta

=0-2(-3)=6

Part (c)

Step 1🖩 On the GDCpolyRoots for k = 1, then add the squares

r:=polyRoots(x^3−3x+1,x)›enter💻 Enter›sum(r^2)›enter💻 Enter

📍 Where to find it on the TI-Nspire

polyRoots(is inmenu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›2: Real Roots of Polynomial

⌨️ store a value: type := (colon, then equals) on the keyboard

sum(is inmenu💻 click menu on keypad›6: Statistics›3: List Math›5: Sum of Elements

Store the list as r, then sum(r^2) squares each root and adds them.

Step 2✍️ By handRoots

x=-1.88,\ 0.347,\ 1.53

; sum of squares

=6

✔A1A1R1

Part (d)

Step 1✍️ By handThe line y = −k must cut between the turning points

x^{3}-3x+k=0\Leftrightarrow x^{3}-3x=-k

. The horizontal line

y=-k

cuts the cubic three times when it lies strictly between the heights

-2

and

2

-2\lt -k\lt 2

-2\lt k\lt 2

A1A1

Part (e)

Step 1🖩 On the GDCcPolyRoots shows the complex pair

menu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›3: Complex Roots of Polynomial›cPolyRoots(x^3−3x+3,x)›enter💻 Enter

Step 2✍️ By handInterpret

x=-2.10,\ 1.05\pm0.565i

The sum and product formulas hold for all roots, real or complex, so (b) still gives 6;R1

the imaginary parts cancel because the roots are a conjugate pairR1

Final answer (a)

0,\ -k,\ -3

(b)

0-2(-3)=6

(c)

-1.88,\ 0.347,\ 1.53

; squares add to 6 (d)

-2\lt k\lt 2

(e)

-2.10,\ 1.05\pm0.565i

; conjugate imaginary parts cancel

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 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.

🎯 What you must know

Degree $n$ polynomial: at most $n$ real roots and at most $n-1$ turning points. Odd degree: the ends go in opposite directions; even degree: same direction.
Remainder theorem: the remainder when $p(x)$ is divided by $(x-a)$ is $p(a)$ . Divided by $(ax-b)$ : the remainder is $p\!\left(\dfrac{b}{a}\right)$ .
Factor theorem: $(x-a)$ is a factor of $p(x)$ $\Leftrightarrow p(a)=0$ .
To factorise a cubic: find one root by trying $\pm$ factors of the constant term, then divide (long division or comparing coefficients) to get a quadratic.
Sum and product of roots of $a_nx^{n}+a_{n-1}x^{n-1}+\dots+a_0=0$ : sum $=-\dfrac{a_{n-1}}{a_n}$ , product $=\dfrac{(-1)^{n}a_0}{a_n}$ . These include complex roots.
Useful identity: $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta$ ; for three roots $\sum\alpha^{2}=\left(\sum\alpha\right)^{2}-2\sum\alpha\beta$ .
A repeated (double) root: the graph touches the $x$ -axis there. A single root: the graph crosses it.

Sum and product of the roots

\displaystyle\sum\alpha=-\dfrac{a_{n-1}}{a_n},\qquad \prod\alpha=\dfrac{(-1)^{n}a_0}{a_n}

📘 In the formula booklet

Remainder theorem

p(x)=(x-a)q(x)+p(a)

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

All real roots of a polynomial

menu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›1: Find Roots of Polynomial…›polyRoots(2x^3+3x^2−11x−6,x)›enter💻 Enter

📍 Where to find it on the TI-Nspire

polyRoots(is inmenu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›2: Real Roots of Polynomial

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 3: Algebra.
Choose 3: Polynomial Tools.
Choose 1: Find Roots of Polynomial….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.
Type polyRoots(2x^3+3x^2−11x−6,x).
Press enter.

The menu path

Exact rational roots

Worked example: Question 7 — polyRoots

Complex roots as well

menu💻 click menu on keypad›3: Algebra›3: Polynomial Tools›3: Complex Roots of Polynomial›cPolyRoots(x^3−3x+3,x)›enter💻 Enter

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 3: Algebra.
Choose 3: Polynomial Tools.
Choose 3: Complex Roots of Polynomial.
Type cPolyRoots(x^3−3x+3,x).
Press enter.

The menu path

Worked example: Question 12 — cPolyRoots shows the complex pair

Turning points from the graph

menu💻 click menu on keypad›6: Analyze Graph›2: Minimum

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 2: Minimum.
Move the pointer to the LEFT of the lowest point and press click, then to the RIGHT and press click. The minimum point and its coordinates appear.

The menu path

Worked example: Question 8 — Graph p, then Minimum (twice) and Maximum

⚠️ Where students lose marks

Using $p(-2)$ for division by $(x-2)$ . The remainder is $p(2)$ — the value that makes the divisor zero.
Sign of the sum of roots: it is $-\dfrac{a_{n-1}}{a_n}$ , with a minus.
Forgetting $(-1)^{n}$ in the product: for a cubic the product is $-\dfrac{a_0}{a_3}$ .
Missing a zero coefficient: in $3x^{4}-5x^{3}+2x-7$ there is no $x^{2}$ term — write $0x^{2}$ when you divide or use polyRoots with a list.
Giving only real roots when the question says “all roots” — check with cPolyRoots.

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · AA 2.12 — Polynomial functions; factor and remainder theorems; sum and product of roots

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

2.12 Polynomial functions; factor and remainder theorems; sum and product of roots

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks