Stop 1 · Leaving Cert Higher Level

Algebra 1

Built in the order it actually makes sense to learn it — starting from basic operations, then division and factorising, and building up to identities, patterns and functions. Each section has a short video, the notes that matter, a worked example, then five questions that build in difficulty.

Algebra — Exam Questions by Topic →
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In this module

1 · Basic Operations 2 · Algebraic Division 3 · Factorising 4 · Algebraic Fractions 5 · Solving Equations 6 · Rearranging Formulae 7 · Simultaneous Equations 8 · Binomial Expansions 9 · Identities 10 · Functions & Roots 11 · Patterns
Section 1
Basic Operations

Everything in algebra starts here. A polynomial is just a string of terms with whole-number powers, like 5x3 − 3x2 + 4x − 6. Before you can factorise or solve anything, you need to add, subtract and expand these fluently — and know the two special products that appear on every paper.

Video — Basic OperationsVideo slot, ready for your embed.
The degree is the highest power. The coefficient is the number in front of a term. The constant is the term with no x.
You can only add or subtract like terms.
Perfect square: (x ± a)2 = x2 ± 2ax + a2
Difference of two squares: (x − a)(x + a) = x2 − a2
EXAMPLE Expand and simplify  (2x − 3)(x2 + 4x − 5)
Multiply every term in the second bracket by each term in the first.
2x(x² + 4x − 5) = 2x³ + 8x² − 10x
−3(x² + 4x − 5) = −3x² − 12x + 15
Add and collect like terms:
2x³ + 5x² − 22x + 15
Line up like terms as you go — most lost marks here are just dropped signs.
Practice · 5 questions
Q1 For  5x3 − 2x2 + 7x − 4, write down the degree, the coefficient of x2, and the constant term.
Degree = 3 (highest power)
Coefficient of x² = −2 (take the sign with it)
Constant = −4
Q2 Simplify  3(2x2 − x + 4) − 2(x2 + 5x − 1)
3(2x² − x + 4) = 6x² − 3x + 12
−2(x² + 5x − 1) = −2x² − 10x + 2
Collect: 4x² − 13x + 14
The −2 flips every sign in the second bracket, including the −1 → +2.
Q3 Expand  (3x − 2)2
Use (x − a)² = x² − 2ax + a² with 3x and 2:
(3x)² − 2(3x)(2) + (2)² = 9x² − 12x + 4
(3x − 2)² is not 9x² − 4. Squaring a bracket always gives three terms.
Q4 9x2 + px + 25 is a perfect square and p > 0. Find p.
9x² = (3x)² and 25 = (5)², so it must be (3x + 5)².
(3x + 5)² = 9x² + 2(3x)(5) + 25 = 9x² + 30x + 25
Comparing: p = 30
(3x − 5)² would give p = −30, rejected since p > 0.
Q5 Expand and simplify  (x + 2)(x − 3)(2x + 1)
Take two brackets at a time.
(x + 2)(x − 3) = x² − x − 6
Now multiply by (2x + 1):
2x(x² − x − 6) = 2x³ − 2x² − 12x
+1(x² − x − 6) = x² − x − 6
Add: 2x³ − x² − 13x − 6
Keep this answer — you'll divide this exact cubic in the next section.
Section 2
Algebraic Division

Division is expansion run backwards. Sometimes you can just cancel; when you can't, you use long division. Either way, a remainder of zero tells you that what you divided by is a factor — which is the doorway into factorising.

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If the bottom divides every term: split the fraction up and cancel term by term.
If the bottom is a factor: factorise the top, then cancel the whole bracket.
Otherwise: use long division. The answer is the quotient, and anything left over is the remainder.
Remainder = 0 ⟹ it was a factor.
EXAMPLE Simplify  (6x3 − 8x2 + 4x) ÷ 2x
2x divides evenly into every term, so split it up:
6x³/2x = 3x²  ·  −8x²/2x = −4x  ·  4x/2x = 2
3x² − 4x + 2
Divide the numbers, then subtract the powers. Every term must be divided — a common slip is forgetting the last one.
Practice · 5 questions
Q1 Simplify  (10x3 + 5x2) ÷ 5x
10x³/5x = 2x²  and  5x²/5x = x
2x² + x
Q2 Simplify  12a2b ÷ 3ab
12 ÷ 3 = 4  ·  a² ÷ a = a  ·  b ÷ b = 1
4a
Q3 Simplify  (2x2 + 5x − 3) ÷ (2x − 1)
Factorise the top first: 2x² + 5x − 3 = (2x − 1)(x + 3)
The (2x − 1) cancels:
x + 3
Always try factorising before reaching for long division — it's far quicker when it works.
Q4 Divide  x3 − 8 by (x − 2).
Long division (leave gaps for the missing x² and x terms):
x³ ÷ x = → x²(x − 2) = x³ − 2x² → subtract → 2x²
2x² ÷ x = 2x → 2x(x − 2) = 2x² − 4x → subtract → 4x
4x ÷ x = 4 → 4(x − 2) = 4x − 8 → subtract → 0
x² + 2x + 4, remainder 0
This is the difference of two cubes in disguise: x³ − 8 = (x − 2)(x² + 2x + 4).
Q5 Divide  2x3 − x2 − 13x − 6 by (x − 3), and hence factorise it fully.
Long division:
2x³ ÷ x = 2x² → 2x²(x − 3) = 2x³ − 6x² → subtract → 5x² − 13x
5x² ÷ x = 5x → 5x(x − 3) = 5x² − 15x → subtract → 2x − 6
2x ÷ x = 2 → 2(x − 3) = 2x − 6 → subtract → 0
Quotient: 2x² + 5x + 2, remainder 0 → so (x − 3) is a factor.
Now factorise the quotient: 2x² + 5x + 2 = (2x + 1)(x + 2)
2x³ − x² − 13x − 6 = (x − 3)(2x + 1)(x + 2)
This is exactly the cubic you expanded in Section 1, Q5 — division has undone it.
Section 3
Factorising

Factorising is expanding in reverse, and it's the single most-used skill on Paper 1. Work the checklist in order every time: take out the HCF first, then count the terms.

Video — FactorisingVideo slot, ready for your embed.
1. Always take out the HCF first.
2 terms → difference of two squares a2 − b2 = (a − b)(a + b), or cubes:
  a3 − b3 = (a − b)(a2 + ab + b2)  ·  a3 + b3 = (a + b)(a2 − ab + b2)
3 terms → quadratic: find two numbers that multiply to a×c and add to b.
4 terms → grouping, in pairs.
EXAMPLE Factorise  2x2 − 7x − 15
a×c = (2)(−15) = −30. Need two numbers multiplying to −30, adding to −7 → −10 and +3.
Split the middle term:   2x² − 10x + 3x − 15
Group:   2x(x − 5) + 3(x − 5)
(2x + 3)(x − 5)
The bracket must match on both halves — if it doesn't, your two numbers are wrong.
Practice · 5 questions
Q1 Factorise  6x2 − 9x
HCF of 6x² and 9x is 3x → 3x(2x − 3)
Q2 Factorise  x2 − 49
Two terms, both squares → difference of two squares.
x² − 49 = x² − 7² = (x − 7)(x + 7)
Q3 Factorise  ab − 3a + 2b − 6
Four terms → group in pairs.
a(b − 3) + 2(b − 3)
Both have (b − 3): (a + 2)(b − 3)
Q4 Factorise  3x2 + 5x − 2
a×c = (3)(−2) = −6. Two numbers → −6 product, +5 sum → +6 and −1.
3x² + 6x − x − 2
3x(x + 2) − 1(x + 2)
(3x − 1)(x + 2)
Q5 Factorise  8x3 − 27
Both terms are cubes: 8x³ = (2x)³ and 27 = 3³.
Using a³ − b³ = (a − b)(a² + ab + b²) with a = 2x, b = 3:
(2x − 3)(4x² + 6x + 9)
The formula is in your Formulae & Tables booklet — know which page.
Section 4
Algebraic Fractions

Algebraic fractions follow exactly the same rules as ordinary fractions — the only difference is that you usually have to factorise first before anything cancels. That's why this comes after factorising.

Video — Algebraic FractionsVideo slot, ready for your embed.
Add/subtract: get a common denominator first.
Multiply: straight across. Divide: flip the second and multiply.
Simplify: factorise top and bottom, then cancel whole brackets — never individual terms.
EXAMPLE Simplify  (x2 − 9) ÷ (x2 + x − 12)
Factorise both:
Top: x² − 9 = (x − 3)(x + 3)
Bottom: x² + x − 12 = (x + 4)(x − 3)
Cancel the common bracket (x − 3):
(x + 3) / (x + 4)
You can only cancel a bracket that appears as a factor of both top and bottom.
Practice · 5 questions
Q1 Simplify  2/x + 3/(2x)
Common denominator 2x:   4/(2x) + 3/(2x) = 7/(2x)
Q2 Simplify  6x2y ÷ 3xy
6 ÷ 3 = 2, x² ÷ x = x, y ÷ y = 1 → 2x
Q3 Express as a single fraction:  3/(x + 1) − 2/(x − 2)
Common denominator (x + 1)(x − 2):
[3(x − 2) − 2(x + 1)] / [(x + 1)(x − 2)]
Top: 3x − 6 − 2x − 2 = x − 8
(x − 8) / [(x + 1)(x − 2)]
The minus applies to the whole second numerator — bracket it before expanding.
Q4 Simplify  (x2 − 4) ÷ (x2 − x − 6)
Top: (x − 2)(x + 2)
Bottom: (x − 3)(x + 2)
Cancel (x + 2): (x − 2) / (x − 3)
Q5 Simplify  (1/x − 1/3) ÷ (x − 3)
Numerator first: 1/x − 1/3 = (3 − x)/(3x)
So the whole thing = (3 − x) / [3x(x − 3)]
Now notice 3 − x = −(x − 3):
= −(x − 3) / [3x(x − 3)] → cancel (x − 3)
−1/(3x)
Spotting 3 − x = −(x − 3) is the whole question. It comes up constantly.
Section 5
Solving Equations

Now that you can factorise, you can solve. Solving means finding the values of x that make the equation true — and the method depends entirely on the type. With surds, you must always check your answers.

Video — Solving EquationsVideo slot, ready for your embed.
Linear: gather x's one side, numbers the other.
Quadratic: get = 0, then factorise — or use x = [−b ± √(b2 − 4ac)] / 2a.
Surd: isolate the root, square both sides — then check every answer in the original. Squaring can invent roots that don't work.
EXAMPLE Solve  2x2 − 5x − 3 = 0
a×c = (2)(−3) = −6. Two numbers: product −6, sum −5 → −6 and +1.
2x² − 6x + x − 3 = 0
2x(x − 3) + 1(x − 3) = 0
(2x + 1)(x − 3) = 0
x = −½  or  x = 3
Practice · 5 questions
Q1 Solve  5x − 3 = 32
5x = 35 → x = 7
Q2 Solve  3(x − 2) = x + 8
3x − 6 = x + 8
2x = 14 → x = 7
Q3 Solve  x2 − 7x + 12 = 0
Two numbers: product 12, sum −7 → −3 and −4
(x − 3)(x − 4) = 0
x = 3 or x = 4
Q4 Solve  2x2 + 3x − 5 = 0
a×c = −10 → numbers +5 and −2
2x² + 5x − 2x − 5 = 0
x(2x + 5) − 1(2x + 5) = 0
(2x + 5)(x − 1) = 0
x = −5/2 or x = 1
Q5 Solve  √(3x + 1) = x − 3
Square both sides:   3x + 1 = (x − 3)² = x² − 6x + 9
Rearrange:   0 = x² − 9x + 8
Factorise:   (x − 1)(x − 8) = 0 → x = 1 or x = 8
Now check both in the original:
x = 1: LHS = √4 = 2, RHS = 1 − 3 = −2.  2 ≠ −2 → reject.
x = 8: LHS = √25 = 5, RHS = 8 − 3 = 5 ✓
x = 8 only
This is the whole point of surd equations. Squaring created a false root — students who skip the check lose the marks.
Section 6
Rearranging Formulae

"Make x the subject" means rearrange until x sits alone on one side. It's the same skill as solving, just with letters instead of numbers — and you'll use it constantly in calculus, trigonometry and every time you open the Formulae & Tables booklet.

Video — Rearranging FormulaeVideo slot, ready for your embed.
Whatever you do to one side, do to the other. Undo operations in reverse order: get rid of + and − first, then × and ÷, then powers and roots.
If your variable appears twice, gather those terms on one side and factorise it out.
EXAMPLE Make r the subject of  A = πr2
Divide both sides by π:   A/π = r²
Square root both sides:   r = √(A/π)
r is a length here, so we take the positive root only.
Practice · 5 questions
Q1 Make a the subject of  v = u + at
v − u = at → a = (v − u)/t
Q2 Make x the subject of  y = 3x − 7
y + 7 = 3x → x = (y + 7)/3
Q3 Make h the subject of  V = ⅓πr2h
Multiply both sides by 3: 3V = πr²h
Divide by πr²: h = 3V/(πr²)
Q4 Make l the subject of  T = 2π√(l/g)
Divide by 2π:   T/(2π) = √(l/g)
Square both sides:   T²/(4π²) = l/g
Multiply by g:   l = gT²/(4π²)
Squaring hits the whole side — (2π)² = 4π², not 2π².
Q5 Make x the subject of  y = (2x + 1)/(x − 3)
Multiply both sides by (x − 3):   y(x − 3) = 2x + 1
Expand:   yx − 3y = 2x + 1
Gather the x terms on one side:   yx − 2x = 1 + 3y
Factorise out x:   x(y − 2) = 1 + 3y
x = (1 + 3y)/(y − 2)
x appearing twice is the whole difficulty. Gather, then factorise — always.
Section 7
Simultaneous Equations

Two unknowns need two equations; three unknowns need three. You're finding the values that satisfy all of them at once — graphically, where the lines cross.

Video — Simultaneous EquationsVideo slot, ready for your embed.
Elimination: add or subtract the equations to cancel a variable.
Substitution: rearrange one equation for a variable, sub it into the other. Usually easiest when a coefficient is already 1.
Always check your pair in both original equations — it's free marks.
EXAMPLE Solve  2x + 3y = 12 and x − y = 1
Second equation rearranges easily:   x = y + 1
Substitute into the first:   2(y + 1) + 3y = 12
2y + 2 + 3y = 12 → 5y = 10 → y = 2
Then x = 2 + 1 = 3
x = 3, y = 2
Check: 2(3) + 3(2) = 12 ✓ and 3 − 2 = 1 ✓
Practice · 5 questions
Q1 Solve  x + y = 7, x − y = 1
Add the two: 2x = 8 → x = 4
Then 4 + y = 7 → y = 3
x = 4, y = 3
Q2 Solve  2x + y = 8, x = 3
Sub x = 3: 2(3) + y = 8 → 6 + y = 8 → x = 3, y = 2
Q3 Solve  3x + 2y = 16, 5x − y = 5
From the second: y = 5x − 5
Sub in: 3x + 2(5x − 5) = 16
3x + 10x − 10 = 16 → 13x = 26 → x = 2
y = 5(2) − 5 = 5
x = 2, y = 5
Q4 Solve  x/2 + y/3 = 4, x − y = −2
Clear the fractions in the first — multiply by 6:   3x + 2y = 24
From the second: x = y − 2
3(y − 2) + 2y = 24 → 3y − 6 + 2y = 24 → 5y = 30 → y = 6
x = 6 − 2 = 4
x = 4, y = 6
Always clear fractions first — multiply by the LCM of the denominators.
Q5 Solve  x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2
Strategy: eliminate z to get two equations in x and y.
Eq1 + Eq3:   2x + 3y = 8
Eq2 + Eq3:   3x + y = 5 → y = 5 − 3x
Sub into 2x + 3y = 8:   2x + 3(5 − 3x) = 8
2x + 15 − 9x = 8 → −7x = −7 → x = 1
y = 5 − 3 = 2,   z = 6 − 1 − 2 = 3
x = 1, y = 2, z = 3
Check in all three: 1+2+3=6 ✓, 2−2+3=3 ✓, 1+4−3=2 ✓
Section 8
Binomial Expansions

Back to expanding — but faster. Expanding (a + b)2 by hand is fine. (a + b)5 is not, so we use Pascal's triangle or the nCr formula to get the coefficients without multiplying it all out.

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Pascal's triangle — each number is the sum of the two above it: 1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1
In (a + b)n, powers of a count down and powers of b count up, and the two powers always add to n.
EXAMPLE Expand  (x + 2)4
Row 4 of Pascal: 1, 4, 6, 4, 1
x⁴ + 4x³(2) + 6x²(2)² + 4x(2)³ + (2)⁴
= x⁴ + 8x³ + 6x²(4) + 4x(8) + 16
x⁴ + 8x³ + 24x² + 32x + 16
Raise the whole term to the power — (2)³ = 8, not 2×3.
Practice · 5 questions
Q1 Write down the row of Pascal's triangle used to expand (a + b)4.
1, 4, 6, 4, 1
Row n always has n + 1 numbers and starts and ends with 1.
Q2 Evaluate  5C2.
⁵C₂ = 5! / (2! × 3!) = 120 / (2 × 6) = 10
Q3 Expand  (x + 3)3
Row 3: 1, 3, 3, 1
x³ + 3x²(3) + 3x(3)² + (3)³
x³ + 9x² + 27x + 27
Q4 Expand  (2x − 1)4
Row 4: 1, 4, 6, 4, 1, with a = 2x and b = −1:
(2x)⁴ + 4(2x)³(−1) + 6(2x)²(−1)² + 4(2x)(−1)³ + (−1)⁴
16x⁴ − 32x³ + 24x² − 8x + 1
With a minus, the signs alternate. That's your check.
Q5 Find the coefficient of x3 in the expansion of  (2x + 3)5.
Don't expand the lot — go straight to the term you need.
The x³ term uses (2x)³, so it pairs with 3² :
⁵C₃ (2x)³ (3)² = 10 × 8x³ × 9 = 720x³
Coefficient = 720
The two powers must add to 5: (2x)³ and 3² → 3 + 2 = 5 ✓
Section 9
Identities

An equation is true for particular values of x. An identity (written ) is true for every value of x — and that's exactly what lets you compare coefficients.

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Because an identity holds for all x, the two sides must match term by term.
Method 1 — compare coefficients: match the x2 terms, the x terms, and the constants.
Method 2 — substitute: pick clever values of x that kill off brackets.
EXAMPLE Find a and b if  x2 + 6x + 11 ≡ (x + a)2 + b
Expand the right: (x + a)² + b = x² + 2ax + a² + b
Compare x terms: 2a = 6 → a = 3
Compare constants: a² + b = 11 → 9 + b = 11 → b = 2
a = 3, b = 2
This is completing the square in disguise — you'll meet it again in Algebra 2.
Practice · 5 questions
Q1 Explain the difference between x + 3 = 7 and 2(x + 1) ≡ 2x + 2.
The first is an equation — true only when x = 4.
The second is an identity — true for every value of x, because both sides are the same expression written differently.
Q2 Find a, b if  x2 + 4x + 7 ≡ (x + a)2 + b
x² + 2ax + a² + b
2a = 4 → a = 2
a² + b = 7 → 4 + b = 7 → b = 3
a = 2, b = 3
Q3 Find a, h, k if  3x2 + 12x + 5 ≡ a(x + h)2 + k
Compare x²: a = 3
3(x + h)² = 3x² + 6hx + 3h². Compare x: 6h = 12 → h = 2
Constants: 3h² + k = 5 → 12 + k = 5 → k = −7
a = 3, h = 2, k = −7
Q4 Find A, B, C if  2x2 − 5x + 3 ≡ A(x − 1)2 + B(x − 1) + C
Expand: A(x² − 2x + 1) + B(x − 1) + C = Ax² + (−2A + B)x + (A − B + C)
x²: A = 2
x: −2A + B = −5 → −4 + B = −5 → B = −1
constant: A − B + C = 3 → 2 + 1 + C = 3 → C = 0
A = 2, B = −1, C = 0
Quicker alternative: let x = 1 to kill both brackets and get C instantly.
Q5 Find a, b, c if  x3 + 2x2 − x + 5 ≡ (x + 1)(x2 + ax + b) + c
Expand: (x + 1)(x² + ax + b) = x³ + ax² + bx + x² + ax + b
= x³ + (a + 1)x² + (a + b)x + b, then + c on the constant.
x²: a + 1 = 2 → a = 1
x: a + b = −1 → 1 + b = −1 → b = −2
constant: b + c = 5 → −2 + c = 5 → c = 7
a = 1, b = −2, c = 7
c is the remainder when you divide by (x + 1). Sanity check: f(−1) = −1 + 2 + 1 + 5 = 7 ✓
Section 10
Functions & Roots

Write a polynomial as a function — f(x) — and everything you've learned pays off. Evaluating is just substitution; finding roots is just solving f(x) = 0, which is just factorising.

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To evaluate f(2), replace every x with 2.
Roots (or zeros) are where f(x) = 0 — exactly where the graph cuts the x-axis.
A polynomial of degree n has at most n roots. So a cubic cuts the x-axis at most 3 times.
EXAMPLE Find the roots of  f(x) = x3 − 4x
Set f(x) = 0:   x³ − 4x = 0
Take out the common factor:   x(x² − 4) = 0
x² − 4 is a difference of two squares:   x(x − 2)(x + 2) = 0
So x = 0, x = 2, x = −2
Never divide both sides by x — you'd lose the root x = 0. Always factorise.
Practice · 5 questions
Q1 f(x) = 2x2 − 3x + 1. Find f(2).
f(2) = 2(2)² − 3(2) + 1 = 8 − 6 + 1 = 3
Q2 For the same f(x) = 2x2 − 3x + 1, find f(−1).
f(−1) = 2(−1)² − 3(−1) + 1 = 2 + 3 + 1 = 6
(−1)² = +1, and −3(−1) = +3. Two sign traps in one line.
Q3 Find the roots of  f(x) = x2 − 5x + 6.
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0
x = 2, x = 3
Q4 Factorise and give the roots of  f(x) = x3 − 9x.
x³ − 9x = x(x² − 9) = x(x − 3)(x + 3)
Roots: x = 0, 3, −3
Q5 A cubic function has roots −1, 2 and 3. Write it in the form f(x) = x3 + ax2 + bx + c.
Each root gives a factor: f(x) = (x + 1)(x − 2)(x − 3)
First: (x − 2)(x − 3) = x² − 5x + 6
Then: (x + 1)(x² − 5x + 6) = x³ − 5x² + 6x + x² − 5x + 6
f(x) = x³ − 4x² + x + 6
Check: f(2) = 8 − 16 + 2 + 6 = 0 ✓. Always test one root back.
Section 11
Patterns

Given a list of numbers, you're expected to spot the type of pattern and write a formula for the nth term. The differences tell you everything — and this is the doorway into Sequences & Series later in the course.

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Constant 1st difference → linear, f(n) = an + b  (a = the difference)
Constant 2nd difference → quadratic, f(n) = an2 + bn + c  (2nd difference = 2a)
Constant ratio → exponential/geometric, f(n) = arn−1
EXAMPLE Find f(n) for the pattern  2, 7, 12, 17, …
Differences: 7 − 2 = 5, 12 − 7 = 5, 17 − 12 = 5 → constant → linear.
So f(n) = 5n + b.
f(1) = 2:   5(1) + b = 2 → b = −3
f(n) = 5n − 3
Check f(3) = 15 − 3 = 12 ✓. Always test a term you weren't given.
Practice · 5 questions
Q1 Write the next two terms:  3, 7, 11, 15, …
Difference is +4 each time → 19, 23
Q2 Find f(n) for  3, 7, 11, 15, …
Constant difference 4 → f(n) = 4n + b
f(1) = 3 → 4 + b = 3 → b = −1
f(n) = 4n − 1
Q3 Find f(n) for  1, 4, 9, 16, …
1st differences: 3, 5, 7 — not constant.
2nd differences: 2, 2 — constant → quadratic.
These are the square numbers: f(n) = n²
Q4 What type of pattern is  2, 6, 18, 54, …? Write f(n).
Differences 4, 12, 36 — not constant. But ratios: 6/2 = 3, 18/6 = 3, 54/18 = 3 → constant ratio.
Geometric / exponential, first term 2, ratio 3:
f(n) = 2 × 3n−1
Check f(3) = 2 × 9 = 18 ✓. Watch the n − 1 — f(1) must give 2.
Q5 Find f(n) for  3, 8, 15, 24, 35, …
1st differences: 5, 7, 9, 11 → not constant.
2nd differences: 2, 2, 2 → constant → quadratic, f(n) = an² + bn + c.
2nd difference = 2a → 2a = 2 → a = 1
f(1) = 3:   1 + b + c = 3 → b + c = 2
f(2) = 8:   4 + 2b + c = 8 → 2b + c = 4
Subtract: b = 2, so c = 0.
f(n) = n² + 2n
Check f(4) = 16 + 8 = 24 ✓ and f(5) = 25 + 10 = 35 ✓

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