Expanding and factorising: a week of ready-to-teach maths

Five lessons for Year 10 Algebra. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus a sheet of algebra tiles, twelve match cards that pair a quadratic with its factors, a mini-check and every answer with its working.

Start here: five minutes to Monday

1. Skim the week at a glance on the next page.
2. Print the five days. Each day is two A4 sheets: a plan and a worksheet.
3. Print the two cut-out sheets, one copy per pair. Cut the tiles before Day 2 and the 12 match cards before Day 3, and keep each set in an envelope.
4. No calculators are needed, except to check a factorisation by substituting a number.
5. Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.

The idea in plain words

Expanding and factorising are one fact read in two directions. The distributive law says a factor outside a bracket multiplies every term inside: 3(x + 4) = 3x + 12. Read from left to right that is expanding; read from right to left it is factorising. Two brackets multiply the same way, every term by every term, which an area grid shows as four rectangles. Factorising a quadratic means finding the two brackets again, by finding two numbers that multiply to the constant and add to the middle coefficient. The exponent laws keep the books for powers of a letter. And factorising earns its keep in solving: when a product is zero, one of its factors must be zero.

One day, one lesson

The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.

How we write these expressions

We write x² for x squared and 3x for 3 × x. A bracket that repeats is written as a square, (x − 5)², and negative numbers use a true minus sign, −5. Every factorisation is checked by expanding it back, and every solution by substituting it into the equation.

The week at a glance

One lesson a day for a week. Each day stands on the day before, so run them in order.

How the week builds

Day 1 sends one factor to every term; Day 2 sends every term to every term, and the area grid shows why; Day 3 runs both days backwards; Day 4 does the same bookkeeping for powers; and Day 5 puts factorising to work on equations. It builds on Year 9, where students expanded binomial products, factorised monic quadratics and extended the exponent laws to variables. It prepares the quadratic models of growth and decay later in Year 10, and the quadratic work of senior mathematics.

Materials for the week (one trip)

• From the classroom: scissors, envelopes for the tiles and cards, pencils, this pack printed.
• Cut-out sheet 1, one per pair: algebra tiles, 4 large squares, 10 strips and 20 small squares (Days 2 and 3).
• Cut-out sheet 2, one per pair: 12 match cards (Day 3).
• If you have them: commercial algebra tiles work just as well. No other equipment to buy.

Words for the word wall

expand, factorise, distributive law, term, coefficient, constant, like terms, binomial, quadratic, highest common factor, difference of two squares, perfect square, exponent, base, exponent laws, null factor law, solution

Dear families

This week in maths, Year 10 is learning to expand and factorise: to multiply out brackets, and to put them back again. It is the algebra behind a mental-arithmetic trick you may already use, and it is the key to solving equations with a squared term.

Try this at home

• Ask: what is 7 × 23? Try 7 × 20 + 7 × 3. (161: that is the distributive law.)
• Ask: why is 21 × 19 the same as 20² − 1? (It is (20 + 1)(20 − 1) = 400 − 1 = 399.)
• Ask: which two numbers multiply to 12 and add to 7? (3 and 4, so x² + 7x + 12 = (x + 3)(x + 4).)
• Ask: if two numbers multiply to make zero, what must be true? (One of them is zero.)

My two-directions week

Fill one row a day. Write an expression from class, its other form, and how you checked it.

One factor meets every term

The distributive law sends the factor outside a bracket to every term inside. Students expand single brackets, with negatives and letters outside, expand and simplify sums, and find the slip in three worked lines.

We are learning to

• expand a bracket by multiplying every term inside by the factor outside,
• keep track of the signs when the factor outside is negative,
• expand two brackets in a sum and collect like terms.

Success criteria

• I can expand −2(x − 4) and 3x(x + 2) correctly.
• I can expand and simplify a sum of two brackets and check it by substituting a number.

You need

The worksheet, one per student. No calculators.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Dropping the sign: −(x − 3) is −x + 3; the minus meets both terms.
• Mixing x² and 2x: x × x is x², while x + x is 2x.
• Collecting unlike terms: 3x + 2 stays as it is; it is not 5x.

Answers

• Part A: 5(x + 3) = 5x + 15; −2(x − 4) = −2x + 8; 3x(x + 2) = 3x² + 6x; 4(2y − 3) = 8y − 12; −a(2a − 5) = −2a² + 5a; 2x(5x − 1) = 10x² − 2x.
• Part B: 3(x + 2) + 2(x − 1) = 5x + 4; 4(a + 3) − 2(a − 5) = 2a + 22; x(x + 3) − 2(x + 3) = x² + x − 6; 2y(y − 1) + 3(y + 4) = 2y² + y + 12.
• Part C: 3(x + 4) = 3x + 12 (the 3 did not meet the 4); −2(x − 5) = −2x + 10 (−2 × (−5) is +10, not −10); x(x + 2) = x² + 2x (the x did not meet the 2).
• Part D: perimeter 2(2x + 3) + 2(x − 1) = 6x + 4; area (2x + 3)(x − 1) = 2x² + x − 3. 3(x + 4) − 3(x − 4) = 24: the x terms cancel, so it is 24 for every x.

One factor meets every term

A. Expand

Multiply every term inside the bracket by the factor outside. The first one is done for you.

B. Expand and simplify

Expand each bracket, then collect like terms. Check one answer by putting 2 in for the letter in both forms.

C. Spot the slip

Each line has one slip. Say what went wrong and write the right expansion.

D. Stretch

Expanding as an area

Two brackets multiply every term by every term. An area grid shows the four products as four rectangles, and algebra tiles let students build them. Students expand with grids and tiles, then with negatives, and see why (x + 3)² is not x² + 9.

We are learning to

• expand a product of two brackets with an area grid,
• expand without the grid, negative terms included,
• explain why the square of a bracket has a middle term.

Success criteria

• I can expand (x + 5)(x − 2) and say where each of the four products comes from.
• I can show with a grid why (x + 3)² is x² + 6x + 9.

You need

The algebra tiles (cut-out sheet 1), one set per pair, cut out before the lesson; the worksheet, one per student.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after the tiles; do Part B and Part D next time.

Watch for these ideas

• “(x + 2)(x + 3) = x² + 6.” The two middle pieces, 3x and 2x, are part of the area too.
• “(x + 3)² = x² + 9.” The square has four pieces, not two: x² + 3x + 3x + 9.
• “(x − 3)(x − 4) ends in −12.” −3 × (−4) is +12.

Answers

• Part A: (x + 2)(x + 5) = x² + 2x + 5x + 10 = x² + 7x + 10; (x + 4)(x + 1) = x² + 4x + 1x + 4 = x² + 5x + 4; (x + 6)(x + 1) = x² + 6x + 1x + 6 = x² + 7x + 6.
• Part B: (x + 5)(x − 2) = x² + 3x − 10; (x − 3)(x − 4) = x² − 7x + 12; (3x − 2)(x + 4) = 3x² + 10x − 8; (x − 6)(x + 6) = x² − 36.
• Part C: (x + 2)(x + 3): 1 large square, 5 strips, 6 small squares; (x + 1)(x + 4): 1 large square, 5 strips, 4 small squares. The strips match because 2 + 3 = 1 + 4; the small squares differ because 2 × 3 ≠ 1 × 4.
• Part D: (x + 3)² = x² + 6x + 9. With x = 2: (2 + 3)² = 25, but 2² + 3² = 13. The missing 12 is the two strips, 6x.

Expanding as an area

A. Area grids

Write the area of each piece, then add the pieces. The first grid is done for you.

B. Without the grid

Multiply every term in the first bracket by every term in the second, then collect like terms.

C. Build it with tiles

Build each rectangle with the tiles, count the tiles, and write the expansion.

D. Stretch: the square of a bracket

Factorising, expanding reversed

Factorising runs Days 1 and 2 backwards: find the factor every term shares, or the two brackets a quadratic came from. Students take out highest common factors, factorise quadratics by finding two numbers, and match twelve cards into pairs.

We are learning to

• take out the highest common factor of every term, letters included,
• factorise x² + bx + c by finding two numbers that multiply to c and add to b,
• recognise a difference of two squares and a perfect square.

Success criteria

• I can factorise 12y³ − 18y² fully and check it by expanding.
• I can factorise x² + 2x − 15 and x² − 49.

You need

The match cards (cut-out sheet 2), one set per pair; the Day 2 tiles for Part D; the worksheet, one per student.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Taking out a factor that is not the highest: 2(3x + 6) still has a common factor of 3.
• Sign slips: x² + 2x − 15 needs two numbers that multiply to −15, so one of them is negative.
• “x² + 9 = (x + 3)(x − 3).” A difference of two squares needs a minus: x² − 9.

Answers

• Part A: 8a − 20 = 4(2a − 5); 5x² + 15x = 5x(x + 3); 12y³ − 18y² = 6y²(2y − 3); 9m² + 6m = 3m(3m + 2).
• Part B: x² + 7x + 10 = (x + 5)(x + 2); x² + 2x − 15 = (x + 5)(x − 3); x² − 9x + 20 = (x − 4)(x − 5); x² − 49 = (x + 7)(x − 7); x² − 10x + 25 = (x − 5)².
• Part C: x² + 7x + 12 = (x + 3)(x + 4); x² − 9 = (x + 3)(x − 3); x² − 6x + 9 = (x − 3)²; x² + x − 12 = (x + 4)(x − 3); 3x² + 6x = 3x(x + 2); x² − x − 12 = (x + 3)(x − 4). Letters: the card key with the mini-check answers.
• Part D: 2x² + 7x + 3 = (2x + 1)(x + 3): 2 large squares, 7 strips and 3 small squares make a rectangle 2x + 1 by x + 3.

Factorising, expanding reversed

A. The highest common factor

Take out the highest common factor of every term. Check by expanding. The first one is done for you.

B. Two numbers

Find two numbers that multiply to the constant and add to the coefficient of x. Then write the factors.

C. Match the cards

Match each expanded card to its factorised card. Write the letters, and check one pair by expanding.

D. Stretch: tiles in reverse

Exponent laws in algebra

The exponent laws keep the books for powers of a letter: add the exponents to multiply, subtract to divide, multiply to raise a power to a power, and give every factor in a bracket the power. Students simplify terms with numbers and letters together, then find the slip in four worked lines.

We are learning to

• multiply and divide terms by working the numbers and each letter separately,
• raise a term to a power, giving the power to every factor,
• spot and explain the common slips with exponents.

Success criteria

• I can simplify (3x²)³ and 20m⁷ ÷ 4m².
• I can explain why x³ × x⁴ is x⁷, not x¹².

You need

The worksheet, one per student. No calculators.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• “5x⁰ = 0.” Any base that is not zero, to the power 0, is 1, so 5x⁰ = 5.
• Forgetting that x means x¹: x × x⁴ is x⁵.
• “20m⁷ ÷ 4m² = 16m⁵.” The numbers divide; only the exponents subtract.

Answers

• Part A: a⁴ × a⁶ = a¹⁰; 3x² × 5x⁴ = 15x⁶; y⁹ ÷ y³ = y⁶; 20m⁷ ÷ 4m² = 5m⁵; (p³)⁴ = p¹²; (3x²)³ = 27x⁶; (2a³b)² = 4a⁶b²; 5x⁰ = 5.
• Part B: (4x³y² × 3xy⁵) ÷ 6x²y⁴ = 2x²y³; (2x²)³ × x ÷ 4x⁵ = 2x².
• Part C: x³ × x⁴ = x⁷, not x¹²: multiplied the exponents instead of adding them; (2x³)² = 4x⁶, not 2x⁶: left the 2 unsquared; x⁸ ÷ x² = x⁶, not x⁴: divided the exponents instead of subtracting; 3x² × 2x³ = 6x⁵, not 5x⁵: added the numbers in front instead of multiplying.
• Part D: by the law x⁵ ÷ x⁵ = x⁰, and by cancelling it is 1, so x⁰ = 1. Likewise x² ÷ x⁵ = x⁻³ by the law and 1/x³ by cancelling, so x⁻³ = 1/x³.

Exponent laws in algebra

A. Simplify

Work the numbers first, then each letter. The first one is done for you.

B. Two laws at once

C. Spot the slip

Each line has one slip. Name the law that was broken and write the right answer.

D. Stretch: zero and below

Solving by factorising

If a product is zero, one of its factors is zero: the null factor law. Students solve equations that are already factorised, factorise first, rearrange to zero first, solve two problems, and finish with the self-check quiz as a class game.

We are learning to

• use the null factor law to solve a factorised equation,
• rearrange a quadratic equation to equal zero, then factorise it,
• check each solution and reject one that does not fit the situation.

Success criteria

• I can solve x² − x − 6 = 0 and check both answers.
• I can explain why x² = 5x + 14 must be rearranged before it is factorised.

You need

The worksheet, one per student; the unit’s self-check quiz on the board for the game.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; do Parts B to D and the quiz next time.

Watch for these ideas

• Solving before rearranging: x(x − 2) = 3 does not mean x = 3 or x − 2 = 3.
• Sign flips: (x + 2)(x + 3) = 0 gives x = −2 or x = −3, not 2 and 3.
• Dividing by x: dividing 2x² = 10x by x loses the solution x = 0.

Answers

• Part A: (x − 3)(x + 5) = 0: x = 3 or x = −5; x(x − 7) = 0: x = 0 or x = 7; x² + 8x + 12 = 0: x = −2 or x = −6; x² − x − 6 = 0: x = 3 or x = −2; x² − 16 = 0: x = 4 or x = −4; 2x² − 10x = 0: x = 0 or x = 5.
• Part B: x² = 5x + 14 becomes x² − 5x − 14 = 0, so (x + 2)(x − 7) = 0, so x = −2 or x = 7; x² + 9 = 6x becomes x² − 6x + 9 = 0, so (x − 3)² = 0, so x = 3.
• Part C: x(x + 3) = 40 becomes x² + 3x − 40 = 0, so (x + 8)(x − 5) = 0, so x = −8 or x = 5; a width cannot be negative, so the bed is 5 m by 8 m. n(n + 1) = 132 becomes n² + n − 132 = 0, so (n + 12)(n − 11) = 0, so n = −12 or n = 11: the numbers are 11 and 12 (or −12 and −11).
• Part D: x = 3 works by luck: 3 × 1 = 3. But x = 5 gives 15, not 3. Right: x² − 2x − 3 = 0, so (x + 1)(x − 3) = 0, so x = −1 or x = 3.

Solving by factorising

A. Solve

Factorise if you need to, then set each factor to zero. Check your answers. The first one is done for you.

B. Rearrange first

C. Two problems

D. Stretch: Kai’s shortcut

Algebra tiles

For Days 2 and 3, one set per pair. A large square is x by x, a strip is x by 1, and a small square is 1 by 1. The strip is not a whole number of small squares long: x can be any length.

4 large squares (x²)

10 strips (x)

20 small squares (1)

Expanded and factorised

For Day 3, one set per pair. Six cards show a quadratic expanded; six show it factorised. Match each pair, then check one by expanding.

Teacher note: the key is with the mini-check answers.

What we know: Expanding and factorising

Work on your own. No calculator. Show your working.

1. Expand −4(2x − 3). ______________________
2. Expand and simplify (x + 6)(x − 2). ______________________
3. Factorise fully 10x² − 15x. ______________________
4. Factorise x² − 3x − 28. ______________________
5. Simplify (3a²b)³ ÷ 9a⁴b. ______________________
6. Circle the expansion of (x − 5)²:   x² − 25  ·  x² − 10x + 25  ·  x² + 25
7. Solve x² + 2x − 24 = 0, and check both answers. ______________________
8. The side of a square garden bed is made 2 m longer, and its new area is 49 m². Write an equation and find the original side. ______________________

Answers and marking guide

Answers with working

1. −4(2x − 3) = −8x + 12: the −4 meets both terms, and −4 × (−3) = +12.
2. (x + 6)(x − 2) = x² − 2x + 6x − 12 = x² + 4x − 12.
3. 10x² − 15x = 5x(2x − 3): the highest common factor is 5x.
4. Two numbers that multiply to −28 and add to −3: 4 and −7, so x² − 3x − 28 = (x + 4)(x − 7).
5. (3a²b)³ = 27a⁶b³, and 27a⁶b³ ÷ 9a⁴b = 3a²b².
6. x² − 10x + 25: (x − 5)² = x² − 5x − 5x + 25. The middle term is the two strips of the area grid.
7. (x + 6)(x − 4) = 0, so x = −6 or x = 4. Check: both make the left side 0.
8. (x + 2)² = 49, so x² + 4x − 45 = 0, so (x + 9)(x − 5) = 0. A side cannot be negative, so x = 5 m (check: 7² = 49).

A quick three-level guide

Eight questions, four ideas. A student at standard answers most questions with working, and checks a factorisation by expanding and a solution by substituting.

Card key

Match cards (Day 3): G and A (x² + 7x + 12 = (x + 3)(x + 4)), D and J (x² − 9 = (x + 3)(x − 3)), E and L (x² − 6x + 9 = (x − 3)²), B and H (x² + x − 12 = (x + 4)(x − 3)), K and F (3x² + 6x = 3x(x + 2)), I and C (x² − x − 12 = (x + 3)(x − 4)).

Weekly class record

Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.

The five columns are the five days: one bracket, two brackets, factorising, the exponent laws, and solving by factorising.