If any of my solutions look wrong, please refer to the mark scheme. You can exit full-screen mode for the question paper and mark scheme by clicking the icon in the bottom-right corner or by pressing Esc on your keyboard.
GCSE November 2023 Edexcel Foundation Paper 1
Mark Scheme Legend
- M: Method mark (for a correct method or partial method)
- P: Process mark (for a correct process in problem solving)
- A: Accuracy mark (for a correct answer, depends on previous method marks)
- B: Unconditional accuracy mark (no method needed)
- C: Communication mark (for clear statements)
- oe: Or Equivalent
- cao: Correct Answer Only
- ft: Follow Through
Table of Contents
- Question 1 (Range)
- Question 2 (Arithmetic: Subtraction)
- Question 3 (Algebra: Simplify)
- Question 4 (Geometry: Angles)
- Question 5 (Fractions of Amount)
- Question 6 (Measures: Capacity)
- Question 7 (Scale Diagrams)
- Question 8 (Data: Bar Charts)
- Question 9 (Probability)
- Question 10 (Proportion)
- Question 11 (Geometry: Enlargement)
- Question 12 (Algebra: Substitution)
- Question 13 (Money Problems)
- Question 14 (Algebra: Linear Graphs)
- Question 15 (Percentages: Interest)
- Question 16 (Percentages: Comparisons)
- Question 17 (Fractions: Division)
- Question 18 (Decimals: Multiplication)
- Question 19 (Indices)
- Question 20 (Prime Factors & HCF)
- Question 21 (Statistics: Mean)
- Question 22 (Geometry: Construction)
- Question 23 (Geometry: Triangles)
- Question 24 (Algebra: Equations)
- Question 25 (Density)
- Question 26 (Estimation)
- Question 27 (Algebra: Expanding/Factorising)
Question 1
(1 mark)Here is a list of numbers.
2 4 4 7 8
Work out the range of these numbers.
Worked Solution
Step 1: Understanding Range
What is the range?
The range is a measure of spread. It tells us the difference between the largest number and the smallest number in a list.
The formula is: \[ \text{Range} = \text{Highest Value} – \text{Lowest Value} \]
Step 2: Identify Values and Calculate
Identify the highest and lowest numbers in the list:
List: 2, 4, 4, 7, 8
Highest value = 8
Lowest value = 2
Calculate the difference:
\[ 8 – 2 = 6 \]Final Answer:
6
✓ (Total: 1 mark)
Question 2
(1 mark)Work out \( 120 – 89 \)
Worked Solution
Step 1: Set up Column Subtraction
Why use column method?
This ensures we handle place value correctly, especially when we need to “borrow” or exchange numbers.
1 2 0 - 8 9 ───────
Step 2: Calculate with Borrowing
Units: \( 0 – 9 \) isn’t possible, so we borrow from the tens.
Tens: After borrowing, we have \( 11 – 8 \).
01112 10
- 8 9
─────────
3 1
Verification (Check): Add the answer back to the number being subtracted:
\( 31 + 89 = 120 \). It works.
Final Answer:
31
✓ (Total: 1 mark)
Question 3
(1 mark)Simplify \( 3 \times a \times 4 \)
Worked Solution
Step 1: Group Numbers and Variables
How algebra works:
We can multiply the numbers together first, regardless of the order.
The calculation is \( (3 \times 4) \times a \).
Step 2: Simplify
Multiply the numbers:
\[ 3 \times 4 = 12 \]Now attach the variable:
\[ 12 \times a = 12a \]Note: In algebra, we remove the multiplication sign to make it simpler.
Final Answer:
\( 12a \)
✓ (Total: 1 mark)
Question 4
(1 mark)Measure the size of the angle marked \( x \).
Worked Solution
Step 1: Using a Protractor
How to measure:
- Place the center point of the protractor on the vertex (corner) of the angle.
- Align the baseline of the protractor (the zero line) with the bottom horizontal line.
- Read the scale starting from 0° on the right-hand side.
Step 2: Reading the Scale
Following the curve of the angle upwards from 0°, the line passes through 40°.
The angle is acute (less than 90°), so 40° makes sense.
Examiner’s Note: The mark scheme allows answers in the range 38° to 42°.
Final Answer:
40°
✓ (Total: 1 mark)
Question 5
(1 mark)Work out \( \frac{1}{5} \) of 300.
Worked Solution
Step 1: Interpret the Fraction
What does this mean?
Finding \( \frac{1}{5} \) of a number is the same as dividing that number by 5.
We need to calculate: \[ 300 \div 5 \]
Step 2: Calculate
We can ignore the last zero for a moment:
\( 30 \div 5 = 6 \)
Now put the zero back:
\[ 300 \div 5 = 60 \]Final Answer:
60
✓ (Total: 1 mark)
Question 6
(3 marks)There are 3 litres of oil in a can.
Jermaine uses 700 millilitres of the oil.
Work out the amount of oil left in the can.
Give your answer in millilitres.
Worked Solution
Step 1: Convert Units
Why convert?
The total oil is in litres, but the amount used is in millilitres. We must have the same units to subtract.
We know that \( 1 \text{ litre} = 1000 \text{ millilitres} \).
Step 2: Calculate Amount Left
Subtract the used amount from the total:
\[ 3000 – 700 \]2 10 0 0 - 7 0 0 ────────── 2 3 0 0
Final Answer:
2300 millilitres
✓ (Total: 3 marks)
Question 7
(2 marks)Matt is drawing a scale diagram.
\( 1 \text{ cm} \) represents \( 5 \text{ m} \).
(a) He draws a line \( 3 \text{ cm} \) long.
What real distance does the line represent?
The real distance between two points is \( 20 \text{ m} \).
(b) What is the distance between the two points on the scale diagram?
Worked Solution
Part (a): Convert Diagram to Real
Strategy:
The scale says every \( 1 \text{ cm} \) is worth \( 5 \text{ m} \).
If we have \( 3 \text{ cm} \), we need 3 lots of \( 5 \text{ m} \).
Answer (a): 15 m
Part (b): Convert Real to Diagram
Strategy:
We need to fit \( 20 \text{ m} \) onto the paper. Since \( 5 \text{ m} \) fits into \( 1 \text{ cm} \), we need to find how many 5s are in 20.
Answer (b): 4 cm
Final Answer:
(a) 15 m
(b) 4 cm
✓ (Total: 2 marks)
Question 8
(6 marks)Miss Bailey asked 24 students where they each wanted to go on a school trip.
Here are the results.
farm castle farm farm
castle farm castle castle
castle farm castle museum
museum farm castle museum
museum museum castle castle
(a) Complete the frequency table.
| Place | Tally | Frequency |
|---|---|---|
| castle | ||
| farm | ||
| museum |
(b) Write down the place that is the mode.
(c) Draw a bar chart to show the results.
Worked Solution
Part (a): Counting (Frequency Table)
Method: Count each item in the list carefully. Crossing them off as you go helps avoid mistakes.
- Castle: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 (Frequency: 11)
- Farm: 1, 2, 3, 4, 5, 6, 7 (Frequency: 7)
- Museum: 1, 2, 3, 4, 5, 6 (Frequency: 6)
Check: \( 11 + 7 + 6 = 24 \). This matches the total students.
Part (b): Finding the Mode
What is the mode?
The mode is the most common or popular item (highest frequency).
Castle has the highest frequency (11).
Answer: castle
Part (c): Drawing the Bar Chart
Rules for a good bar chart:
- Linear scale on the vertical axis (0, 2, 4…).
- Label the axes (Frequency, Place).
- Bars must have equal width.
- Equal gaps between bars.
Final Answer:
(a) Table completed (11, 7, 6)
(b) Castle
(c) Bar chart drawn correctly
✓ (Total: 6 marks)
Question 9
(3 marks)Selina has a bag of 22 counters.
- 5 of the counters are blue.
- 9 of the counters are red.
- 8 of the counters are pink.
Selina takes at random a counter from the bag.
Write down the probability that Selina
(i) takes a red counter,
(ii) does not take a pink counter,
(iii) takes a white counter.
Worked Solution
Part (i): Probability of Red
Formula: \[ P(\text{Event}) = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} \]
Red counters = 9
Total counters = 22
\[ P(\text{Red}) = \frac{9}{22} \]Part (ii): Probability of Not Pink
Method 1: Add the others
Not pink means Blue or Red.
Blue + Red = \( 5 + 9 = 14 \)
\[ P(\text{Not Pink}) = \frac{14}{22} \]Method 2: Subtract from Total
Total – Pink = \( 22 – 8 = 14 \)
Answer: \( \frac{14}{22} \) (can simplify to \( \frac{7}{11} \))
Part (iii): Probability of White
Reasoning:
Are there any white counters? No.
This is an impossible event.
Final Answer:
(i) \( \frac{9}{22} \)
(ii) \( \frac{14}{22} \) or \( \frac{7}{11} \)
(iii) 0
✓ (Total: 3 marks)
Question 10
(3 marks)Here are the ingredients needed to make 20 peanut butter cookies.
Makes 20 cookies
250 g peanut butter
200 g sugar
2 small eggs
Derek wants to make 60 cookies.
He has 900 g of peanut butter.
Does Derek have enough peanut butter to make 60 cookies?
You must show how you get your answer.
Worked Solution
Step 1: Find the Scale Factor
Strategy:
We need to go from 20 cookies to 60 cookies.
How many times does 20 go into 60?
So, we need 3 times the ingredients.
Step 2: Calculate Peanut Butter Needed
Original amount: 250 g
Amount needed:
\[ 250 \times 3 = 750 \text{ g} \]Step 3: Compare and Conclude
What this tells us:
Derek needs 750 g. He has 900 g.
Since 900 is greater than 750, he has enough.
Conclusion: Yes, he has enough.
Final Answer:
Yes (with working shown)
✓ (Total: 3 marks)
Question 11
(2 marks)On the grid, draw an enlargement of the triangle with a scale factor of 3.
Worked Solution
Step 1: Understand Scale Factor 3
What does Scale Factor 3 mean?
Every side length of the original shape must be multiplied by 3.
Original Base: 2 units
New Base: \( 2 \times 3 = 6 \) units
Original Height: 3 units
New Height: \( 3 \times 3 = 9 \) units
Step 2: Draw the Enlarged Triangle
Drawing:
Draw a horizontal line 6 squares long.
From the middle of that line, go up 9 squares to find the top vertex.
Connect the corners.
Final Answer:
Triangle drawn with base 6 units and height 9 units.
✓ (Total: 2 marks)
Question 12
(6 marks)(a) \( P = 2g + 4h \)
(i) Work out the value of \( P \) when \( g = 3 \) and \( h = 5 \).
(ii) Work out the value of \( g \) when \( P = 38 \) and \( h = 3 \).
(b) \( V = 3r – q \)
Work out the value of \( V \) when \( r = -3 \) and \( q = 2 \).
Worked Solution
Part (a)(i): Substitution
Substitute: Replace \( g \) with 3 and \( h \) with 5.
Part (a)(ii): Solving an Equation
Substitute and Solve: Replace \( P \) with 38 and \( h \) with 3, then find \( g \).
Subtract 12 from both sides:
\[ 38 – 12 = 2g \] \[ 26 = 2g \]Divide by 2:
\[ g = 13 \]Part (b): Negative Numbers
Watch the signs: Be careful substituting the negative value.
First, multiply:
\[ 3 \times -3 = -9 \]Now subtract:
\[ V = -9 – 2 \] \[ V = -11 \]Final Answer:
(a)(i) \( P = 26 \)
(a)(ii) \( g = 13 \)
(b) \( V = -11 \)
✓ (Total: 6 marks)
Question 13
(4 marks)Chloe is making scrunchies.
Chloe has a large number of hair bands.
Each hair band costs 8p.
She buys 100g of wool for £3.
Chloe uses 1 hair band and 5g of wool to make each scrunchy.
She makes as many scrunchies as she can.
Work out the total cost of each scrunchy that she makes.
Give your answer in pence.
Worked Solution
Step 1: Work out the cost of wool per gram
Convert units: First, change £3 to pence so everything is in the same unit.
£3 = 300p
Now find the cost of 1g of wool.
Step 2: Calculate cost of wool for one scrunchy
She uses 5g of wool for one scrunchy.
Step 3: Add the cost of the hair band
We need the total cost (Wool + Hair band).
Wool cost: 15p
Hair band cost: 8p
\[ 15 + 8 = 23\text{p} \]Final Answer:
23p
✓ (Total: 4 marks)
Question 14
(3 marks)On the grid, draw the graph of \( y = 4x – 1 \) for values of \( x \) from -2 to 2.
Worked Solution
Step 1: Create a Table of Values
Method: Substitute values of \( x \) into the equation \( y = 4x – 1 \).
| x | calculation | y |
|---|---|---|
| -2 | \( 4(-2) – 1 = -8 – 1 \) | -9 |
| -1 | \( 4(-1) – 1 = -4 – 1 \) | -5 |
| 0 | \( 4(0) – 1 = 0 – 1 \) | -1 |
| 1 | \( 4(1) – 1 = 4 – 1 \) | 3 |
| 2 | \( 4(2) – 1 = 8 – 1 \) | 7 |
Step 2: Plot and Connect
Plot the coordinates: (-2, -9), (-1, -5), (0, -1), (1, 3), (2, 7).
Draw a straight line through them.
Final Answer:
Straight line drawn through \( (0, -1) \) with gradient 4.
✓ (Total: 3 marks)
Question 15
(4 marks)Steve is buying a car.
The car costs £12000.
Steve pays 25% of the cost as a deposit.
He pays the rest of the cost in 20 equal monthly payments.
How much is each monthly payment?
Worked Solution
Step 1: Calculate the Deposit
Method: Find 25% of £12000.
25% is the same as \( \frac{1}{4} \), so we divide by 4.
Deposit = £3000
Step 2: Calculate the “Rest of the Cost”
Total Cost – Deposit = Amount left to pay
\[ 12000 – 3000 = 9000 \]Step 3: Calculate Monthly Payments
Divide the remaining £9000 by 20 months.
Divide both by 10 to make it easier:
\[ 900 \div 2 = 450 \]Final Answer:
£450
✓ (Total: 4 marks)
Question 16
(2 marks)Shah takes an exam.
The exam is out of 60 marks.
Shah needs to score at least 70% of the marks to pass the exam.
He scores 45 marks.
Show that Shah passes the exam.
Worked Solution
Step 1: Choose a Method
We can either:
- Calculate 70% of 60 marks (the pass mark) and compare it to 45.
- Convert Shah’s score (45 out of 60) to a percentage and compare it to 70%.
Method 1 is usually easier without a calculator.
Step 2: Calculate 70% of 60 (Method 1)
First, find 10%:
\[ 10\% \text{ of } 60 = 60 \div 10 = 6 \]Now, find 70% (multiply by 7):
\[ 70\% = 6 \times 7 = 42 \text{ marks} \]Step 3: Compare and Conclude
Comparison: Shah needed 42 marks. He got 45 marks.
Therefore, Shah passed.
Final Answer:
Shah passed because \( 45 > 42 \)
✓ (Total: 2 marks)
Question 17
(3 marks)Work out \( \frac{3}{5} \div \frac{1}{6} \)
Give your answer as a mixed number.
Worked Solution
Step 1: KCF Method (Keep, Change, Flip)
Rule: Dividing by a fraction is the same as multiplying by its reciprocal (flip it).
- Keep the first fraction: \( \frac{3}{5} \)
- Change \( \div \) to \( \times \)
- Flip the second fraction: \( \frac{1}{6} \rightarrow \frac{6}{1} \)
Step 2: Multiply and Simplify
Top times top, bottom times bottom:
\[ \frac{3 \times 6}{5 \times 1} = \frac{18}{5} \]Step 3: Convert to Mixed Number
How many times does 5 go into 18?
\( 5, 10, 15… \) (3 times)
Remainder is \( 18 – 15 = 3 \).
Final Answer:
\( 3 \frac{3}{5} \)
✓ (Total: 3 marks)
Question 18
(3 marks)Work out \( 6.3 \times 2.4 \)
Worked Solution
Step 1: Remove Decimals
Strategy: It’s easier to multiply whole numbers.
\( 6.3 \times 10 = 63 \)
\( 2.4 \times 10 = 24 \)
We will calculate \( 63 \times 24 \), then divide by 100 later (because \( 10 \times 10 = 100 \)).
Step 2: Long Multiplication
6 3
× 2 4
─────
2 5 2 (4 × 63)
1 2 6 0 (20 × 63)
───────
1 5 1 2
Step 3: Put Decimal Point Back
We had 2 decimal places in total in the question (\( .3 \) and \( .4 \)).
We need 2 decimal places in the answer.
1512 becomes 15.12
Final Answer:
15.12
✓ (Total: 3 marks)
Question 19
(4 marks)(a) (i) Write down the value of \( 5^0 \)
(ii) Write down the value of \( 5^{-2} \)
(b) Write \( \frac{2^5 \times 2^4}{2^3} \) in the form \( 2^n \) where \( n \) is an integer.
Worked Solution
Part (a)(i): Zero Power
Rule: Anything to the power of 0 is 1.
Answer: 1
Part (a)(ii): Negative Power
Rule: A negative power creates a reciprocal (1 over…).
\( 5^{-2} = \frac{1}{5^2} \)
Answer: \( \frac{1}{25} \)
Part (b): Index Laws
Multiplication Rule: Add powers. \( 2^5 \times 2^4 = 2^{5+4} = 2^9 \)
Division Rule: Subtract powers. \( \frac{2^9}{2^3} = 2^{9-3} \)
So the answer is \( 2^6 \).
Final Answer:
(a)(i) 1
(a)(ii) \( \frac{1}{25} \)
(b) \( 2^6 \)
✓ (Total: 4 marks)
Question 20
(4 marks)(a) Write 156 as a product of its prime factors.
(b) Find the highest common factor (HCF) of 156 and 130.
Worked Solution
Part (a): Factor Tree for 156
Break the number down into pairs of factors until you reach prime numbers (circle them).
156 is even, so divide by 2:
\( 156 \div 2 = 78 \)
78 is even, so divide by 2:
\( 78 \div 2 = 39 \)
39 divides by 3:
\( 39 \div 3 = 13 \)
13 is prime.
Prime factors: 2, 2, 3, 13
Answer: \( 2 \times 2 \times 3 \times 13 \) or \( 2^2 \times 3 \times 13 \)
Part (b): HCF of 156 and 130
First, find prime factors of 130.
\( 130 = 13 \times 10 = 13 \times 2 \times 5 \)
Primes of 130: 2, 5, 13
Primes of 156: 2, 2, 3, 13
Identify common factors:
- Both have a 2
- Both have a 13
Multiply the common factors:
\[ 2 \times 13 = 26 \]Final Answer:
(a) \( 2^2 \times 3 \times 13 \)
(b) 26
✓ (Total: 4 marks)
Question 21
(4 marks)The mean length of 5 sticks is 4.2 cm.
Nawal measured the length of one of the sticks as 7 cm.
(a) Work out the mean length of the other 4 sticks.
Nawal made a mistake.
The stick was not 7 cm long.
It was 17 cm long.
(b) How does this affect your answer to part (a)?
Worked Solution
Part (a): Calculate Mean of Remainder
Step 1: Find the Total Length
Mean = Total ÷ Count. Therefore, Total = Mean × Count.
\( 4 \times 5 = 20 \)
\( 0.2 \times 5 = 1 \)
Total = 21 cm
Step 2: Remove the known stick
Subtract the 7 cm stick to find the total of the other 4.
Step 3: Calculate new mean
Divide the remaining total by 4.
Part (b): Effect of the Mistake
Reasoning:
We assume the initial Mean of 5 (4.2 cm) is a fixed fact about the group total.
Total length = 21 cm.
If the stick is actually 17 cm (larger), it takes up more of the total.
Remaining Total = \( 21 – 17 = 4 \).
New Mean = \( 4 \div 4 = 1 \).
Answer: The mean of the other sticks would be lower (it would be 1 cm).
Final Answer:
(a) 3.5 cm
(b) It reduces the answer (to 1 cm).
✓ (Total: 4 marks)
Question 22
(2 marks)The point \( P \) lies on the line \( AB \).
Use ruler and compasses to construct an angle of 90° at \( P \).
You must show all your construction lines.
Worked Solution
Method: Constructing a Perpendicular at a Point
- Place compass point on \( P \). Draw arcs on the line on both sides of \( P \) (labeled \( C \) and \( D \) in the diagram below).
- Open the compass wider. Place point on \( C \) and draw an arc above \( P \).
- Keep the same compass width. Place point on \( D \) and draw an arc to cross the first one.
- Draw a straight line from \( P \) through the crossing point.
Final Answer:
Correct perpendicular construction with visible arcs.
✓ (Total: 2 marks)
Question 23
(4 marks)The diagram shows an isosceles triangle \( ABD \) and the straight line \( ABC \).
\( BA = BD \)
\( x : y = 2 : 1 \)
Work out the value of \( w \).
Worked Solution
Step 1: Use Isosceles Triangle Properties
Fact: \( BA = BD \). This means triangle \( ABD \) is isosceles.
The angles opposite the equal sides are equal.
Opposite \( BD \) is angle \( A \) (which is \( x \)).
Opposite \( BA \) is angle \( D \).
Therefore, angle \( D = x \).
Step 2: Form an Equation
Angles in a triangle add up to 180°.
\[ x + x + y = 180 \] \[ 2x + y = 180 \]Step 3: Use the Ratio
We are given \( x : y = 2 : 1 \).
This means \( x \) is twice as big as \( y \).
\( x = 2y \)
Substitute \( x = 2y \) into the triangle equation:
\[ 2(2y) + y = 180 \] \[ 4y + y = 180 \] \[ 5y = 180 \]Solve for \( y \):
\[ y = 180 \div 5 = 36 \]Step 4: Calculate w
Angles on a straight line add up to 180°.
Angles \( y \) and \( w \) are on the straight line \( ABC \).
Final Answer:
\( w = 144 \)
✓ (Total: 4 marks)
Question 24
(5 marks)Mano has three shelves of books.
- There are \( x \) books on shelf A.
- There are \( (3x + 1) \) books on shelf B.
- There are \( (2x – 5) \) books on shelf C.
There is a total of 44 books on the three shelves.
All the books have the same mass.
The books on shelf B have a total mass of 7500 g.
Work out the total mass of the books on shelf A.
Worked Solution
Step 1: Form an Equation for Total Books
Add the expressions for all shelves and set equal to 44.
Collect like terms:
\[ (x + 3x + 2x) + (1 – 5) = 44 \] \[ 6x – 4 = 44 \]Step 2: Solve for x
Add 4 to both sides:
\[ 6x = 48 \]Divide by 6:
\[ x = 8 \]Step 3: Find details for Shelf B
Calculate number of books on Shelf B using \( x = 8 \).
These 25 books weigh 7500 g.
Mass of 1 book = \( 7500 \div 25 \)
Tip: \( 75 \div 25 = 3 \), so \( 7500 \div 25 = 300 \) g.
Step 4: Calculate Mass of Shelf A
Shelf A has \( x \) books.
Books on A = 8
Total Mass = \( 8 \times 300 \)
\[ 8 \times 3 = 24 \] \[ 8 \times 300 = 2400 \text{ g} \]Final Answer:
2400 g
✓ (Total: 5 marks)
Question 25
(2 marks)A piece of glass has a mass of 27 g and a volume of 10 cm\(^3\).
Work out the density of the piece of glass.
Give your answer in g/cm\(^3\).
Worked Solution
Step 1: Use Density Formula
Formula: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
Step 2: Calculate
Dividing by 10 moves the decimal point one place to the left.
Final Answer:
2.7 g/cm\(^3\)
✓ (Total: 2 marks)
Question 26
(3 marks)Work out an estimate for \( \frac{5.7 \times 8.2}{0.26} \)
Worked Solution
Step 1: Round to 1 Significant Figure
Rule: To estimate, round every number to one significant figure (the first non-zero digit).
- \( 5.7 \rightarrow 6 \)
- \( 8.2 \rightarrow 8 \)
- \( 0.26 \rightarrow 0.3 \)
Alternative: You could round 0.26 to 0.25 if you spot that it makes division easier, but 0.3 is the standard 1 s.f. rule.
Step 2: Calculate with Rounded Numbers
Multiply the top:
\[ \frac{48}{0.3} \]Step 3: Divide
To divide by a decimal, multiply top and bottom by 10 to clear the decimal point.
Now divide 480 by 3:
\[ 48 \div 3 = 16 \] \[ 480 \div 3 = 160 \]Final Answer:
160 (Answers between 160 and 200 are accepted)
✓ (Total: 3 marks)
Question 27
(3 marks)(a) Expand and simplify \( (3x + 2)(2x – 5) \)
(b) Factorise \( x^2 – 16 \)
Worked Solution
Part (a): Expand Double Brackets
Method: Multiply every term in the first bracket by every term in the second. (FOIL method).
1. Firsts: \( 3x \times 2x = 6x^2 \)
2. Outers: \( 3x \times -5 = -15x \)
3. Inners: \( 2 \times 2x = 4x \)
4. Lasts: \( 2 \times -5 = -10 \)
Combine them:
\[ 6x^2 – 15x + 4x – 10 \]Simplify: Collect the middle terms (\( x \) terms).
Result: \( 6x^2 – 11x – 10 \)
Part (b): Difference of Two Squares
Recognition: We have a squared term (\( x^2 \)) minus a square number (16).
The pattern is: \( a^2 – b^2 = (a – b)(a + b) \)
\( \sqrt{x^2} = x \)
\( \sqrt{16} = 4 \)
So we get two brackets, one with plus, one with minus.
\[ (x – 4)(x + 4) \]Final Answer:
(a) \( 6x^2 – 11x – 10 \)
(b) \( (x – 4)(x + 4) \)
✓ (Total: 3 marks)