Edexcel International GCSE Maths Expanding and Factorising Past Paper Questions and Solutions (4MA1 Higher)

Part (a)

1. Subtract the indices when dividing: 8 − 2 = 6

Answer: e⁶

Part (b)

1. x × x = x², x × 1 = x, −3 × x = −3x, −3 × 1 = −3
2. x − 3x = −2x

Answer: x² − 2x − 3

Part (a)

1. 3x(2x + 3) = 6x² + 9x and −x(3x + 5) = −3x² − 5x
2. Collect like terms: 6x² − 3x² + 9x − 5x

Answer: 3x² + 4x

Part (b)

1. Add d to both sides: p + d = at
2. Divide by a.

Answer: t = ^{p + d}⁄_a

Part (c)

1. Simplify the left side: w^{5 + n − 3} = w^{n + 2}
2. n + 2 = 10

Answer: n = 8

Part (a)

1. 3c³ × c = 3c⁴ (add the indices) and 3c³ × 4 = 12c³

Answer: 3c⁴ + 12c³

Part (b)(i)

1. Two numbers that multiply to −9 and add to +8: +9 and −1.

Answer: (x + 9)(x − 1)

Part (b)(ii)

1. Set each bracket equal to zero.

Answer: x = −9 or x = 1

Part (a)

1. Cube each part: 2³ = 8, (c⁴)³ = c¹², (d⁷)³ = d²¹

Answer: 8c¹²d²¹

Part (b)

1. y⁰ = 1, so 5y⁰ = 5 × 1. (Only the y is raised to the power 0, not the 5.)

Answer: 5

Part (c)

1. Highest common factor of the numbers: 4. Lowest power of a: a². Lowest power of b: b.
2. Take out 4a²b: 16a²b³ ÷ 4a²b = 4b² and 20a³b ÷ 4a²b = 5a

Answer: 4a²b(4b² + 5a)

Part (d)(i)

1. Two numbers that multiply to −22 and add to +9: +11 and −2.

Answer: (x + 11)(x − 2)

Part (d)(ii)

1. Set each bracket to zero.

Answer: x = −11 or x = 2

Part (a)

1. Multiply both terms in the bracket by 4x: 4x × x = 4x² and 4x × (−5) = −20x

Answer: 4x² − 20x

Part (b)

1. Two numbers that multiply to +20 and add to −9: −4 and −5.

Answer: (y − 4)(y − 5)

Part (a)

1. When dividing powers of the same letter, subtract the indices: 9 − 2 = 7

Answer: g⁷

Part (b)

1. Multiply each term inside the bracket by 5k². Add the indices when multiplying powers.
2. 5k² × k³ = 5k⁵ and 5k² × 4 = 20k²

Answer: 5k⁵ + 20k²

Part (c)(i)

1. Find two numbers that multiply to −63 and add to −2: −9 and +7.

Answer: (x − 9)(x + 7)

Part (c)(ii)

1. If (x − 9)(x + 7) = 0, one of the brackets must equal 0.
2. x − 9 = 0 gives x = 9, and x + 7 = 0 gives x = −7

Answer: x = 9 or x = −7

Part (d)

1. Get the y terms on one side: add 2y to both sides, giving 7 < 5y − 12
2. Add 12 to both sides: 19 < 5y
3. Divide by 5: 3.8 < y

Answer: y > 3.8

Part (a)

1. The highest common factor of both terms is 3c².

Answer: 3c²(6cd² − 7)

Part (b)(i)

1. Two numbers that multiply to −18 and add to −3: −6 and +3.

Answer: (y − 6)(y + 3)

Part (b)(ii)

1. (y − 6)(y + 3) = 0, so one of the brackets must be 0.

Answer: y = 6 or y = −3

Part (a)

1. Add the indices when multiplying: 6 + 10 = 16

Answer: a¹⁶

Part (b)

1. Subtract the indices when dividing: 30 − 12 = 18

Answer: c¹⁸

Part (c)(i)

1. Two numbers that multiply to +21 and add to −10: −3 and −7.

Answer: (y − 3)(y − 7)

Part (c)(ii)

1. Set each bracket equal to zero.

Answer: y = 3 or y = 7

Part (a)

1. 2⁴ = 16, (x³)⁴ = x¹², (y⁵)⁴ = y²⁰

Answer: 16x¹²y²⁰

Part (b)(i)

1. Two numbers that multiply to −36 and add to +5: +9 and −4.

Answer: (x + 9)(x − 4)

Part (b)(ii)

1. Set each bracket to zero.

Answer: x = −9 or x = 4

Part (a)

1. HCF of 25 and 45 is 5. Lowest power of a is a⁴, of c is c³. (d and h are not in both terms.)
2. Take out 5a⁴c³: 25a⁴c⁷d ÷ 5a⁴c³ = 5c⁴d and 45a⁹c³h ÷ 5a⁴c³ = 9a⁵h

Answer: 5a⁴c³(5c⁴d + 9a⁵h)

Part (b)

1. Expand the left: (2x + 5)² = 4x² + 20x + 25
2. Expand the right: (2x + 3)(2x − 1) = 4x² − 2x + 6x − 3 = 4x² + 4x − 3
3. The 4x² terms cancel: 20x + 25 = 4x − 3, so 16x = −28

Answer: x = −1.75

Part (a)

1. When you multiply powers of the same letter, add the indices: 4 + 5 = 9

Answer: x⁹

Part (b)

1. 4³ = 64 and (y²)³ = y⁶

Answer: 64y⁶

Part (c)

1. Two numbers that multiply to +12 and add to −7: −3 and −4.

Answer: (n − 3)(n − 4)

Part (a)

1. This is a difference of two squares: (3x)² − (2y)²

Answer: (3x − 2y)(3x + 2y)

Part (b)

1. Common denominator 8x: ^7x⁄_8x − ^{2(x + 3)}⁄_8x
2. Numerator: 7x − 2x − 6 = 5x − 6 (the minus sign applies to both terms in the bracket)

Answer: ^{5x − 6}⁄_8x

Part (a)

1. Common denominator (x − 2)(x + 1): ^{4(x + 1) − 3(x − 2)}⁄_{(x − 2)(x + 1)}
2. Numerator: 4x + 4 − 3x + 6 = x + 10

Answer: ^{x + 10}⁄_{(x − 2)(x + 1)}

Part (b)

1. (x − 5)(x − 3) = x² − 8x + 15
2. Multiply by 2x.

Answer: 2x³ − 16x² + 30x

Part (a)

1. (n − 4)(3n + 5) = 3n² + 5n − 12n − 20 = 3n² − 7n − 20
2. Multiply by n.

Answer: 3n³ − 7n² − 20n

Part (b)

1. The common denominator is 4x: ¹²⁄_4x + ^{2(x + 2)}⁄_4x + ^x⁄_4x
2. Numerator: 12 + 2x + 4 + x = 3x + 16

Answer: ^{3x + 16}⁄_4x

Part (a)

1. Multiply the first and last numbers: 6 × (−4) = −24. Find two numbers that multiply to −24 and add to −5: −8 and +3.
2. Split the middle term: 6y² − 8y + 3y − 4
3. Factorise in pairs: 2y(3y − 4) + 1(3y − 4)
4. Take out the common bracket (3y − 4).

Answer: (2y + 1)(3y − 4)

Part (b)

1. The lowest common denominator of 4x and 3x is 12x.
2. Multiply the first fraction top and bottom by 3, and the second by 4: ^{3(2x + 1)}⁄_12x + ^{4(7 − 5x)}⁄_12x
3. Expand the numerators: 6x + 3 + 28 − 20x
4. Collect like terms: 6x − 20x = −14x and 3 + 28 = 31

Answer: ^{31 − 14x}⁄_12x

1. The lowest common denominator of 4 and 6x is 12x.
2. ⁵⁄₄ = ^{5 × 3x}⁄_12x = ^15x⁄_12x and ^{x − 3}⁄_6x = ^{2(x − 3)}⁄_12x
3. Add the numerators: 15x + 2x − 6 = 17x − 6

Answer: ^{17x − 6}⁄_12x

1. Expand the two brackets first: (3x + 1)(2x − 3) = 6x² − 9x + 2x − 3 = 6x² − 7x − 3
2. Multiply every term by 4x: 4x × 6x² = 24x³, 4x × (−7x) = −28x², 4x × (−3) = −12x

Answer: 24x³ − 28x² − 12x

Part (a)

1. (x + 2)(3x − 4) = 3x² − 4x + 6x − 8 = 3x² + 2x − 8
2. Multiply by 5x.

Answer: 15x³ + 10x² − 40x

Part (b)

1. The negative power flips the fraction: (^y20⁄_16w⁸)^3⁄₄
2. Power ³⁄₄ means 4th root, then cube: ∜16 = 2 and 2³ = 8; (w⁸)^3⁄₄ = w⁶; (y²⁰)^3⁄₄ = y¹⁵

Answer: ^y15⁄_8w⁶

1. (2x − 5)² = (2x − 5)(2x − 5) = 4x² − 10x − 10x + 25 = 4x² − 20x + 25
2. Multiply every term by 3x.

Answer: 12x³ − 60x² + 75x

1. First two brackets: (4x + 1)(x − 3) = 4x² − 12x + x − 3 = 4x² − 11x − 3
2. Multiply by (5x + 6): 20x³ + 24x² − 55x² − 66x − 15x − 18
3. Collect like terms.

Answer: 20x³ − 31x² − 81x − 18

1. Both terms have a common factor of 2: 2(25g² − 9)
2. 25g² − 9 is a difference of two squares: (5g)² − 3² = (5g − 3)(5g + 3)

Answer: 2(5g − 3)(5g + 3)

Part (a)

1. Multiply two brackets first: (3x + 1)(2 − x) = 6x − 3x² + 2 − x = −3x² + 5x + 2
2. Now multiply by (4 + x): (−3x² + 5x + 2)(4 + x) = −12x² − 3x³ + 20x + 5x² + 8 + 2x
3. Collect like terms: −3x³ + (−12 + 5)x² + (20 + 2)x + 8

Answer: −3x³ − 7x² + 22x + 8

Part (b)

1. Simplify inside the bracket first by subtracting indices: a^{3 − 9}b^{1 − 5} = a⁻⁶b⁻⁴
2. Raising to the power −½ means multiplying each index by −½.
3. a^{−6 × −½} = a³ and b^{−4 × −½} = b²

Answer: a³b²

Part (a)

1. Multiply the two brackets that make a difference of two squares first: (3x − 1)(3x + 1) = 9x² − 1
2. Then multiply by (x + 2): (9x² − 1)(x + 2) = 9x³ + 18x² − x − 2

Answer: 9x³ + 18x² − x − 2

Part (b)

1. Simplify inside the bracket: ^2x5⁄_8xy² = ^x4⁄_4y²
2. A power of −2 means flip the fraction and square it: (^4y2⁄_x⁴)²
3. 4² = 16, (y²)² = y⁴, (x⁴)² = x⁸

Answer: ^16y4⁄_x⁸

1. Factorise: 9x² − 4 = (3x − 2)(3x + 2) and 3x² − 13x − 10 = (3x + 2)(x − 5)
2. ¹⁄_{3x − 2} × ^{(3x − 2)(3x + 2)}⁄_{(3x + 2)(x − 5)} = ¹⁄_{x − 5} (everything else cancels)
3. ¹⁄_{x − 5} − ⁷⁄_{x − 1} = ^{(x − 1) − 7(x − 5)}⁄_{(x − 5)(x − 1)}
4. Numerator: x − 1 − 7x + 35 = 34 − 6x

Answer: ^{34 − 6x}⁄_{(x − 5)(x − 1)}

1. Substitute y into 9y + 5: 9 × ⁵⁄_{5a − 2} + 5 = ⁴⁵⁄_{5a − 2} + ^{5(5a − 2)}⁄_{5a − 2} = ^{25a + 35}⁄_{5a − 2}
2. So x = 5 ÷ ^{25a + 35}⁄_{5a − 2} = ^{5(5a − 2)}⁄_{25a + 35}
3. Factorise the denominator: 25a + 35 = 5(5a + 7). Cancel the 5.

Answer: x = ^{5a − 2}⁄_{5a + 7}

1. Factorise: x² − 4 = (x − 2)(x + 2) and 4x² − 7x − 2 = (4x + 1)(x − 2)
2. Dividing by a fraction means multiplying by its reciprocal: (x − 2)(x + 2) × ^x⁄_{(4x + 1)(x − 2)} = ^{x(x + 2)}⁄_{4x + 1}
3. Subtract 2x using the common denominator 4x + 1: ^{x2 + 2x − 2x(4x + 1)}⁄_{4x + 1} = ^{x2 + 2x − 8x2 − 2x}⁄_{4x + 1}
4. Simplify the numerator: −7x²

Answer: ^−7x2⁄_{4x + 1}