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A4

Simplify and manipulate algebraic expressions (incl. surds and algebraic fractions): like terms, common factors, expanding two or more binomials, factorising quadratics incl. ax² + bx + c, indices

Simplifying expressions

Worked answers, methods and verified real exam appearances for A4 on Edexcel GCSE Maths 1MA1.

Explanation

  • Simplify by collecting only like terms and applying index laws only to matching bases. Expand brackets by multiplying every required pair of terms, then collect.
  • When factorising, first remove any common factor and check that re-expansion reproduces every term.
  • Foundation questions can include expanding two binomials and factorising x2+bx+cx^2+bx+c.
  • Higher tier: manipulation extends to surds, algebraic fractions, products of more binomials and quadratics ax2+bx+cax^2+bx+c.
  • Never cancel terms across addition; factorise complete numerators and denominators first, then retain values excluded by the original denominator.

Worked example

Factorise x2+7x+12x^2+7x+12 fully.

  1. 1.Find two numbers with product 1212 and sum 77: 33 and 44.
  2. 2.Write the factors (x+3)(x+4)(x+3)(x+4).
  3. 3.Check by expanding: x2+4x+3x+12=x2+7x+12x^2+4x+3x+12=x^2+7x+12.

Answer: (x+3)(x+4)(x+3)(x+4).

Common mistakes

  • Don't collect unlike terms such as 3x+2x23x+2x^2 to make 5x35x^3.
  • Don't miss a cross-term when expanding two brackets.
  • Don't cancel terms across addition in an algebraic fraction.

Exam tip

After factorising, expand your answer mentally; it must reproduce the original expression exactly.

Worked practice

Q1
Tier 1 · Easy

1

Simplify 7a+3b−2a+5b7a+3b-2a+5b.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • 5a+8b5a+8b
2Collect the like aa-terms and the like bb-terms separately: (7a−2a)+(3b+5b)=5a+8b(7a-2a)+(3b+5b)=5a+8b.
Q2
Tier 2 · Standard

2

Factorise 6x2+x−26x^2+x-2.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • (3x+2)(2x−1)(3x+2)(2x-1)
3The product of the leading and constant coefficients is 6×(−2)=−126\times(-2)=-12. Split the middle term using 4x−3x4x-3x: 6x2+4x−3x−2=2x(3x+2)−(3x+2)=(3x+2)(2x−1)6x^2+4x-3x-2=2x(3x+2)-(3x+2)=(3x+2)(2x-1).
Q3
Tier 3 · Hard

3

Simplify x2−9x2+x−6\frac{x^2-9}{x^2+x-6}, stating every value of xx excluded from the original expression.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • x−3x−2\frac{x-3}{x-2}
  • x≠−3x\ne-3 and x≠2x\ne2
4Factorise both parts: x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3) and x2+x−6=(x+3)(x−2)x^2+x-6=(x+3)(x-2). Cancelling the common factor gives (x−3)/(x−2)(x-3)/(x-2). The original denominator is zero at x=−3x=-3 or x=2x=2, so both values remain excluded.
Q4
Tier 1 · Easy

4

Simplify 5(2x−3)+155(2x-3)+15.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • 10x10x
2Expand the bracket first: 5(2x−3)+15=10x−15+155(2x-3)+15=10x-15+15. The constant terms cancel, leaving 10x10x.
Q5
Tier 2 · Standard

5

A square has side length (x+3)(x+3) cm, and a rectangle has length (x+6)(x+6) cm and width xx cm. Work out how much greater the area of the square is than the area of the rectangle. Simplify your answer.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • 9 cm29\,\text{cm}^2
3The difference in areas is (x+3)2−x(x+6)(x+3)^2-x(x+6). Expanding gives x2+6x+9−(x2+6x)=9x^2+6x+9-(x^2+6x)=9, so the square is greater by 9 cm29\,\text{cm}^2.
Q6
Tier 3 · Hard

6

Factorise fully 2x2−4x−702x^2-4x-70.

(2)

(Total for Question 6 is 2 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • 2(x−7)(x+5)2(x-7)(x+5) (or 2(x+5)(x−7)2(x+5)(x-7))
2First take out the common factor 22: 2x2−4x−70=2(x2−2x−35)2x^2-4x-70=2(x^2-2x-35). The two numbers with product −35-35 and sum −2-2 are −7-7 and 55, so the full factorisation is 2(x−7)(x+5)2(x-7)(x+5).
Q7
Tier 2 · Standard

7

Leah says that (3p2q)2=6p4q2(3p^2q)^2=6p^4q^2. Explain Leah's error and simplify (3p2q)2(3p^2q)^2 correctly.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • Leah has doubled the coefficient instead of squaring it; (3p2q)2=9p4q2(3p^2q)^2=9p^4q^2
3The power 22 applies to every factor in the bracket. The coefficient is 32=93^2=9, while (p2)2=p4(p^2)^2=p^4 and q2=q2q^2=q^2. Therefore (3p2q)2=9p4q2(3p^2q)^2=9p^4q^2.
Q8
Tier 3 · Hard

8

Simplify (x+4)(x−3)−x(x−1)(x+4)(x-3)-x(x-1). Hence work out the value of the original expression when x=9x=9.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • 2x−122x-12
  • 66
4Expand both products: (x+4)(x−3)=x2+x−12(x+4)(x-3)=x^2+x-12 and x(x−1)=x2−xx(x-1)=x^2-x. Subtracting gives x2+x−12−(x2−x)=2x−12x^2+x-12-(x^2-x)=2x-12. When x=9x=9, this is 18−12=618-12=6.
Q9
Tier 3 · Hard

9

Higher only: The identity (2x+a)(3x+b)≡6x2+x−15(2x+a)(3x+b)\equiv6x^2+x-15 holds for every value of xx. Work out the integers aa and bb. Hence factorise 6x2+x−156x^2+x-15.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • a=−3a=-3 and b=5b=5
  • (2x−3)(3x+5)(2x-3)(3x+5)
4Expanding gives 6x2+(2b+3a)x+ab6x^2+(2b+3a)x+ab. Therefore ab=−15ab=-15 and 2b+3a=12b+3a=1. The integer pair a=−3a=-3, b=5b=5 satisfies both conditions. Hence 6x2+x−15=(2x−3)(3x+5)6x^2+x-15=(2x-3)(3x+5).
Q10
Tier 3 · Hard

10

Higher only: Given that p=x+1xp=x+\dfrac1x, show that x2+1x2=p2−2x^2+\dfrac1{x^2}=p^2-2. You may assume x≠0x\ne0.

(3)

(Total for Question 10 is 3 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • x2+1x2=p2−2x^2+\dfrac1{x^2}=p^2-2
3Square the given expression: p2=(x+1x)2=x2+2+1x2p^2=\left(x+\dfrac1x\right)^2=x^2+2+\dfrac1{x^2}. Subtracting 22 from both sides gives p2−2=x2+1x2p^2-2=x^2+\dfrac1{x^2}, as required.

Verified exam appearances

SeriesPaperQuestionMarksCalculatorTierLinks
2022-111HQ195Non-calculatorHigherQPMS
2019-111FQ263Non-calculatorFoundationQPMS
2023-112HQ93AllowedHigherQPMS
2019-113HQ224AllowedHigherQPMS
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2022-063HQ73AllowedHigherQPMS
2024-112FQ194AllowedFoundationQPMS
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2023-063HQ123AllowedHigherQPMS
2021-111FQ154Non-calculatorFoundationQPMS
2023-113HQ234AllowedHigherQPMS
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2024-113FQ253AllowedFoundationQPMS
2024-112HQ173AllowedHigherQPMS
2019-113HQ153AllowedHigherQPMS
2022-062HQ15AllowedHigherQPMS
2019-112HQ153AllowedHigherQPMS
2024-062FQ133AllowedFoundationQPMS
2021-111HQ153Non-calculatorHigherQPMS
2023-063FQ31AllowedFoundationQPMS
2019-063HQ186AllowedHigherQPMS
2019-113HQ104AllowedHigherQPMS
2023-113FQ132AllowedFoundationQPMS
2024-113HQ93AllowedHigherQPMS
2022-113HQ145AllowedHigherQPMS
2021-112FQ202AllowedFoundationQPMS
2024-111FQ154Non-calculatorFoundationQPMS
2023-062FQ41AllowedFoundationQPMS
2024-111HQ135Non-calculatorHigherQPMS
2022-063HQ145AllowedHigherQPMS
2019-113FQ224AllowedFoundationQPMS
2023-063HQ14AllowedHigherQPMS
2022-062FQ215AllowedFoundationQPMS
2023-063FQ204AllowedFoundationQPMS
2022-111HQ164Non-calculatorHigherQPMS
2023-063HQ214AllowedHigherQPMS
2022-063HQ194AllowedHigherQPMS

Other points in A Algebra · notation and manipulation

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