GCSE Identities and argument Practice Questions

Free GCSE Identities and argument practice questions with full step-by-step worked solutions. Covers expanding brackets, collecting like terms, even numbers, algebraic expressions. Practise exam-style problems and check your method.

expanding bracketscollecting like termseven numbersalgebraic expressionsodd numbersidentities
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Expand 3(x+2)3(x+2).
Show worked solution

Worked solution

  1. Multiply the first term

    3×x=3x3 \times x = 3x

    Multiply the 3 outside the bracket by the x inside.

  2. Multiply the second term

    3×2=63 \times 2 = 6

    Multiply the 3 outside the bracket by the 2 inside.

  3. Write the expanded expression

    3(x+2)=3x+63(x+2) = 3x + 6

    Add the two results to give the expanded form.

Answer
3x+63x + 6
Question 2
2 markseasy
Expand 5(2x+1)5(2x+1).
Show worked solution

Worked solution

  1. Multiply the first term

    5×2x=10x5 \times 2x = 10x

    Multiply the 5 by 2x.

  2. Multiply the second term

    5×1=55 \times 1 = 5

    Multiply the 5 by 1.

  3. Write the expansion

    10x+510x + 5

    Combine the two results.

Answer
10x+510x + 5
Question 3
2 marksintermediate
2(x+a)≡2x+102(x+a) \equiv 2x + 10 for all values of xx. Find the value of aa.
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Worked solution

  1. Expand the left side

    2(x+a)=2x+2a2(x+a) = 2x + 2a

    Multiply 2 into the bracket.

  2. Compare with 2x + 10

    2x+2a≡2x+102x + 2a \equiv 2x + 10

    Line the two sides up.

  3. Match the x terms

    2x=2x2x = 2x

    The x terms already agree.

  4. Match the constants

    2a=102a = 10

    Set the constant terms equal.

  5. Solve for a

    a=5a = 5

    Divide both sides by 2.

  6. State the value

    a=5a = 5

    So a=5a = 5.

Answer
a=5a = 5
Question 4
4 markshard
2x2+bx+c≡(2x−3)(x+4)2x^2 + bx + c \equiv (2x-3)(x+4) for all values of xx. Find bb and cc.
Show worked solution

Worked solution

  1. Set up the expansion

    (2x−3)(x+4)(2x-3)(x+4)

    Multiply each pair of terms.

  2. Multiply the first terms

    2x×x=2x22x \times x = 2x^2

    2x times x gives 2x squared.

  3. Multiply the outer terms

    2x×4=8x2x \times 4 = 8x

    2x times 4 gives 8x.

  4. Multiply the inner terms

    −3×x=−3x-3 \times x = -3x

    -3 times x gives -3x.

  5. Multiply the last terms

    −3×4=−12-3 \times 4 = -12

    -3 times 4 gives -12.

  6. Write all terms

    2x2+8x−3x−122x^2 + 8x - 3x - 12

    List every product.

  7. Group the x terms

    8x−3x=5x8x - 3x = 5x

    Add the two x terms.

  8. Write the simplified expression

    2x2+5x−122x^2 + 5x - 12

    Put the pieces together.

  9. Compare with 2x22x^2 + bx + c

    b=5,  c=−12b = 5, \; c = -12

    Match the coefficients.

  10. State

    b=5,  c=−12b = 5, \; c = -12

    So b=5b = 5 and c=−12c = -12.

Answer
b=5,c=−12b = 5, c = -12
Question 5
6 markschallenging
A student wants to prove that the product of two consecutive integers is always even. Which argument is fully correct?
Show worked solution

Worked solution

  1. Write two consecutive integers

    n,  n+1n, \; n+1

    Let n be any integer.

  2. Write their product

    n(n+1)n(n+1)

    Multiply the two integers.

  3. Consider whether n is even or odd

    n(n+1)n(n+1)

    Split into two cases.

  4. Case 1: n is even

    n=2kn = 2k

    Write n as 2 times an integer k.

  5. Then the product has a factor of 2

    2k(n+1)2k(n+1)

    The product is a multiple of 2.

  6. So the product is even

    2k(n+1)2k(n+1)

    An even factor makes the product even.

  7. Case 2: n is odd

    n=2k+1n = 2k + 1

    Write n as an odd number.

  8. Then n+1 is even

    n+1=2k+2=2(k+1)n + 1 = 2k + 2 = 2(k+1)

    One more than an odd number is even.

  9. So the product has a factor of 2

    n×2(k+1)n \times 2(k+1)

    Again the product is a multiple of 2.

  10. So the product is even

    n×2(k+1)n \times 2(k+1)

    An even factor makes the product even.

  11. Combine the cases

    n(n+1)n(n+1)

    In both cases the product is even.

  12. State the shortcut

    n or n+1 is evenn \text{ or } n+1 \text{ is even}

    Of any two consecutive integers, one is always even.

  13. Check n=4n = 4

    4×5=204 \times 5 = 20

    20 is even.

  14. Check n=7n = 7

    7×8=567 \times 8 = 56

    56 is even.

  15. Conclude

    n(n+1)n(n+1)

    So the product of two consecutive integers is always even.

Answer
n(n+1) is always even because one of two consecutive integers is even

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