GCSE Expanding binomials Practice Questions

Free GCSE Expanding binomials practice questions with full step-by-step worked solutions. Covers expanding double brackets, FOIL, squaring a binomial, perfect square. Practise exam-style problems and check your method.

expanding double bracketsFOILsquaring a binomialperfect squaredifference of two squaresnegative terms
GCSE Foundation70 questionsStep-by-step solutions
Question 1
2 markseasy
Which of the following is the correct expansion of (x+2)(x+3)(x + 2)(x + 3)?
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Worked solution

  1. Set up the expansion (FOIL)

    (x+2)(x+3)(x + 2)(x + 3)

    Every term in the first bracket must be multiplied by every term in the second.

  2. Multiply out the pairs

    =x2+3x+2x+6= x^2 + 3x + 2x + 6

    Multiply the First, Outer, Inner and Last terms to get four terms.

  3. Collect like terms

    =x2+5x+6= x^2 + 5x + 6

    Combine the x-terms 3x and 2x to get 5x.

Answer
x2+5x+6x^2 + 5x + 6
Question 2
2 markseasy
Expand and simplify (x+2)(x+7)(x + 2)(x + 7).
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Worked solution

  1. Set up the expansion (FOIL)

    (x+2)(x+7)(x + 2)(x + 7)

    Every term in the first bracket must be multiplied by every term in the second.

  2. Multiply out the pairs

    =x2+7x+2x+14= x^2 + 7x + 2x + 14

    Multiply the First, Outer, Inner and Last terms to get four terms.

  3. Collect like terms

    =x2+9x+14= x^2 + 9x + 14

    Combine the x-terms 7x and 2x to get 9x.

Answer
x2+9x+14x^2 + 9x + 14
Question 3
2 marksintermediate
Expand and simplify (x1)(x9)(x - 1)(x - 9).
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Worked solution

  1. Set up the expansion (FOIL)

    (x1)(x9)(x - 1)(x - 9)

    Every term in the first bracket must be multiplied by every term in the second.

  2. First and Outer terms

    xx=x2,x(9)=9xx \cdot x = x^2, \quad x \cdot (-9) = -9x

    Multiply the first terms in each bracket, then the outer terms.

  3. Inner and Last terms

    (1)x=x,(1)(9)=9(-1) \cdot x = -x, \quad (-1) \cdot (-9) = 9

    Multiply the inner terms, then the last terms — watch the signs.

  4. Write out the four terms

    =x29xx+9= x^2 - 9x - x + 9

    Bring the four products together before simplifying.

  5. Collect like terms

    =x210x+9= x^2 - 10x + 9

    Combine the x-terms -9x and -x to get -10x.

  6. State the expansion

    =x210x+9= x^2 - 10x + 9

    This is the expansion written in standard form.

Answer
x210x+9x^2 - 10x + 9
Question 4
4 markshard
Expand and simplify (3x+4)(2x1)(3x + 4)(2x - 1).
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Worked solution

  1. Set up the expansion (FOIL)

    (3x+4)(2x1)(3x + 4)(2x - 1)

    Every term in the first bracket must be multiplied by every term in the second.

  2. First terms

    3x2x=6x23x \cdot 2x = 6x^2

    Multiply the first term of each bracket.

  3. Outer terms

    3x(1)=3x3x \cdot (-1) = -3x

    Multiply the two outer terms.

  4. Inner terms

    42x=8x4 \cdot 2x = 8x

    Multiply the two inner terms.

  5. Last terms

    4(1)=44 \cdot (-1) = -4

    Multiply the last term of each bracket.

  6. Write out all four terms

    (3x+4)(2x1)=6x23x+8x4(3x + 4)(2x - 1) = 6x^2 - 3x + 8x - 4

    List every product from FOIL.

  7. Collect like terms

    =6x2+5x4= 6x^2 + 5x - 4

    Combine the x-terms -3x and 8x to get 5x.

  8. Identify the leading term

    3x2x=6x23x \cdot 2x = 6x^2

    The highest power is x squared with coefficient 6.

  9. Check by substituting x=1x = 1

    (7)(1)=7(7)(1) = 7

    The original product and the expansion both give 7 when x=1x = 1, so the working checks out.

  10. State the final answer

    =6x2+5x4= 6x^2 + 5x - 4

    The expansion in expanded and simplified form.

Answer
6x2+5x46x^2 + 5x - 4
Question 5
6 markschallenging
Expand and simplify (2x+1)(2x+3)(x1)(2x + 1)(2x + 3)(x - 1).
Show worked solution

Worked solution

  1. Plan the expansion

    (2x+1)(2x+3)(x1)(2x + 1)(2x + 3)(x - 1)

    Expand two of the brackets first, then multiply the result by the third.

  2. Multiply the first two brackets

    (2x+1)(2x+3)=4x2+6x+2x+3(2x + 1)(2x + 3) = 4x^2 + 6x + 2x + 3

    Use FOIL on the first pair of brackets.

  3. Simplify the quadratic

    =4x2+8x+3= 4x^2 + 8x + 3

    Combine the x-terms 6x and 2x to get 8x.

  4. Set up the next multiplication

    (4x2+8x+3)(x1)(4x^2 + 8x + 3)(x - 1)

    Now multiply this quadratic by the remaining bracket (x - 1).

  5. Multiply the x-squared term

    4x2x=4x3,4x2(1)=4x24x^2 \cdot x = 4x^3, \quad 4x^2 \cdot (-1) = -4x^2

    Distribute the x-squared term across the bracket.

  6. Multiply the x term

    8xx=8x2,8x(1)=8x8x \cdot x = 8x^2, \quad 8x \cdot (-1) = -8x

    Distribute the middle term of the quadratic over the bracket.

  7. Multiply the constant term

    3x=3x,3(1)=33 \cdot x = 3x, \quad 3 \cdot (-1) = -3

    Distribute the constant term across the bracket.

  8. Write out all six terms

    =4x34x2+8x28x+3x3= 4x^3 - 4x^2 + 8x^2 - 8x + 3x - 3

    Collect every product before simplifying.

  9. Group the x-cubed term

    4x34x^3

    There is only one x-cubed term: 4x34x^3.

  10. Group the x-squared terms

    4x2+8x2=4x2-4x^2 + 8x^2 = 4x^2

    Add the two x-squared terms together.

  11. Group the x terms

    8x+3x=5x-8x + 3x = -5x

    Add the two x terms together.

  12. Group the constant term

    3-3

    The only constant term is -3.

  13. Write the expansion in standard form

    =4x3+4x25x3= 4x^3 + 4x^2 - 5x - 3

    Arrange the terms from the highest power of x to the lowest.

  14. Check by substituting x=1x = 1

    (3)(5)(0)=0(3)(5)(0) = 0

    Both the original product and the expansion give 0 when x=1x = 1.

  15. State the final answer

    =4x3+4x25x3= 4x^3 + 4x^2 - 5x - 3

    This is the fully expanded and simplified expression.

Answer
4x3+4x25x34x^3 + 4x^2 - 5x - 3

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