Hard GCSE Expanding binomials Questions

Challenging, exam-style GCSE Expanding binomials questions with worked solutions. Stretch yourself on the hardest coefficients, FOIL, negative terms, difference of two squares problems.

coefficientsFOILnegative termsdifference of two squaressquaring a binomialexpand and simplify
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Expand and simplify (2x+1)(2x+3)(x−1)(2x + 1)(2x + 3)(x - 1).
Show worked solution

Worked solution

  1. Plan the expansion

    (2x+1)(2x+3)(x−1)(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 6x6x and 2x2x to get 8x8x.

  4. Set up the next multiplication

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

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

  5. Multiply the x-squared term

    4x2⋅x=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

    8x⋅x=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

    3⋅x=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

    =4x3−4x2+8x2−8x+3x−3= 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 −3-3.

  13. Write the expansion in standard form

    =4x3+4x2−5x−3= 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 00 when x=1x = 1.

  15. State the final answer

    =4x3+4x2−5x−3= 4x^3 + 4x^2 - 5x - 3

    This is the fully expanded and simplified expression.

Answer
4x3+4x2−5x−34x^3 + 4x^2 - 5x - 3
Question 2
5 markschallenging
Expand and simplify (x+1)(x+2)(x−3)(x + 1)(x + 2)(x - 3).
Show worked solution

Worked solution

  1. Plan the expansion

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

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

  2. Multiply the first two brackets

    (x+1)(x+2)=x2+2x+x+2(x + 1)(x + 2) = x^2 + 2x + x + 2

    Use FOIL on the first pair of brackets.

  3. Simplify the quadratic

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

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

  4. Set up the next multiplication

    (x2+3x+2)(x−3)(x^2 + 3x + 2)(x - 3)

    Now multiply this quadratic by the remaining bracket (x - 3).

  5. Multiply the x-squared term

    x2⋅x=x3,x2⋅(−3)=−3x2x^2 \cdot x = x^3, \quad x^2 \cdot (-3) = -3x^2

    Distribute the x-squared term across the bracket.

  6. Multiply the x term

    3x⋅x=3x2,3x⋅(−3)=−9x3x \cdot x = 3x^2, \quad 3x \cdot (-3) = -9x

    Distribute the middle term of the quadratic over the bracket.

  7. Multiply the constant term

    2⋅x=2x,2⋅(−3)=−62 \cdot x = 2x, \quad 2 \cdot (-3) = -6

    Distribute the constant term across the bracket.

  8. Write out all six terms

    =x3−3x2+3x2−9x+2x−6= x^3 - 3x^2 + 3x^2 - 9x + 2x - 6

    Collect every product before simplifying.

  9. Group the x-cubed term

    x3x^3

    There is only one x-cubed term: x3x^3.

  10. Group the x-squared terms

    −3x2+3x2=0-3x^2 + 3x^2 = 0

    Add the two x-squared terms together.

  11. Group the x terms

    −9x+2x=−7x-9x + 2x = -7x

    Add the two x terms together.

  12. Group the constant term

    −6-6

    The only constant term is −6-6.

  13. Write the expansion in standard form

    =x3−7x−6= x^3 - 7x - 6

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

  14. Check by substituting x=1x = 1

    (2)(3)(−2)=−12(2)(3)(-2) = -12

    Both the original product and the expansion give −12-12 when x=1x = 1.

  15. State the final answer

    =x3−7x−6= x^3 - 7x - 6

    This is the fully expanded and simplified expression.

Answer
x3−7x−6x^3 - 7x - 6
Question 3
5 markschallenging
Expand and simplify (x−4)(x+4)(x+1)(x - 4)(x + 4)(x + 1).
Show worked solution

Worked solution

  1. Plan the expansion

    (x−4)(x+4)(x+1)(x - 4)(x + 4)(x + 1)

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

  2. Multiply the first two brackets

    (x−4)(x+4)=x2+4x−4x−16(x - 4)(x + 4) = x^2 + 4x - 4x - 16

    Use FOIL on the first pair of brackets.

  3. Simplify the quadratic

    =x2−16= x^2 - 16

    The x-terms 4x4x and −4x-4x cancel, so there is no x-term.

  4. Set up the next multiplication

    (x2−16)(x+1)(x^2 - 16)(x + 1)

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

  5. Multiply the x-squared term

    x2⋅x=x3,x2⋅1=x2x^2 \cdot x = x^3, \quad x^2 \cdot 1 = x^2

    Distribute the x-squared term across the bracket.

  6. Multiply the x term

    0⋅x=0,0⋅1=00 \cdot x = 0, \quad 0 \cdot 1 = 0

    The quadratic has no x-term, so this row contributes nothing.

  7. Multiply the constant term

    −16⋅x=−16x,−16⋅1=−16-16 \cdot x = -16x, \quad -16 \cdot 1 = -16

    Distribute the constant term across the bracket.

  8. Write out all six terms

    =x3+x2+0+0−16x−16= x^3 + x^2 + 0 + 0 - 16x - 16

    Collect every product before simplifying.

  9. Group the x-cubed term

    x3x^3

    There is only one x-cubed term: x3x^3.

  10. Group the x-squared terms

    x2+0=x2x^2 + 0 = x^2

    Add the two x-squared terms together.

  11. Group the x terms

    0−16x=−16x0 - 16x = -16x

    Add the two x terms together.

  12. Group the constant term

    −16-16

    The only constant term is −16-16.

  13. Write the expansion in standard form

    =x3+x2−16x−16= x^3 + x^2 - 16x - 16

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

  14. Check by substituting x=1x = 1

    (−3)(5)(2)=−30(-3)(5)(2) = -30

    Both the original product and the expansion give −30-30 when x=1x = 1.

  15. State the final answer

    =x3+x2−16x−16= x^3 + x^2 - 16x - 16

    This is the fully expanded and simplified expression.

Answer
x3+x2−16x−16x^3 + x^2 - 16x - 16
Question 4
6 markschallenging
Expand and simplify (2x+3)(3x−1)(x+2)(2x + 3)(3x - 1)(x + 2).
Show worked solution

Worked solution

  1. Plan the expansion

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

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

  2. Multiply the first two brackets

    (2x+3)(3x−1)=6x2−2x+9x−3(2x + 3)(3x - 1) = 6x^2 - 2x + 9x - 3

    Use FOIL on the first pair of brackets.

  3. Simplify the quadratic

    =6x2+7x−3= 6x^2 + 7x - 3

    Combine the x-terms −2x-2x and 9x9x to get 7x7x.

  4. Set up the next multiplication

    (6x2+7x−3)(x+2)(6x^2 + 7x - 3)(x + 2)

    Now multiply this quadratic by the remaining bracket (x + 2).

  5. Multiply the x-squared term

    6x2⋅x=6x3,6x2⋅2=12x26x^2 \cdot x = 6x^3, \quad 6x^2 \cdot 2 = 12x^2

    Distribute the x-squared term across the bracket.

  6. Multiply the x term

    7x⋅x=7x2,7x⋅2=14x7x \cdot x = 7x^2, \quad 7x \cdot 2 = 14x

    Distribute the middle term of the quadratic over the bracket.

  7. Multiply the constant term

    −3⋅x=−3x,−3⋅2=−6-3 \cdot x = -3x, \quad -3 \cdot 2 = -6

    Distribute the constant term across the bracket.

  8. Write out all six terms

    =6x3+12x2+7x2+14x−3x−6= 6x^3 + 12x^2 + 7x^2 + 14x - 3x - 6

    Collect every product before simplifying.

  9. Group the x-cubed term

    6x36x^3

    There is only one x-cubed term: 6x36x^3.

  10. Group the x-squared terms

    12x2+7x2=19x212x^2 + 7x^2 = 19x^2

    Add the two x-squared terms together.

  11. Group the x terms

    14x−3x=11x14x - 3x = 11x

    Add the two x terms together.

  12. Group the constant term

    −6-6

    The only constant term is −6-6.

  13. Write the expansion in standard form

    =6x3+19x2+11x−6= 6x^3 + 19x^2 + 11x - 6

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

  14. Check by substituting x=1x = 1

    (5)(2)(3)=30(5)(2)(3) = 30

    Both the original product and the expansion give 3030 when x=1x = 1.

  15. State the final answer

    =6x3+19x2+11x−6= 6x^3 + 19x^2 + 11x - 6

    This is the fully expanded and simplified expression.

Answer
6x3+19x2+11x−66x^3 + 19x^2 + 11x - 6
Question 5
5 markschallenging
Expand and simplify (x+5)(x−1)(x+2)(x + 5)(x - 1)(x + 2).
Show worked solution

Worked solution

  1. Plan the expansion

    (x+5)(x−1)(x+2)(x + 5)(x - 1)(x + 2)

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

  2. Multiply the first two brackets

    (x+5)(x−1)=x2−x+5x−5(x + 5)(x - 1) = x^2 - x + 5x - 5

    Use FOIL on the first pair of brackets.

  3. Simplify the quadratic

    =x2+4x−5= x^2 + 4x - 5

    Combine the x-terms -x and 5x5x to get 4x4x.

  4. Set up the next multiplication

    (x2+4x−5)(x+2)(x^2 + 4x - 5)(x + 2)

    Now multiply this quadratic by the remaining bracket (x + 2).

  5. Multiply the x-squared term

    x2⋅x=x3,x2⋅2=2x2x^2 \cdot x = x^3, \quad x^2 \cdot 2 = 2x^2

    Distribute the x-squared term across the bracket.

  6. Multiply the x term

    4x⋅x=4x2,4x⋅2=8x4x \cdot x = 4x^2, \quad 4x \cdot 2 = 8x

    Distribute the middle term of the quadratic over the bracket.

  7. Multiply the constant term

    −5⋅x=−5x,−5⋅2=−10-5 \cdot x = -5x, \quad -5 \cdot 2 = -10

    Distribute the constant term across the bracket.

  8. Write out all six terms

    =x3+2x2+4x2+8x−5x−10= x^3 + 2x^2 + 4x^2 + 8x - 5x - 10

    Collect every product before simplifying.

  9. Group the x-cubed term

    x3x^3

    There is only one x-cubed term: x3x^3.

  10. Group the x-squared terms

    2x2+4x2=6x22x^2 + 4x^2 = 6x^2

    Add the two x-squared terms together.

  11. Group the x terms

    8x−5x=3x8x - 5x = 3x

    Add the two x terms together.

  12. Group the constant term

    −10-10

    The only constant term is −10-10.

  13. Write the expansion in standard form

    =x3+6x2+3x−10= x^3 + 6x^2 + 3x - 10

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

  14. Check by substituting x=1x = 1

    (6)(0)(3)=0(6)(0)(3) = 0

    Both the original product and the expansion give 00 when x=1x = 1.

  15. State the final answer

    =x3+6x2+3x−10= x^3 + 6x^2 + 3x - 10

    This is the fully expanded and simplified expression.

Answer
x3+6x2+3x−10x^3 + 6x^2 + 3x - 10

Unlock 29 more Expanding binomials questions

Create a free account to work through every GCSE Expanding binomials question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Expanding binomials practice

Practice QuestionsWorked Solutions

Related Algebra topics

Algebraic fractionsEquation of a circleCoordinatesCubic and reciprocal graphsExponential and trig graphsFactorising ax2 + bx + c