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Algebra

Algebra

Expressions, substitution and expanding brackets

Core + Extended 7 min read

On this page
  1. In this note
  2. Formula bank
  3. The words
  4. Substitution
  5. Simplifying
  6. Expanding brackets
  7. Ways they test it
  8. Mistakes to avoid
  9. Summary
  10. Practice
  11. Can I do this?
  12. Quiz

In this note

Syllabus 2.1 and 2.2: expressions, substitution and expanding
  • 2.1Substitute numbers (including negatives and decimals) into expressions and formulas.
  • 2.2Simplify: collect like terms, multiply and divide terms with powers.
  • 2.2Expand one, two or three brackets, including squares like \((3x-2y)^{2}\).

Next: Linear equations and rearranging. Factorising undoes expanding: Factorization.

Formula bank

Single bracketLearn it\(a(b+c)=ab+ac\)the outside term multiplies EVERY inside term.Use it when one bracket with something in front.
Two bracketsLearn it\((a+b)(c+d)=ac+ad+bc+bd\)every term × every term (4 products), then collect.Use it when two brackets multiplied.
Squared bracketLearn it\((a+b)^{2}=a^{2}+2ab+b^{2}\), \((a-b)^{2}=a^{2}-2ab+b^{2}\)always a middle term.Use it when a bracket squared.
Index lawsLearn it\(x^{m}\times x^{n}=x^{m+n}\), \(x^{m}\div x^{n}=x^{m-n}\)Use it when multiplying or dividing terms.

The words

WordMeaningExample
termone part, separated by + or −\(5ab\), \(-3x^{2}\)
expressionterms, no = sign\(4a^{2}+5ab-2\)
equationtrue for SOME values\(2x+3=11\) (only \(x=4\))
formulaa rule linking quantities\(v=u+at\)
identitytrue for EVERY value (≡)\((x+1)^{2}\equiv x^{2}+2x+1\)
like termssame letters with the same powers\(5ab\) and \(-8ab\); not \(a^{2}\) and \(a\)

Substitution

General methodHow to substitute into a formula

  1. Write the formula out again.
  2. Replace each letter with its value IN BRACKETS, especially negatives.
  3. Work it out with BIDMAS: powers, then × and ÷, then + and −.

Example \(s=ut+\tfrac12at^{2}\). Find \(s\) when \(u=5\), \(a=-3\), \(t=2\).

Put each value in brackets; square only \(t\).

  1. \(5(2)+\tfrac12(-3)(2)^{2}\)
  2. \(=10-6\)

Answer 4

Try it yourself

1
\(v=u+at\). Find \(v\) when \(u=30\), \(a=-2\), \(t=7\).
Show answer

Answer 16

2
\(c=7b-a\). Find \(c\) when \(a=-4\), \(b=2\).
Hint
\(7(2)-(-4)\).
Show answer

Answer 18

3
Find \(2x^{2}-5x\) when \(x=-3\).
Hint
\(2(-3)^{2}=18\).
Show answer

Answer 33

Simplifying

General methodHow to simplify

  1. Collect like terms: add the numbers in front of terms with exactly the same letters and powers.
  2. Multiplying terms: multiply the numbers, add the powers of each letter.
  3. Dividing terms: divide the numbers, subtract the powers.

Example Simplify \(4a^{2}+5ab-2+3a^{2}-8ab+7\).

Collect the \(a^{2}\) terms, the \(ab\) terms and the numbers.

  1. \(4a^{2}+3a^{2}=7a^{2}\)
  2. \(5ab-8ab=-3ab\)
  3. \(-2+7=5\)

Answer \(7a^{2}-3ab+5\)

Try it yourself

1
Simplify \(5x-3y-2x+7y\).
Show answer

Answer \(3x+4y\)

2
Simplify \(3x\times5x^{2}\).
Show answer

Answer \(15x^{3}\)

3
Simplify \(\dfrac{12a^{5}}{3a^{2}}\).
Show answer

Answer \(4a^{3}\)

Expanding brackets

Expanding = multiplying out. Each term in one bracket multiplies every term in the other. A grid keeps track: for \((3x+y)(x-4y)\) the four boxes are \(3x^{2}\), \(-12xy\), \(xy\), \(-4y^{2}\), which collect to \(3x^{2}-11xy-4y^{2}\).

General methodHow to expand and simplify

  1. One bracket: multiply the outside term by each inside term (watch the signs: − × − = +).
  2. Two brackets: multiply every term in the first by every term in the second (4 products).
  3. A square: write it as two brackets, \((x-5)^{2}=(x-5)(x-5)\). There is always a middle term.
  4. Three brackets: expand two of them first, then multiply by the third.
  5. Collect like terms. Check by putting x = 1 into the question and your answer.

Example Expand and simplify \((2x-3)(x+5)\).

Four products, then collect the two x terms.

  1. \(2x^{2}+10x-3x-15\)
  2. \(=2x^{2}+7x-15\)

Answer \(2x^{2}+7x-15\)

Try it yourself

1
Expand and simplify \((x+4)(x-6)\).
Show answer

Answer \(x^{2}-2x-24\)

2
Expand and simplify \((2x+1)^{2}\).
Hint
\((2x+1)(2x+1)\).
Show answer

Answer \(4x^{2}+4x+1\)

3
Expand and simplify \(4(x+3)-2(x-1)\).
Hint
\(-2\times-1=+2\).
Show answer

Answer \(2x+14\)

Ways they test it

Type 1 Substitute into a formula

"\(s=ut+\tfrac12at^{2}\). Find s when u = 3.5, t = 6, a = −1.2." (2 marks)

Spot it values given for letters.

  1. Brackets round each value.
  2. BIDMAS.

For example \(3.5(6)+\tfrac12(-1.2)(6)^{2}\)

  1. \(21-21.6\)

Answer −0.6

Type 2 Simplify

"Simplify \(3p-t-p-4t\)." (1 mark)

Spot it one expression, no brackets.

  1. Collect like terms.

For example \(3p-p-t-4t\)

    Answer \(2p-5t\)

    Type 3 Expand and simplify

    "Expand and simplify \(3(2x-y)-2(x-4y)\)." (2 marks)

    Spot it brackets, a minus in front of one.

    1. The minus multiplies BOTH terms.
    2. Collect.

    For example \(6x-3y-2x+8y\)

      Answer \(4x+5y\)

      Type 4 Three brackets

      "Expand and simplify \((x-1)(x+3)(2x-5)\)." (3 marks)

      Spot it three brackets: a cubic answer.

      1. Expand two.
      2. Multiply by the third (6 products).
      3. Collect.

      For example \((x^{2}+2x-3)(2x-5)\)

      1. \(2x^{3}-5x^{2}+4x^{2}-10x-6x+15\)

      Answer \(2x^{3}-x^{2}-16x+15\)

      Type 5 Identity

      "\((x+a)(x-1)(3x+2)\equiv3x^{3}+bx^{2}+cx+8\). Find a, b and c." (3 marks)

      Spot it the ≡ sign and unknown letters in the answer.

      1. Compare the number terms first.
      2. Expand and compare each power of x.

      For example constant: \(-2a=8\)

      1. \(a=-4\)
      2. \((x-4)(x-1)(3x+2)=3x^{3}-13x^{2}+2x+8\)

      Answer \(a=-4,\ b=-13,\ c=2\)

      Mistakes to avoid

      • ✗ \(-2(x-4y)=-2x-8y\)✓ \(-2x+8y\)− × − = +.
      • ✗ \((x-5)^{2}=x^{2}+25\)✓ \(x^{2}-10x+25\)There is always a middle term.
      • ✗ \(7b-a\) with \(a=-4\): \(14-4\)✓ 14 + 4 = 18Put negatives in brackets.
      • ✗ \(4a^{2}+3a=7a^{3}\)✓ can't be combinedNot like terms.

      Summary

      • Substitute in brackets, then BIDMAS.
      • Like terms = same letters with the same powers.
      • Expanding: every term × every term; a squared bracket has a middle term.
      • Three brackets: two first, then the third.

      If you see …

      • "expand and simplify"multiply out, then collect like terms.
      • a minus in front of a bracketchange the sign of every term inside.

      Practice

      Practice: easy, medium, hard

      Easy

      1
      Simplify \(3p-t-p-4t\).
      Show answer

      Answer \(2p-5t\)

      2
      Expand \(3x(2x-4y)\).
      Show answer

      Answer \(6x^{2}-12xy\)

      3
      Expand and simplify \((x+5)(x-2)\).
      Show answer

      Answer \(x^{2}+3x-10\)

      Medium

      1
      \(s=ut+\tfrac12at^{2}\). Find \(s\) when \(u=3.5\), \(t=6\), \(a=-1.2\).
      Guide me

      What is it asking? s

      What do we know? brackets round each value

      Plan square only t

      Show working
      1. \(3.5(6)+\tfrac12(-1.2)(36)\)
      2. \(21-21.6=-0.6\)

      Answer −0.6

      2
      Expand and simplify \(3(2x-y)-2(x-4y)\).
      Guide me

      What is it asking? an expression

      What do we know? the −2 multiplies both terms

      Plan collect

      Show working
      1. \(6x-3y-2x+8y\)
      2. \(4x+5y\)

      Answer \(4x+5y\)

      3
      Expand and simplify \((3x-2y)^{2}\).
      Guide me

      What is it asking? three terms

      What do we know? two brackets

      Plan middle: −6xy twice

      Show working
      1. \((3x-2y)(3x-2y)\)
      2. \(9x^{2}-6xy-6xy+4y^{2}\)

      Answer \(9x^{2}-12xy+4y^{2}\)

      Hard

      1
      Expand and simplify \((x-1)(x+3)(2x-5)\).
      Guide me

      What is it asking? a cubic

      What do we know? two brackets first

      Plan then × (2x − 5)

      Show working
      1. \((x^{2}+2x-3)(2x-5)\)
      2. \(2x^{3}-5x^{2}+4x^{2}-10x-6x+15\)
      3. \(2x^{3}-x^{2}-16x+15\)

      Answer \(2x^{3}-x^{2}-16x+15\)

      PartAnswer / workingMarksGuidance
      \(x^{2}+2x-3\)M1
      six correct termsM1
      \(2x^{3}-x^{2}-16x+15\)A1
      Total: 3 marks
      2
      Expand and simplify \((x-2)(x+3)(2x+1)\).
      Guide me

      What is it asking? a cubic

      What do we know? \((x-2)(x+3)=x^{2}+x-6\)

      Plan × \((2x+1)\)

      Show working
      1. \((x^{2}+x-6)(2x+1)\)
      2. \(2x^{3}+x^{2}+2x^{2}+x-12x-6\)

      Answer \(2x^{3}+3x^{2}-11x-6\)

      PartAnswer / workingMarksGuidance
      \(x^{2}+x-6\)M1
      six correct termsM1
      \(2x^{3}+3x^{2}-11x-6\)A1
      Total: 3 marks
      3
      \((x+a)(x-1)(3x+2)\equiv3x^{3}+bx^{2}+cx+8\). Find \(a\), \(b\) and \(c\).
      Guide me

      What is it asking? three numbers

      What do we know? constant term: \(a\times(-1)\times2=8\)

      Plan then expand and compare

      Show working
      1. \(-2a=8\Rightarrow a=-4\)
      2. \((x^{2}-5x+4)(3x+2)=3x^{3}-13x^{2}+2x+8\)

      Answer \(a=-4,\ b=-13,\ c=2\)

      PartAnswer / workingMarksGuidance
      \(a=-4\)B1
      expansionM1
      \(b=-13,\ c=2\)A1
      Total: 3 marks

      Can I do this?

      Anything unticked: revise that section first. All ticked: take the quiz.

      Quiz

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