Expanding, index laws and algebraic fractions
What this lesson covers
- An expression can be rewritten in many equivalent forms, and each rewrite must keep its value for every value of the variable.
- Index laws follow from what a power means: a product of repeated factors.
What simplifying is allowed to do
- A rewrite must give the same value for every x.
- One value that disagrees proves it wrong.
- One value that agrees proves nothing yet.
| Expression | At x = 2 | At x = 5 |
|---|---|---|
| 2(x + 3) − x | 8 | 11 |
| x + 6 | 8 | 11 |
| x² and 2x | 4 and 4 | 25 and 10 |
| Equivalent? | first pair: yes, by expanding | second pair: no |
A squared bracket
- Squared means the bracket times itself.
- Four products, then the two middle terms join.
- x squared plus six x plus nine.
| Step | Working | Why |
|---|---|---|
| (x + 3)² | (x + 3)(x + 3) | squared means times itself |
| Four products | x² + 3x + 3x + 9 | every term by every term |
| Collect | x² + 6x + 9 | the middle two join |
| Pattern | middle is 2 × 3x | last is 3 squared |
Why (x + 3)² is not x² + 9
- At x = 0 both give nine. That proves nothing.
- At x = 2 one gives 25 and the other 13.
- One mismatch is enough: they are not equivalent.
| Test | (x + 3)² | x² + 9 |
|---|---|---|
| x = 2 | 25 | 13 |
| x = 0 | 9 | 9 |
| Verdict | one mismatch is enough | not equivalent |
| What went missing | the middle term, 6x |
A coefficient and a minus inside
- Write it as two brackets first.
- Minus five times minus five is plus twenty-five.
- Check with x = 1: both sides give nine.
| Step | Working | Note |
|---|---|---|
| (2x − 5)² | (2x − 5)(2x − 5) | |
| Four products | 4x² − 10x − 10x + 25 | minus times minus |
| Collect | 4x² − 20x + 25 | |
| Check x = 1 | (−3)² = 9 | 4 − 20 + 25 = 9 |
Expanding, then subtracting
- Expand each product on its own.
- Keep the second in a bracket: the minus is for all of it.
- The x squared terms cancel, leaving 8x − 17.
| Step | Working | Result |
|---|---|---|
| First product | x² + 2x − 8 | |
| Second product | (x² − 6x + 9) | keep the bracket |
| Subtract it | every sign inside flips | −x² + 6x − 9 |
| Collect | the squares cancel | 8x − 17 |
| Check x = 1 | −5 − 4 = −9 | 8 − 17 = −9 |
Three brackets, two at a time
- Multiply the first two brackets.
- Then every term of that by every term of the third.
- Collect, then check with x = 1.
| Step | Working |
|---|---|
| Two at a time: (x + 1)(x − 2) | x² − x − 2 |
| x² times (x + 3) | x³ + 3x² |
| −x − 2 times (x + 3) | −x² − 3x − 2x − 6 |
| Collect like terms | x³ + 2x² − 5x − 6 |
| Check x = 1: 2 × (−1) × 4 = −8 | 1 + 2 − 5 − 6 = −8 |
Where the index laws come from
- A power counts repeated factors.
- Multiplying joins the lists, so the indices add.
- Dividing cancels factors, so the indices subtract.
| Power | Means | Count |
|---|---|---|
| a³ | a × a × a | 3 factors |
| a² | a × a | 2 factors |
| a³ × a² | a × a × a × a × a | a⁵ |
| (a³)² | a³ × a³ | a⁶ |
| a⁵ ÷ a² | two pairs cancel | a³ |
Multiplying with coefficients
- Numbers multiply as numbers.
- Each letter adds its own indices.
- A letter on its own has index one.
| Part | Working | Gives |
|---|---|---|
| Numbers | 3 × 4 | 12 |
| The a terms | a² × a¹ | a³ |
| The b terms | b¹ × b³ | b⁴ |
| Together | multiply, then add powers | 12a³b⁴ |
A power of a bracket
- The square applies to everything inside.
- So the two is squared as well.
- 4x⁶, not 2x⁶.
| Step | Working | Gives |
|---|---|---|
| (2x³)² | bracket times itself | |
| Numbers | 2 × 2 | 4 |
| Powers | x³ × x³ | x⁶ |
| Answer | 4x⁶ | not 2x⁶ |
Dividing with coefficients
- Numbers divide as numbers.
- Each letter subtracts its indices.
- Three x cubed y.
| Part | Working | Gives |
|---|---|---|
| Numbers | 12 ÷ 4 | 3 |
| The x terms | x⁵ ÷ x² | x³ |
| The y terms | y² ÷ y¹ | y |
| Answer | divide, then subtract powers | 3x³y |
Zero and negative indices
- Any non-zero number divided by itself is one, so x⁰ = 1.
- Three factors of x left below: x⁻³ = 1/x³.
- The negative index belongs to x only, not the 3.
| Division | Working | Result |
|---|---|---|
| x⁴ ÷ x⁴ | non-zero ÷ itself is 1 | the rule says x⁰ |
| So | x⁰ = 1 | for any x ≠ 0 |
| x² ÷ x⁵ | three factors of x below | 1/x³ |
| The rule says | x⁻³ | same as 1/x³ |
| 6x² ÷ 2x⁵ | 3x⁻³ | 3/x³: the 3 stays up |
An algebraic fraction
- Factorise first. Only whole factors cancel.
- x + 3 is on top and bottom, so it divides out.
- x − 3, as long as x is not minus three.
| Step | Working | Reason |
|---|---|---|
| Factorise top | (x + 3)(x − 3) | difference of squares |
| Common factor | (x + 3) top and bottom | divide both by it |
| What is left | x − 3 | for x ≠ −3 |
| Wrong move | cancel the x terms first | x² is a term, not a factor |
| Check x = 5 | 16 ÷ 8 = 2 | 5 − 3 = 2 |
Say why each step is allowed
- Factorise the top and the bottom, and say how.
- Name the common factor before you cancel it.
- Finish with a check at a number.
| Step | Working | Reason |
|---|---|---|
| Top | (x + 2)(x + 3) | two numbers: × 6, + 5 |
| Bottom | (x + 2)(x − 2) | difference of squares |
| Cancel (x + 2) | (x + 3)/(x − 2) | x ≠ 2 or −2 |
| Check x = 3 | 30 ÷ 5 = 6 | 6 ÷ 1 = 6 |
What to take away
- Seven lines, and the first is why the checks work.
| Equivalent forms | equal for every value of x |
| (x + a)² | x² + 2ax + a², never x² + a² |
| Minus a bracket | flips every sign inside it |
| Multiply or divide powers | add or subtract the indices |
| Power of a power | multiply the indices |
| x⁰ = 1 and x⁻ⁿ = 1/xⁿ | both come from dividing |
| Algebraic fractions | factorise, then cancel factors |
Your turn: four questions
- Pause here and write all four out, with reasons.
- Question 1: keep the second product in a bracket.
- Question 3: factorise the top and the bottom first.
| Question | |
|---|---|
| 1 | Expand and simplify (x − 4)² − (x + 1)(x − 3). |
| 2 | Simplify 10a³b ÷ 5a⁵b, using positive indices. |
| 3 | Simplify (x² − 16)/(x² + x − 12). |
| 4 | Use x = 1 to show (x − 5)² ≠ x² − 25, then expand it. |
Show answer
1. −6x + 19. (x − 4)² is x² − 8x + 16, and (x + 1)(x − 3) is x² − 2x − 3. Subtracting the whole bracket flips its signs, the x² terms cancel, and −8x + 2x = −6x, 16 + 3 = 19. 2. 2/a². Ten divided by five is 2, a³ ÷ a⁵ is a⁻², and b ÷ b is b⁰ = 1. A negative index means one over, so 2a⁻² = 2/a². 3. (x − 4)/(x − 3). The top is (x + 4)(x − 4), the bottom is (x + 4)(x − 3), and the common factor (x + 4) cancels, for x ≠ −4 and x ≠ 3. 4. At x = 1, (x − 5)² = 16 but x² − 25 = −24, so they are not equivalent. Correctly, (x − 5)² = x² − 10x + 25, which is 16 at x = 1.
The answers
- Minus six x plus nineteen.
- Two over a squared.
- x − 4 over x − 3.
- Sixteen against minus 24. It is x² − 10x + 25.
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