Mathematics notes, chapter 2
Factorisation and Algebraic Fractions
Textbook Mathematics Form 2 (DLP), pages 18–41
What this chapter covers
- Expand one or two algebraic expressions, simplify combined operations including expansion and solve problems (2.1)
- Relate expansion to factors and factorisation and list the factors of an algebraic product (2.2)
- Factorise expressions using the HCF, difference of squares, perfect squares, cross multiplication and grouping, and solve problems (2.2)
- Add and subtract algebraic fractions after expanding and factorising (2.3)
- Multiply and divide algebraic fractions and perform combined operations involving expansion and factorisation (2.3)
2.1Expansion
Expansion means writing the product of one or two expressions in brackets as a sum of terms. Every term inside a bracket must be multiplied by every term outside it.
Expansion with one bracket
Use the Distributive Law: a(b + c) = ab + ac and a(b − c) = ab − ac. Watch the signs: (+)×(+) = +, (+)×(−) = −, (−)×(+) = −, (−)×(−) = +.
Example 1 (textbook p. 21–22): expand
(a) 6(3 + 4w) = 18 + 24w(b) 3r(r − 2s) = 3r2 − 6rs
(c) −5b(a + 3) = −5ab − 15b
(d) −2y3(9y − 3z + 6x) = −6y2 + 2yz − 4xy
Forgetting the sign in front of the bracket. In (c) the −5b multiplies both terms, so +3 becomes −15b, not +15b.
Expansion with two brackets
Multiply each term in the first bracket by each term in the second bracket, then collect the like terms. A rectangle shows why: the area of (x + 3) by (x + 2) is the sum of four smaller areas.
Example 2 (textbook p. 23–24): expand
(a) (y + 1)(y − 3) = y(y − 3) + 1(y − 3) = y2 − 3y + y − 3 = y2 − 2y − 3(b) (4 + 3r)(2 + r) = 8 + 4r + 6r + 3r2 = 3r2 + 10r + 8
(c) (3r + 4s)(r − 2s) = 3r2 − 6rs + 4rs − 8s2 = 3r2 − 2rs − 8s2
(d) (3p + 2)2 = (3p + 2)(3p + 2) = 9p2 + 6p + 6p + 4 = 9p2 + 12p + 4
Write the answer from the highest power to the lowest power, and remember that sr = rs, so 4sr and −6rs are like terms.
Three results worth remembering
| Expansion | Result |
|---|---|
| (a + b)2 | a2 + 2ab + b2 |
| (a − b)2 | a2 − 2ab + b2 |
| (a + b)(a − b) | a2 − b2 |
(a + b)2 is not a2 + b2. The two middle terms 2ab are always there. Test with a = 3, b = 2: (3 + 2)2 = 25 but 32 + 22 = 13.
Expand (a) (2x + 3)(x − 5) (b) (x − 7)2
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(a) 2x2 − 10x + 3x − 15 = 2x2 − 7x − 15
(b) x2 − 14x + 49
(b) x2 − 14x + 49
Combined operations including expansion
Follow BODMAS: brackets (expand them), then multiplication and division, then addition and subtraction. Collect like terms last.
Example 3 (textbook p. 25): simplify
(a) (3w − 2)(4w − 1) − 10w = 12w2 − 3w − 8w + 2 − 10w = 12w2 − 21w + 2(b) (r − 3t)2 + 4rt = r2 − 6rt + 9t2 + 4rt = r2 − 2rt + 9t2
(c) (x + y)(x − y) + x(x − 2y) = x2 − y2 + x2 − 2xy = 2x2 − 2xy − y2
Solving problems with expansion
Example 4 (textbook p. 25): area of a carpet
A carpet is (3r − 2) m long and (r + 1) m wide. Area = (3r − 2)(r + 1) = 3r2 + 3r − 2r − 2 = (3r2 + r − 2) m2.Follow the four steps of the textbook: understand the problem, plan the strategy, carry it out, then state the conclusion with units.
2.2Factorisation
Factorisation writes an expression as a product of factors. It is the reverse of expansion: a(a + b) = a2 + ab is expansion read left to right and factorisation read right to left.
Factors and HCF
The factors of 3p are 1, 3, p and 3p. A common factor divides two or more terms exactly. The highest common factor (HCF) is the largest of them.
Example 1 (textbook p. 28): HCF of 6h and 4gh
6h = 2 × 3 × h and 4gh = 2 × 2 × g × h. The common factors are 1, 2, h and 2h, so the HCF is 2h.Choosing a method
| What you see | Method | Example |
|---|---|---|
| Every term shares a factor | Take out the HCF | 8x + 12x2 = 4x(2 + 3x) |
| Two perfect squares with a minus | Difference of squares: a2 − b2 = (a + b)(a − b) | 9m2 − 100 = (3m + 10)(3m − 10) |
| a2 ± 2ab + b2 | Perfect square | x2 − 6x + 9 = (x − 3)2 |
| Three terms ax2 + bx + c | Cross multiplication | m2 − 2m − 8 = (m + 2)(m − 4) |
| Four terms | Group in pairs | pq + qr + ps + rs = (q + s)(p + r) |
Always check by expanding your answer. You should get back the original expression.
Common factor and difference of squares
Example 2 (textbook p. 28–29): factorise
(a) 7m + 21m2 = 7m(1 + 3m)(b) b2 − 1 = (b + 1)(b − 1)
(c) 3y2 − 147 = 3(y2 − 49) = 3(y + 7)(y − 7)
(d) 5k2 − 80 = 5(k2 − 16) = 5(k + 4)(k − 4)
Stopping too early. 3y2 − 147 is not a difference of perfect squares until you take out the HCF 3. Always take out the common number first.
Cross multiplication for ax2 + bx + c
For x2 + 6x + 8: find two numbers whose product is 8 (the constant) and whose sum is 6 (the coefficient of x). They are 2 and 4.
Example 3 (textbook p. 30–31): factorise
(a) m2 − 2m − 8: product −8, sum −2, so −4 and +2. Answer (m + 2)(m − 4).(b) 2m2 + 7m + 6: try (2m + 3)(m + 2). Cross products 4m + 3m = 7m. Correct.
(c) −2y2 − 9y + 5 = (2y − 1)(−y − 5)
When the middle term does not match, change the pairing of the factors and try again. Each attempt is a "trial" in the textbook.
Four terms: grouping
Example 4 (textbook p. 32): factorise pq + qr + ps + rs
Group: (pq + qr) + (ps + rs) = q(p + r) + s(p + r) = (q + s)(p + r).Factorise (a) x2 − 100 (b) x2 + 2x − 15 (c) 6xy + 2y + 9x + 3
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(a) (x + 10)(x − 10)
(b) numbers with product −15 and sum 2 are 5 and −3: (x + 5)(x − 3)
(c) 2y(3x + 1) + 3(3x + 1) = (2y + 3)(3x + 1)
(b) numbers with product −15 and sum 2 are 5 and −3: (x + 5)(x − 3)
(c) 2y(3x + 1) + 3(3x + 1) = (2y + 3)(3x + 1)
Solving problems with factorisation
Example 5 (textbook p. 32): flooded field
A field has area (4x2 + 16x) m2 and width 4x m. Its length is (4x2 + 16x) ÷ 4x = 4x(x + 4) ÷ 4x = (x + 4) m. Two flooded right-angled triangles have base 4x ÷ 2 = 2x, so their area is 2 × 12 × 2x × (x + 4) = 2x2 + 8x. Dry area = 4x2 + 16x − (2x2 + 8x) = (2x2 + 8x) m2.2.3Algebraic expressions and laws of basic arithmetic operations
An algebraic fraction has an algebraic expression in the numerator, the denominator or both. The rules are the same as for number fractions, with one extra step: factorise first.
Expanding and factorising expressions
Before you add or subtract fractions, you may need to expand a bracket and then factorise the result.
Example 1 (textbook p. 34): simplify
(a) 2x2 − 2(4x + 5) = 2x2 − 8x − 10 = 2(x2 − 4x − 5) = 2(x − 5)(x + 1)(b) 4w(w − 2) − 5 = 4w2 − 8w − 5 = (2w − 5)(2w + 1)
Adding and subtracting
Make the denominators the same, using the LCM, then add or subtract the numerators.
Example 2 (textbook p. 34–35): simplify
(a) 34y − 12y = 34y − 24y = 14y(b) x+25w − x−55w = x + 2 − (x − 5)5w = 75w
(c) 14p + 46p: LCM = 12p, so 312p + 812p = 1112p
(d) 5m4r − m14rs: LCM = 28rs, so 35ms28rs − 2m28rs = 35ms − 2m28rs
Subtracting a bracket. In (b) the minus sign changes both terms of (x − 5): −(x − 5) = −x + 5. Write the numerator in a bracket before you subtract.
Simplify a3 + 2a5
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5a15 + 6a15 = 11a15
Multiplying and dividing
Factorise every numerator and denominator, change ÷ to × the upside-down fraction, then cancel the common factors.
Example 3 (textbook p. 36): simplify
(a) 2p + 4p2 − 4 = 2(p + 2)(p + 2)(p − 2) = 2p − 2(b) (h + k)2h2 − k2 × 6k − 3h2k − h = (h + k)(h + k)(h + k)(h − k) × 3(2k − h)2k − h = 3(h + k)h − k
(c) 5aa + 2b ÷ 2ab3a + 6b = 5aa + 2b × 3(a + 2b)2ab = 152b
Cancelling terms instead of factors. x + 22 is not x + 1. Only things that are multiplied can be cancelled, so factorise first. Also a − b = −(b − a), but (a − b)2 = (b − a)2.
Simplify x + 2x2 − 4
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x + 2(x + 2)(x − 2) = 1x − 2
Combined operations
Expand and factorise as needed, then add, subtract, multiply or divide. Simplify each fraction before you combine them.
Example 4 (textbook p. 37, with (b) a similar question): simplify
(a) 9k2 − 12k + 4(3k + 2)(3k − 2) = (3k − 2)(3k − 2)(3k + 2)(3k − 2) = 3k − 23k + 2(b) 2x + 1 + x − 1x2 − 1 = 2x + 1 + x − 1(x + 1)(x − 1) = 2x + 1 + 1x + 1 = 3x + 1
Perfect squares to know by heart: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. They make difference of squares easy to spot.
Simplify x2 − 9x + 3 × 2x − 3
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(x + 3)(x − 3)x + 3 × 2x − 3 = 2