1. Introduction
Expressions such as (x + 5)(x − 2) were met in the module on Brackets. There, the emphasis was on the expansion of such expressions, which in this case would be x2 + 3x − 10. There are many instances when the reverse of this procedure, i.e. factorising, is required. This section begins with some simple examples.
Example 1: Factorise the following expressions.
Solution: This is easy since 7x − x2 = x(7 − x)
Solution: In this case the largest common factor is 2ab so:
Here are some exercises for you to try.
Exercise 1: Factorise the following expressions as far as possible.
Solution: The only common factor of the two terms is x, so:
Solution: Again, the two terms in the expression have only the common factor x, so:
Solution: Here, the only common factor is y, so:
Solution: In this case, the largest common factor is 2ax2, so:
Solution: Here the largest common factor is a2b, so this factorises as:
2a3b + 5a2b2 = a2b(2a + 5b)
Solution: The largest common factor is xy, so:
Click on questions to reveal their solutions
Quiz 1: Which of the expressions below is the full factorisation of the expression: 16a − 2a2?
Explanation: Here 2 is a factor of both terms, but so is a. Thus:
Quiz 2: Which of the expressions below is the full factorisation of the expression: ab2c − a2bc3 + 2abc2?
Explanation: The largest common factor in this case is: a × b × c = abc. Thus:
| ab2c − a2bc3 + 2abc2 | = | (abc × b) − (abc × ac2) + (abc × 2c) |
| = | abc(b − ac2 + 2c) |
2. Further Expressions
Each of the previous expressions may be factored in a single operation. Many examples require more than one such operation. Here is a worked example of this type.
Example 2: Factorise the following expressions as far as possible:
Solution: Note that a is a factor of the first two terms, and b is a factor of the second two. Thus:
| ax + ay + bx + by | = | a(x + y) + b(x + y) |
The expression in this form consists of a sum of two terms, each of which has the common factor (x + y), so it may be further factorised. Thus:
| ax + ay + bx + by | = | a(x + y) + b(x + y) |
| = | (a + b)(x + y) |
Solution: Here 3x is a factor of the first two terms and y is a factor of the second two. Thus:
| 6ax − 3bx + 2ay − by | = | 3x(2a − b) + y(2a − b) |
| = | (3x + y)(2a − b) |
- taking out (2a − b) as a common factor.
Exercise 2: Factorise each of the following as fully as possible.
Solution: We proceed as follows:
| xb + xc + yb + xc | = | x(b + c) + y(b + c) |
| = | (x + y)(b + c) |
Solution:
| ah − ak + bh − bk | = | a(h − k) + b(h − k) |
| = | (a + b)(h − k) |
Solution:
| hs + ht + ks + kt | = | h(s + t) + k(s + t) |
| = | (h + k)(s + t) |
Solution:
| 2mh−2mk + nh−nk | = | 2m(h−k) + n(h−k) |
| = | (2m + n)(h − k) |
Solution:
| 6ax + 2bx + 3ay + by | = | 2x(3a + b) + y(3a + b) |
| = | (2x + y)(3a + b) |
Solution:
| ms + 2mt2 − ns − 2nt2 | = | m(s + 2t2) − n(s + 2t2) |
| = | (m − n)(s + 2t2) |
Click on questions to reveal their solutions
Quiz 3: Which of the following is the factorisation of the expression: 2ax−6ay−bx + 3by ?
Explanation: Noting that 2a is a factor of the first two terms and −b is a factor of the second two, we have:
| 2ax − 6ay − bx + 3by | = | 2a(x − 3y) − b(x − 3y) |
| = | (2a − b)(x − 3y) |
3. Quadratic Expressions
A quadratic expression is one of the form ax2 + bx + c, with a, b, c being some numbers. When faced with a quadratic expression it is often, but not always, possible to factorise it by inspection. To get some insight into how this is done it is worthwhile looking at how such an expression is formed.
Suppose that a quadratic expression can be factored into two linear terms, say (x + d) and (x + e), where d,e are two numbers. Then the quadratic is
| (x + d)(x + e) | = | x2 + xe + xd + de |
| = | x2 + (e + d)x + de | |
| = | x2 + (d + e)x + de |
Notice how it is formed. The coefficient of x is (d + e), which is the sum of the two numbers in the linear terms (x + d) and (x + e). The final term, the one without an x, is the product of those two numbers. This is the information which is used to factorise by inspection.
Example 3: Factorise the following expressions:
Solution: The only possible factors of 7 are 1 and 7, and these do add up to 8, so:
| x2 + 8x + 7 | = | (x + 7)(x + 1) |
Checking this using FOIL (see the module on Brackets):
| F | O | I | L | |||||
| (x + 7)(x + 1) | = | x2 | + | x × 1 | + | x × 7 | + | 7 × 1 |
| = | x2 | + | 8x | + | 7 | |||
Solution: Here the term independent of x (i.e. the one without an x) is negative, so the two numbers must be opposite in sign. The obvious contenders are 3 and −5, or −3 and 5. The first pair can be ruled out as their sum is −2. The second pair sum to +2, which is the correct coefficient for x. Thus:
| y2 + 2y − 15 | = | (y − 3)(y + 5) |
Here are some exercises for you to try.
Exercise 3: Factorise the following into linear factors:
Solution: Since 10 has the factors 5 and 2, and their sum is 7:
| (x + 5)(x + 2) | = | x2 + 2x + 5x + 10 |
| = | x2 + 7x + 10 |
Solution: There are several ways of factorising 12, but on closer inspection the only factors that work are 4 and 3. This leads to the following:
| (x + 4)(x + 3) | = | x2 + 3x + 4x + 12 |
| = | x2 + 7x + 12 |
Solution: There are several different possible factors for 24, but only one pair, 8 and 3, add up to 11. Thus:
| (y + 8)(y + 3) | = | y2 + 3y + 8y + 24 |
| = | y2 + 11y + 24 |
Solution: There are several different possible factors for 24 but only one pair, 6 and 4, add up to 10. Since the coefficient of y is negative, and the constant term is positive, the required numbers this time are −6 and −4.
| (y − 6)(y − 4) | = | y2 − 4y − 6y + (−6)(−4) |
| = | y2 − 10y + 24 |
Solution: The constant term in this case is negative. Since this is the product of the numbers required, they must have opposite signs, i.e. one is positive and one is negative. In that case, the number in front of the x must be the difference of these two numbers. On inspection, 5 and 2 have product 10 and difference 3. Since the x term is negative, the larger number must be negative.
| (z − 5)(z + 2) | = | z2 + 2z − 5z + (−5 × 2) |
| = | z2 − 3z − 10 |
Solution: This is an example of a perfect square. These are mentioned in the module on Brackets. The factors of 16 in this case are −4 and −4.
| (a−4)2 | = | (a−4)(a−4) |
| = | a2 − 4a − 4a + (−4) × (−4) | |
| = | a2 − 8a + 16 |
Click on questions to reveal their solutions
Quiz 4: Which of the following is the factorisation of the expression: z2 − 6z + 8 ?
Explanation: Here the two numbers have product 8, so a possible choice is 2 and 4. However, their sum in this case is 6, whereas the sum required is −6. Taking the pair to be −2 and −4 will give the same product, +8, but with the correct sum. Thus:
| z2 − 6z + 8 | = | 2a(x − 3y) − b(x − 3y) |
| = | (z − 2)(z − 4) |
This can be checked by expanding the brackets.
4. Quiz on Factorisation
Factorise each of the following expressions and choose the solution from the options given