Topic 12 Factorisation 1. How to find the greatest common factors of an algebraic expression. Definition: A factor of a number is an integer that divides the number exactly. So for example, the factors of 12 are 1,2,3,4,6 and 12. If we write down the factors of two numbers then the factors which belong to both groups are called the common factors. For example: The factors of 12 are 1,2,3,4,6, and 12 The factors of 20 are 1,2,4,5,10 and 20 The common factors of 12 and 20 are 1,2 and 4 The Greatest Common Factor (GCF) of two numbers is the largest of the common factors. For example, the greatest common factor of 12 and 20 is 4. It is often more convenient to simplify the process of finding the GCF mentally by going through the factors of each and finding the largest factor that belongs to both. If the only common factor of two numbers is 1 then that we say that the two numbers are mutually prime or prime for short. Example 1: Find the GCF of the following numbers. (a) 20 and 30 (b) 12 and 24 (c) 16 and 44 (d) 4 and 9 (e) 60 and 32 (f) 8,12 and 24 Solution(a): Factors of 20 = 1,2,4,5,10 and 20 Factors of 30 = 1,2,3,5,6,10,15 and 30 GCF of 20 and 30 = 10 Solution(b): Factors of 12 = 1,2,3,4,6 and 12 Factors of 24 = 1,2,3,4,6,8,12 and 24 GCF of 12 and 24 = 12 Solution(c): Factors of 16 = 1,2,4,8,and 16 Factors of 44 = 1,2,4,11,22 and 44 GCF of 16 and 44 = 4 Solution(d): Factors of 4 = 1,2 and 4 Factors of 9 = 1,3 and 9 GCF of 4 and 9 = 1 (Prime) Solution(e): Factors of 60 = 1,2,3,4,5,6,10,12,15,20,30 and 60 Factors of 32 = 1,2,4,8,16 and 32 GCF of 60 and 32 = 4 Solution(f): Factors of 8 = 1,2,4,and 8 Factors of 12 = 1,2,3,4,6 and 12 Factors of 24 = 1,2,3,4,6,8,12 and 24 GCF of 8,12 and 24 = 4 Page 1
To find the factors of an exponential we simply write down all its powers starting at 1. Example 2: Find all the factors of x 4. The factors of x 4 are 1,x,x 2 x 3 and x 4 Example 3: Find all the factors of y 3. The factors of y 3 are 1,y,y 2 and y 3 To find the GCF of two exponentials it will be the smaller of the two powers. Example 3: Find the GCF of x 3 and x 7 The factors of x 3 are 1,x,x 2 and x 3 The factors of x 7 are 1,x,x 2 x 3,x 4,x 5,x 6 and x 7 The GCF of x 3 and x 7 = x 3 (Notice that this is the smaller of the two powers) Example 4: Find the GCF of y 5 and y 6 The factors of y 5 are 1,y,y 2,y 4 and y 5 The factors of y 6 are 1,y,y 2 y 3,y 4,y 5 and y 6 The GCF of y 5 and y 6 = y 6 (Notice that this is the smaller of the two powers) When the expression is a product of two or more exponentials then the factors are the products of all the permutations of each exponential. Example 5: Find the factors of x 3 y 2 The factors of x 3 y 2 are 1,x,x 2, x 3 y,xy,x 2 y,x 3 y y 2,xy 2,x 2 y 2,x 3 y 3 Example 6: Find the factors of b 2 d 5 The factors of b 2 d 5 are 1,b,b 2 d,bd,b 2 d d 2,bd 2,b 2 d 2 d 3,bd 3,b 2 d 3 d 4,bd 4,b 2 d 4 d 5,bd 5,b 2 d 5 When the expression is a multiple of an exponential then its factors are all the permutations of the factors of the number with the powers of the exponential. Page 2
Example 7: Find all the factors of 6x 3 The factors of 6x 3 are 1,x,x 2, x 3 and 2,2x,2x 2,2x 3 and 3,3x,3x 2,3x 3 and 6,6x,6x 2,6x 3 Example 8: Find all the factors of 4y 2 The factors of 4y 2 are 1,y,y 2 and 2,2y,2y 2 and 4,4y,4y 2 The greatest common factor (GCF) of two expressions is the largest factor they both share. There are two methods to finding the GCF of two expressions the first is by writing out all the factors of each expression and then finding the largest result that they both share. This method is very time consuming as the following worked example shows. Example 9: Find the GCF of 12x 3 and 16x 2 The factors of 12x 3 are 1,x,x 2, x 3 2,2x,2x 2,2x 3 3,3x,3x 2,3x 3 4,4x,4x 2,4x 3 6,6x,6x 2,6x 3 12,12x,12x 2,12x 3 The factors of 16x 2 are 1,x,x 2 2,2x,2x 2 4,4x,4x 2 8,8x,8x 2 16,16x,16x 2 The common factors are 1,x,x 2 2,2x,2x 2 4,4x,4x 2 The GCF of 12x 3 and 16x 2 is 4x 2 The second method for finding the GCF of two expressions is to find the GCF of the constants and the exponentials separately and then combine the result. Example 10: Find the GCF of 12x 3 and 16x 2 The GCF of 12 and 16 is 4 The GCF of x 2 and x 3 is x 2 (The smaller of the two powers) The GCF of 12x 2 and 16x 3 is 4x 2 Example 11: Find the GCF of 10y 4 x 2 and 25y 3 x 3 The GCF of 10 and 25 is 5 The GCF of x 2 and x 3 is x 2 The GCF of y 4 and y 3 is y 3 (The smaller of the two powers) (The smaller of the two powers) The GCF of 10y 4 x 2 and 25y 3 x 3 is 5x 2 y 3 Page 3
Exercise 1: 1. Find all the factors of the following numbers. (a) 26 (b) 60 (c) 82 (d) 144 (e) 100 2. Find all the factors of the following exponentials. (a) x 6 (b) x 4 (c) b 3 x 2 (d) c 2 d 2 (e) a 5 b 3. Find all the factors of the following expressions. (a) 6x 3 (b) 4b 4 (c) 20x 2 (d) 10d 3 (e) 12b 4. Find the greatest Common Factors GCF for each of the following. (a) 12 and 20 (b) 16 and 60 (c) 43 and 57 (d) 100 and 45 (e) 16,24 and 30 (f) 18,36 and 9 5. Find the greatest Common Factors GCF for each of the following. (a) 6b 3 and 15b 2 (b) 30c 2 and 20c 6 (c) 2d 3 and 6d 3 (d) 12x 4 and 14x 5 (e) 7x 3 and 14x 4 (f) 18g 10 and 81g 22 (g) 14e 4 and 21e 8 (h) 2k 5 and 3k 5 (i) 8g 2 and 36g 12 6. Find the greatest Common Factors GCF for each of the following. (a) 4y 3 x 5 and 20y 6 x 6 (b) 8b 4 c 2 and 40b 2 c 4 (c) y 5 x 5 and 2y 3 x 3 (d) 14a 3 b 2 and 21a 3 b 2 (e) 5x 4 y 9 and 9x 9 y 4 (f) 9r 2 t 11 and 21r 12 t 12 (g) 6c 7 d 9 and 12c 7 d 6 (h) 4c 2 d 2 and 40c 3 d 2 (i) 10c 4 d 2 e and 8c 2 d 2 e 4 Page 4
2. Factoring by taking out a Common factors. When we expand parenthesis we can use the distributive law so that 5(2x 7) = 10x 35 Factorising an expression simplifies it by reversing the process of expanding parenthesis. So we take 10x 35 = 5(2x 7) this process is called factoring by taking out a common factor This process can be split into two steps. Step 1: Find the common factors of each term in the expression In the above example the common factors of 10x and 35 is 5. Step 2: Write down the common factor and in the parenthesis write down what is left when you divide each term in the expression by the common factor. In the above example we get 10x 35 = 5(.) The first term is 10x when you divide it by the common factor 5 you get 2x The second term is 35 and when you divide it by the common factor 5 you get 7 These two results are now put in the parenthesis to complete the process. So 10x 35 = 5(2x 7) Example 1: Factor 18x 24y The two terms of this expression are 18x and 24y the GCF is 6 18x 24y = 6(3x 4y) Example 2: Factor 3a 12b + 24n The three terms of this expression are 3a, 12b and 24n the GCF is 3 3a 12b + 24n = 3(a 4b + 8n) Example 3: Factor 8x + 7y + 14z The three terms of this expression are 8x, 7y and 14z the GCF is 1 so these terms are prime so it is not possible to factor this expression except in a very trivial way as shown below so we typically just say Prime and leave the expression alone 8x + 7y + 14z = 1(8x + 7y + 14z) Example 4: Factor 4t 3 6t The two terms of this expression are 4t 3 and 6t the common factors are 2 and t so combined the common factor is 2t. So 4t 3 6t = 2t(2t 2 3) Page 5
Example 5: Factor 40x 2 25x The two terms of this expression are 40x 2 and 25x the common factors are 5 and x so combined the common factor is 5x. So 40x 2 25x = 5x(8x 5) Example 6: Factor the expression 3x 3 y 2 + 9x 2 y 3 The 2 terms of this expression are 3x 3 y 2 and 9x 2 y 3 The common factors are 3, x 2 and y 2 so combined the common factor is 3x 2 y 2. So 3x 3 y 2 + 9x 2 y 3 = 3x 2 y 2 (x +3y) Example 7: Factor the expression 3ab 2 + 6a 2 b 2 12a 3 b 3 The 3 terms of this expression are 3ab 2, 6a 2 b 2 and 12a 3 b 3 The common factors are 3, a and b 2 so combined the common factor is 3ab 2. So 3ab 2 + 6a 2 b 2 12a 3 b 3 = 3ab 2 (1 + 2a 4ab) Example 8: Factor the expression 4x 2 y 8xy + 10x The 3 terms of this expression are 4x 2 y, 8xy and 10x The common factors are 2, and x so combined the common factor is 2x. So 4x 2 y 8xy + 10x = 2x(2xy 4y + 5) Example 9: Factor the expression 4a(x + y) + 6b(x + y) The 2 terms of this expression are 4a(x + y) + 6b(x + y) The common factors are 2, and (x + y) and when we combine them we get a GCF of 2(x + y) So 4a(x + y) + 6b(x + y) = 2(x + y)(2a + 3b) It may seem a little strange that (x + y) is a factor but mathematically we can just consider it a single object that both terms contain. Page 6
Example 10: Factor the expression 4(x + 4) 2 + 12(x + 4) The 2 terms of this expression are 4(x + 4) 2 + 12(x + 4) The common factors are 4, and (x + 4) 2 and when we combine them we get a GCF of 4(x + 4) 2 So 4(x + 4) 2 + 12(x + 4) = 4(x + 4)( (x + 4) + 3) = 4(x + 4)( (x + 7) It may seem a little strange that (x + y) is a factor but mathematically we can just consider it a single object that both terms contain. There are two checks that you can make to be sure that you have factored properly. The first is to expand the expression to make sure it is equal to the one given at the start. So in Example 8 we were asked to factor 4x 2 y 8xy + 10x The solution was 2x(2xy 4y + 5) to check that this is correct we expand the parenthesis. 2x(2xy 4y + 5) = 2x(2xy) + 2x( 4y) + 2x(5) = 4x 2 y 8xy + 10x This is the same expression so we have shown that our factoring part of the process was done correctly. The second check is to see if we have used the correct Greatest Common Factor (GCF) For example suppose we were asked to factor the expression 8x 3 y 2 + 12x 4 y 3 Lets assume that we wrongly thought that the GCF was 2x 3 y in which case we would get as our factoring 8x 3 y 2 + 12x 4 y 3 = 2x 3 y(4y + 6xy) We can tell this would not be the full factoring by looking at what is left in the parenthesis in this case we have 4y + 6xy and since this expression can still be factored more as it has a common factor of 2y we know we have factored the original expression wrongly and we should have used the GCF of 4x 3 y 2 this would then give us \the correct factorization is 8x 3 y 2 + 12x 4 y 3 = 4x 3 y 2 (2 + 3x) Page 7
Exercise 2: 1. Find the GCF of the terms of the polynomial 8x 6 + 12x 3 A. x 3 B. 2x 3 C. 4x 3 D 8x 6 2. Factor each of the following expressions. (a) 8x + 8 (b) 6d + 12e (c) 4x + 9 (d) 12 + 24y (e) 12c 24 (f) 10y 15x (g) 12x 10b (h) 15e 20g (i) 6x 16 (j) 5 25g (k) 2x + 4y (l) 4x 12d 3. Factor each of the following expressions. (a) 4x + 12y + 6c (b) 8b + 4c + 8d (c) 42d + 12 8c (d) 12 + 8d 12c (e) 30c 12d + 15e (f) 25x 5c 15 (g) 2x 18 + 24y (h) 50 20g 15h (i) 6x + 3y + 3t (j) 16b 20h 10j (k) 20t 15 + 50p (l) 6x 3 + 9y 4. Factor each of the following expressions. (a) 2ab + 3ac (b) 21b + 3bd (c) 5c + tc (d) 22k + 11gk (e) 30c 9cd (f) 20x 15xc (g) 20xd 10xg (h) 15rt 20ft (i) 30xc + 3yc (j) 15by + 12b 9y (k) 4ab 6ac + 12ad (l) 12x + 6x 2 5. Factor each of the following expressions. (a) 4b 4 8b 3 (b) 6ab + 9ab 3 3a 2 b (c) 2cy 2 + 20c 2 y 3 (d) 12by 2 + 18b 2 y 3 2by (e) 12ax 3 y 2 + 8a 2 x 3 y 2 (f) 6b 3 y 2 + b 3 y 3 9b 3 y (g) 30xyz + 9x 2 y 2 z 2 (h) 4x 2 +12x (i) 15x 3 c 40x 3 c 4 (j) 24t 4 + 4t 3 + 32t (k) 3c 4 d 9c 5 d 5 (l) 5x 4 c 10x 2 c 7 (m) 3cx 2 18c 2 x (n) 5mnp 5 20m 5 n 6 (o) 3x 2 y 4 + 3x 2 y 6 (p) 3a(x + 1) 6b(x + 1) (q) 3a(x + y) 9a 2 (x + y) (r) (x + 7) 2 2(x + 7) Page 8
3. Factoring the Difference of Two Squares When you look at the following expansions you will notice a pattern. (x +3)(x 3) = x 2 3x + 3x 9 = x 2 9 (b +10)(b 10) = b 2 10b + 10b 100 = b 2 100 (x +3)(x 3) = x 2 3x + 3x 9 = x 2 9 (a +b)(a b) = a 2 ab + ab b 2 = a 2 b 2 The general pattern gives us a way to factor the difference of two squares by simply finding the values of a and b. Definition: A difference of two squares is an expression that contain two terms that are subtracted. The two terms will in turn be squares of some other term. When you factor such expressions the answer is always in the form (a + b)(a b) where a and b are the square roots of the first and second terms. a 2 b 2 = (a + b)(a b) Example 1: Factor the expression t 2 4 The 2 terms of this expression are t 2 and 4, their square roots are t and 2. So t 2 4 = (t + 2)(t 2) Example 2: Factor the expression x 2 36 The 2 terms of this expression are x 2 and 36, their square roots are x and 6. So x 2 36 = (x + 6)(x 6) Example 3: Factor the expression 25x 2 16 The 2 terms of this expression are 25x 2 and 16, their square roots are 5x and 4. So 25x 2 16 = (5x + 4)(5x 4) Example 4: Factor the expression 4y 6 9a 4 The 2 terms of this expression are 4y 6 and 9a 4, their square roots are 2y 3 and 3a 2. So 4y 2 9a 2 = (2y 3 + 3a 2 )(2y 3 3a 2 ) Page 9
Exercise 3 1. Factor the following expressions. (a) x 2 25 (b) x 2 49 (c) b 2 100 (d) 144 b 2 (e) x 2 b 2 (f) 4x 2 81 (g) 9x 2 4 (h) 16a 2 25b 2 (i) 1 4a 2 b 4 (j) 9x 2 4y 2 (k) b 6 y 6 (l) 4x 2 9b 2 Page 10
4. Factoring trinomials of the form x 2 + bx +c These expressions contain three terms (hence the name trinomial). The most common type are simple trinomials these are expressions where there is only an x 2 term and have the following general form x 2 + bx + c. When you factor x 2 + bx + c it will always take the form (x..)(x ) The missing terms in the parenthesis are the two numbers which are the factors of the constant term c and whose sum equals the coefficient of x called b. There are 4 basic situations in factoring x 2 + bx + c b and c are both positive b is negative and c is positive b is positive and c is negative b is negative and c is negative The first situation is when a and c are both positive. In this situation the two factors of c will both be positive. Example 1: Factor the expression x 2 + 5x + 6 Solution : In this situation we are looking for two factors of 6 that add up to 5. Since both b and c are positive we need only look at positive values of the factors of 6. The Positive factors of 6 are:- +1 and + 6 Add to 7 so not the correct combination +2 and + 3 Add to 5 so this is correct combination Using these two factors of 6 we get x 2 + 5x + 6 = (x + 2)(x + 3) Example 2: Factor the expression x 2 + 7x + 10 Solution : In this situation we are looking for two factors of 10 that add up to 7. Since both b and c are positive we need only look at positive values of the factors of 10. The Positive factors of 10 are:- +1 and + 10 Add to 11 so this is not the correct combination +2 and + 5 Add to 7 so this is correct combination Using these two factors of 10 we get x 2 + 7x + 10 = (x + 2)(x + 5) Page 11
The second situation is when b is negative and c is positive. In this situation the two factors will both be negative. Example 3: Factor the expression x 2 6x + 8 Solution : In this situation we are looking for two factors of 8 that add up to 6. Since b is negative and c are positive we need only look at the negative factors of 8. The negative factors of 8 are:- 1 and 8 Add to 9 so this is not the correct combination 2 and 4 Add to 6 so this is correct combination Using these two factors of 8 we get x 2 6x + 8 = (x 2)(x 4) Example 4: Factor the expression x 2 7x + 12 Solution : In this situation we are looking for two factors of 12 that add up to 7. Since b is negative and c are positive we need only look at the negative factors of 12. The negative factors of 12 are:- 1 and 12 Add to 9 so this is not the correct combination 2 and 6 Add to 8 so this is not the correct combination 3 and 4 Add to 7 so this is not the correct combination Using these two factors of 12 we get x 2 7x + 12 = (x 3)(x 4) Example 5: Factor the expression b 2 20b + 100 Solution : In this situation we are looking for two factors of 100 that add up to 20. Since b is negative and c are positive we need only look at the negative factors of 100. The negative factors of 100 are:- 1 and 100 Add to 101 so this is not the correct combination 2 and 50 Add to 52 so this is not the correct combination 4 and 25 Add to 29 so this is not the correct combination 5 and 20 Add to 25 so this is not the correct combination 10 and 10 Add to 20 so this is not the correct combination Using these two factors of 100 we get b 2 20b + 100 = (b 10)(b 10) = (b 10) 2 Note: This process does not need to be as laborious as it seems since the factoring is unique once you find a combination that works you can stop. Also with practice you will get an insight into which combinations are more likely to succeed and so you will try these ones first and hopefully reduce the number of failed combinations to a minimum. Page 12
The third situation is when b is positive and c is negative. In this situation the larger factor will be positive and the smaller factor will be negative. Example 6: Factor the expression x 2 + 2x 15 Solution : In this situation we are looking for factors of 15 that add up to 2. Since the large factor is positive and the smaller is factor is negative we only need to look at the following combinations. Factors of 15 are:- 15 and 1 Add to 14 so this is not the correct combination 5 and 3 Add to 2 so this is the correct combination Using these two factors of 15 we get x 2 + 2x 15 = (x + 5)(x 3) Example 7: Factor the expression x 2 + 10x 24 Solution : In this situation we are looking for factors of 24 that add up to 10. Since the large factor is positive and the smaller is factor is negative we only need to look at the following combinations. Factors of 24 are:- 24 and 1 Add to 23 so this is not the correct combination 12 and 2 Add to 10 so this is the correct combination Using these two factors of 24 we get x 2 + 10x 24 = (x + 12)(x 2) Example 8: Factor the expression y 2 + 13y 60 Solution : In this situation we are looking for factors of 60 that add up to 13. Since the large factor is positive and the smaller is factor is negative we only need to look at the following combinations. Factors of 60 are:- 60 and 1 Add to 59 so this is not the correct combination 30 and 2 Add to 28 so this is not the correct combination 15 and 2 Add to 13 so this is the correct combination Using these two factors of 60 we get y 2 + 13y 60 = (y + 15)(y 2) Page 13
The fourth situation is when b is negative and c is positive. In this situation the larger factor will be negative and the smaller factor will be positive. Example 9: Factor the expression y 2 y 12 In this situation we are looking for factors of 12 that add up to 1. Since the large factor is negative and the smaller is factor is positive we only need to look at the following combinations. Factors of 60 are:- 1 and 12 Add to 11 so this is not the correct combination 2 and 6 Add to 4 so this is not the correct combination 3 and 4 Add to 1 so this is the correct combination Using these two factors of 12 we get y 2 y 12 = (y + 3)(y 4) Example 10: Factor the expression k 2 3k 40 In this situation we are looking for factors of 40 that add up to 3. Since the large factor is negative and the smaller is factor is positive we only need to look at the following combinations. Factors of 40 are:- 1 and 40 Add to 41 so this is not the correct combination 2 and 20 Add to 18 so this is not the correct combination 4 and 10 Add to 6 so this is not the correct combination 5 and 8 Add to 3 so this is not the correct combination Using these two factors of 40 we get k 2 3k 40 = (k + 5)(k 8) Exercise 4 1. Factor the following expressions. (a) x 2 + 8x + 7 (b) x 2 7x + 10 (c) x 2 6x + 8 (d) x 2 + 5x + 6 (e) x 2 + 4x + 3 (f) x 2 7x 30 (g) x 2 + 2x 80 (h) x 2 8x + 12 (i) x 2 x 20 (j) x 2 + 2x 63 (k) x 2 + 3x 88 (l) x 2 4x 21 (m) x 2 5x 150 (n) b 2 + 2b 15 (o) a 2 10a + 16 (p) x 2 + 2x 63 (q) x 2 + 3x 88 (r) x 2 4x 21 (s) x 2 5x 150 (t) b 2 + 8t + 15 (u) a 2 10a + 16 (v) x 2 + 2x 63 (w) x 2 + 3x 88 (x) x 2 4x 21 Page 14
5. Factoring trinomials of the form ax 2 + bx + c A trinomial of the type ax 2 + bx + c in which typically there is not single x 2 term but some multiple instead are normally the most difficult to factor as they do not fall neatly into categories as they did for situations like factor x 2 + bx + c. When you factor a complex trinomial of the form ax 2 + bx + c there are 2 main stages these are. 1. Get the x term part of the factor 2. Get the constant term part of the factor. Example 1: Factor the expression 2x 2 + 5x 3 Step 1 Get the x term part of the factor. There are a set of rules you can follow that will help you to factorise a complex trinomial expression the first is look at the number of x 2 the expression contains, this will tell you the possible number of x's that go into the factor, so for example if the expression has 2x 2 can only be formed by 2x times x. So the solution must take the form on the other hand if it were 4x 2 it could be one of two possible situations either(4x...)(x...) or (2x...)(2x...). In this example we have 2x 2 + 5x 3 and we get. 2x 2 + 5x 3 = (2x...)(x...) Step 2 Get the constant term part of the factor. The second stage in the process is to look at the constant term, in this case its value is 3 and try all its factor pairs in order until we find the combination that works. So in this example we can use (1 & 3) and ( 3 & 1) and ( 1 & 3) and (3 & 1 ) These 4 combinations as shown below. We now use FOIL on each one until we get the one that works. Notice that in the factors of 3 the order is also important so we reverse each possible factor pair. (2x + 1)(x 3) this gives 2x 2 5x 3 (2x 1)(x + 3) this gives 2x 2 + 5x 3 This is the correct one. (2x + 3)(x 1) this gives 2x 2 + x 3 (2x 3)(x + 1) this gives 2x 2 x 3 Only one combination actually works and this is our solution. So 2x 2 + 5x 3 = (2x 1)(x + 3) Page 15
Example 2: Factor the expression 5y 2 12y + 4 Step 1 Get the y term part of the factor. There are a set of rules you can follow that will help you to factorise a complex trinomial expression the first is look at the number of y 2 the expression contains, this will tell you the possible number of y's that go into the factor. So for example since 5y 2 can only be formed by 5y times y. So the solution must take the form 5y 2 12y + 4 = (5y...)(y...) Step 2 Get the constant term part of the factor. The second stage in the process is to look at the constant term, in this case its value is + 4 and try all its factor pairs in order until we find the combination that works. So in this example we can use (1& 4) and (4 & 1) and (2 & 2) and ( 1& 4) and ( 4 & 1) and ( 2 & 2) This in turn gives rise to 6 combinations as shown below. We now use FOIL on each one until we get the one that works. Notice that in the factors of 4 the order is also important so we reverse each possible factor pair. (5y + 4)(y + 1) this gives 5y 2 + 9y + 4 (5y + 2)(y + 2) this gives 5y 2 + 12y + 4 (5y 1)(y 4) this gives 5y 2 21y + 4 (5y 4)(y 1) this gives 5y 2 9y + 4 (5y 2)(y 2) this gives 5y 2 12y + 4 This is the correct one. Only one combination actually works and this is our solution. So 5y 2 12y + 4 = (5y 2)(y 2) Notice that the first three choices were bound to give positive values for the y term and so we could have saved ourselves some time by not bothering to check them as they clearly cannot get us the 12y term that we need. Another concept that can reduce your workload is when you get a solution that is almost correct one except for the sign such as when you tested (5y + 2)(y + 2) = 5y 2 + 12y + 4 we got + 12y instead of the 12y we were seeking In these situations it is often a good idea to look at other factor combinations similar to those ones but with different signs so instead of using the factors (2 & 2) we used ( 2 & 2) to get (5y 2)(y 2) = 5y 2 12y + 4 the correct factorisation. Page 16
Example 3: Factor the expression 4x 2 17x 15 Step 1 Get the x term part of the factor. There are a set of rules you can follow that will help you to factorise a complex trinomial expression the first is look at the number of x 2 the expression contains, this will tell you the possible number of x's that go into the factor. So for example since 5x 2 can only be formed by 4x times x or by 2x times 2x. So the solution must take the one of these forms 4x 2 17x 15 = (4x...)(x...) 4x 2 17x 15 = (2x...)(2x...) Step 2 Get the constant term part of the factor. Since we do not know which of the above 2 forms (4x...)(x...) or (2x...)(2x...) Will yield the correct factorization two must pick one and test it first. If we choose to test (2x.)(2x.) then the second stage of the process is to look at the constant term, in this case its value 15 and try all its factor pairs in order until we find the combination that works. Since both terms have 2x we do not need to reverse the factors. So we can use (1& 15) and ( 1 & 15) and (3 & 5) and ( 3 & 5) This in turn gives rise to 4 combinations as shown below. We now use FOIL on each one until we get the one that works. (2x + 1)(2x 15) this gives 4x 2 27x 15 (2x 1)(2x + 15) this gives 4x 2 + 27x 15 (2x + 3)(2x 5) this gives 4x 2 4x 15 (2x 3)(2x + 5) this gives 4x 2 + 4x 15 None of these combination work so we must try the combination (4x...)(x...) So we can use (1& 15) and ( 1 & 15) and (3 & 5) and ( 3 & 5) This in turn gives rise to 4 combinations as shown below. We now use FOIL on each one until we get the one that works. (4x + 1)(x 15) this gives 4x 2 59x 15 (4x 1)(2x + 15) this gives 4x 2 + 59x 15 (4x + 3)(x 5) this gives 4x 2 17x 15 This is the correct one. (4x 3)(x + 5) this gives 4x 2 + 17x 15 Only one combination actually works and this is our solution. So 4x 2 17x 15 = (4x + 3)(x 5) Page 17
Exercise 5: 1. Factor the following expressions. (a) 2x 2 + 15x + 7 (b) 7y 2 2y 24 (c) 16x 2 + 24x + 9 (d) 2x 2 11x + 15 (e) 3p 2 13p + 10 (f) 4x 2 12x 40 (g) 5x² + 3x 8 (h) 2x² + 3x 2 (i) 2x² 11x + 5 (j) 16x² 40x + 25 (k) 4x² + 40x + 25 (l) 4c² 13c + 9 (m) (4x + 1)(x 5) (n) (4b + 3)(4b + 5) (o) (3y 1)(2y 9) (p) 3x² 26x 9 (q) 25x² + 20x + 4 (r) 9d² + 24d + 16 2. Which of the following is the factorization of 3x 2 + 7x 6 A. (3x 2)(x 3) B. (x + 3)(3x + 2) C. (3x 2)(x + 3) D. (3x + 2)(x 3) 3. Which of the following is the factorization of 40p 2 13p 36 A. (8p + 9)(5p + 4) B. ( 8p 9)(5p + 4) C. (8p 9)(5p 4) D. (8p + 9)(5p 4) Page 18
6. Factoring - Special Cases There are 3 special cases of factoring we will deal with in this section. Case 1: Trinomials with a negative x 2 coefficient such as x 2 + 10x 9 Case 2: Trinomials in x and y such as x 2 + 6xy + 5y 2 Case 3: Combined factoring where a common factor and another factoring method are combined such as ax 2 + 4ax + 3a or ax 2 9a Case 1: Trinomials with a negative x 2 coefficient In these situations we take out the negative as a common factor of 1 and the resulting trinomial can be solved in the usual way. Example 1: Factor x 2 + 10x 9 x 2 + 10x 9 = (x 2 10x + 9) Take out 1 as a common factor Example 2: Factor 2x 2 5x + 18 = (x 9)(x 1) Factoring x 2 10x + 9 2x 2 5x + 18 = (2x 2 + 5x 9) Take out 1 as a common factor = (2x 9)(x 1) Factoring 2x 2 + 5x 9 Case 2: Trinomials in x and y In these situations there will be a trinomial in x and y such as x 2 + 6xy + 5y 2 in order to factor these we use a method similar to those used for other trinomials. Example 3: Factor x 2 + 6xy + 5y 2 in order to factor x 2 + 6xy + 5y 2 we need to look for factors of 5 that add up to 6 in this case we get + 5 and + 1 this represents + 5y and + y and so we get the solution: x 2 + 6xy + 5y 2 = (x + 5y)(x + y) Example 4: Factor x 2 + 3xy 10y 2 in order to factor x 2 + 3xy 10y 2 we need to look for factors of0 10 that add up to 3 in this case we get + 2 and 5 this represents + 2y and 5 and so we get the solution: x 2 + 3xy 10y 2 = (x + 2y)(x + 5y) Page 19
Case 3: Combined Factoring In this situation we combine two or more of the methods where the first process involves taking out a common factor followed by one ot the processes described earlier. Example 1: Factor the expression 5x 2 180 First we take a common factor of 5 from the two terms then we factor the difference of two squares that remains. So 5x 2 180 = 5(x 2 36) Take out a common factor of 5 5(x 2 36) = 5(x + 6)(x 6) Factor the difference of two squares 5x 2 180 = 5(x + 6)(x 6) Example 2: Factor the expression 2ax 2 + 12ax + 18a First we take a common factor of 2a from the 3 terms then we factor what remains. 2ax 2 + 12ax + 18a = 2a(x 2 + 6x + 9) Take out a C.F. of 2a = 2a(x + 3)(x + 3) Factor the trinomial x 2 + 6x + 9 = 2a(x + 3) 2 Example 3: Factor the expression 2ab 4 x 2 + 8ab 4 x + 6ab 4 First we take a common factor of 2ab 4 from the 3 terms then we factor what remains. 2ab 4 x 2 + 8ab 4 x + 6ab 4 = 2a b 4 (x 2 + 4x + 3) Take out a C.F. of 2ab 4 = 2a b 4 (x + 3)(x + 1) Factor the trinomial x 2 + 4x + 3 Example 4: Factor the expression x 4 + 10x 3 11x 2 First we take a common factor of x 2 from the 3 terms then we factor what remains. x 4 + 10x 3 11x 2 = x 2 (x 2 + 10x 11) Take out a C.F. of x 2 = 2a(x + 11)(x 1) Factor the trinomial x 2 + 10x 11 Page 20
Exercise 6 1. Factor the following. (a) x 2 + 5x 6 (b) 2x 2 17x + 19 (c) 4x 2 x + 39 2. Factor the following. (a) x 2 + 26xy + 25y 2 (b) x 2 + 8xy + 12y 2 (c) x 2 9xy + 14y 2 (d) x 2 + 3xy 10y 2 (e) x 2 3xy 4y 2 (f) x 2 4xy 12y 2 3. Factor the following. (a) 4x 2 100 (b) 2x 2 50y 2 (c) 11x 2 99c 2 (d) 2 50d 2 (e) 4x 2 4x 8 (f) 2ax 2 2ax 12a (g) 6abx 2 6abx 36ab (h) 2a 2 x 2 + 12a 2 x + 18a 2 (i) abx 2 6abx + 5ab (j) 3xy 2 48xy 4 (k) 6a 3 + 3a 2 30a (l) 4x 4 + 10x 3 6x 2 Page 21
Solutions Exercise 1: 1.(a) 1,2,13,26 (b) 1,2,3,4,5,6,10,12,15,20,30,60 (c) 1,2,41,81 1.(d) 1,2,3,4,6,8,12,16,24,36,48,72,144 (e) 1,2,4,5,10,20,25,50,100 2.(a) 1,x,x 2,x 3,x 4,x 5,x 6 (b) 1,x,x 2,x 3,x 4 (c) 1,x,x 2 b,bx,bx 2 b 2,b 2 x,b 2 x 2 b 3,b 3 x,b 3 x 2 2.(d) 1,d,d 2 c,cd,cd 2 c 2, c 2 d, c 2 d 2 (e) 1,a,a 2,a 3,a 4,a 5 b,ba,ba 2,ba 3,ba 4,ba 5 3.(a) 1,x,x 2,x 3 2,2x,2x 2,2x 3 3,3x,3x 2,3x 3 6,6x,6x 2,6x 3 3.(b) 1,b,b 2,b 3,b 4 2,2b,2b 2,2b 3,2b 4 4,4b,4b 2,4b 3,4b 4 3.(c) 1,x,x 2 2,2x,2x 2 4,4x,4x 2 5,5x,5x 2 10,10x,10x 2 20,20x,20x 2 3.(d) 1,d,d 2,d 3 2,2d,2d 2,2d 3 5,5d,5d 2,5d 3 10,10d,10d 2,10d 3 3.(e) 1,2,3,4,6,12 b,2b,3b,4b,6b,12b 4.(a) 4 (b) 4 (c) 1 (d) 5 (e) 2 (f) 9 5.(a) 3b 2 (b) 10c 2 (c) 2d 3 (d) 2x 4 (e) 7x 3 (f) 9g 10 (g) 7e 4 (h) k 5 (i) 4g 2 6.(a) 4y 3 x 5 (b) 8b 2 c 2 (c) y 3 x 3 (d) 7a 3 b 2 (e) x 4 y 4 (f) 3r 2 t 11 (g) 6c 7 d 6 (h) 4c 2 d 2 (i) 2c 2 d 2 e Exercise 2: 1. C 2.(a) 8(x + 1) (b) 6(d + 2e) (c) 4x + 9 (d) 12(1 + 2y) (e) 12(c 2) 2.(f) 5(2y 3x) (g) 2(6x 5b) (h) 5(3e 4g) (i) 2(x 8)4 (j) 5(1 5g) (k) 2( x + 2) or 2(x 2y) (l) 4( x 3d) or 4(x + 3d) 3.(a) 2(2x + 6y + 3c) (b) 4(2b + c + 2d) (c) 2(21d + 6 4c) 3.(d) 4(3 + 2d 3c) (e) 3(10c 4d + 5e) (f) 5(5x c 3) 3.(g) 2(x 9 + 12y) (h) 5(10 4g 3h) (i) 3(2x + y + t) 3.(j) 2(8b 10h 5j) (k) 5(4t 3 + 10p) (l) 3(2x 1 + 3y) 4.(a) a(2b + 3c) (b) 3b(7 + d) (c) c(5 + t) (d) 11k(2 + 2g) 4.(e) 3c(10 3d) (f) 5x(4 3c) (g) 10x(2d g) (h) 5t(3r 4f) 4.(i) 3c(10x + y) (j) 3(5by + 4b 3y) (k) 2a(2b 3c + 6d) (l) 6x(2 + x) 5.(a) 4b 2 (1 2b) (b) 3ab(2 + 3b 2 a) (c) 2cy 2 (1 + 10y) 5.(d) 2by(6y + 9by 2 1) (e) 4ax 3 y 2 (3 + 2a) (f) b 3 y(6y + y 2 9) 5.(g) 3xyz(10 + 9xyz) (h) 4x(x +3) (i) 5x 3 c(3 8c 3 ) 5.(j) 4t(6t 3 + t 2 + 8) (k) 3c 4 d(1 3cd 4 ) (l) 5x 2 c(1 2c 6 ) 5.(m) 3cx(x 6c) (n) 5mn(p 5 4 m 4 n 5 ) (o) 3x 2 y 4 (1 + y 4 ) 5.(p) 3(x + 1)(a 2b) (q) 3a(x + y)(1 3a) (r) (x + 7)(x +5) Page 22
Exercise 3 1.(a) (x + 5)(x 5) (b) (x + 7)(x 7) (c) (x + 10)(x 10) 1.(d) (12 + b)(12 b) (e) (x + b)(x b ) (f) (2x + 9)(2x 9) 1.(g) (3x + 2)(3x 2) (h) (4a + 5b)(4a 5b) (i) (1 + 2ab 2 )(1 2ab 2 ) 1.(j) (3x + 2y)(3x 2y) (k) (b 3 + y 3 )( b 3 y 3 ) (l) (2x + 3b)(2x 3b) Exercise 4 1.(a) (x + 7)(x + 1) (b) (x 5)(x 2) (c) (x 4)(x 2) (d) (x + 3)(x + 2) 1.(e) (x + 3)(x + 1) (f) (x + 3)(x 10) (g) (x + 10)(x 8) (h) (x 6)(x 2) 1.(i) (x + 4)(x 5) (j) (x + 9)(x 7) (k) (x + 11)(x 8) (l) (x + 3)(x 7) 1.(m) (x + 10)(x 15) (n) (b + 5)(b 3) (o) (a 2)(a 8) (p) (x + 9)(x 7) 1.(q) (x + 11)(x 3) (r) (x + 3)(x 7) (s) (x + 10)(x 15) (t) (b + 3)(b + 5) 1.(u) (a 8)(x 2) (v) (x + 9)(x 7) (s) (x + 11)(x 8) (t) (x + 3)(x 7) Exercise 5: 1.(a) (2x + 1)(x + 7) (b) (7y + 12 )(y 2) (c) (4x + 3)(4x + 3) 1.(d) (2x 5)(x 3) (e) (3p + 2)(p 5) (f) (5x + 6)(x 1) 1.(g) (5x + 8)(x 1) (h) (2x 1)(x + 2) (i) (2x 1)(x 5) 1.(j) (2x 1)(x 5) (k) (4x + 5)(4x + 5) (l) (4c 9)(c 1) 1.(m) (4x + 1)(x 5) (n) (4b + 3)(4b + 5) (o) (3y 1)(2y 9) 1.(p) (3x + 1)(x 9) (q) (5x + 2) 2 (r) (3d + 4) 2 2. C 3. B Exercise 6 1.(a) (x 3)(x 2) (b) (2x + 19)(x + 1) (c) (4x + 13)(x 3) 2. (a) (x + 25y)(x + y) (b) (x + 6y)(x + 2y) (c) (x 2y)(x 7y) 2.(d) (x + 5y)(x 2y) (e) (x + y)(x 4y) (f) (x + 2y)(x 6y) 3.(a) 4(x + 5)(x 5) (b) 2(x + 5y)(x 5y) (c) 11(x + 3y)(x 3y) 3.(d) 2(1 =5d)(1 5d) (e) 4(x + 1)(x 2) (f) 2a(x + 2)(x 3) 3.(g) 6ab(x + 2)(x 3) (h) 2a 2 (x + 3) 2 (i) ab(x 5)(x 1) 3.(j) 3xy 2 (1 + y)(1 y) (k) 3a(2a + 5)(a 2) (l) 2x 2 (2x 1)(x + 3) Page 23