Download presentation

Presentation is loading. Please wait.

Published byLesly Watling Modified over 2 years ago

1
Revision Algebra II Factorisation of Binomials, Trinomials, Sum & Difference of Two Cubics By I Porter

2
2 Factors A factor of a term or number divides into that term or number, without a remainder. For examples, the factors of 18 are 1, 2, 3, 6, 9 and 18. 6x 2 are 1, 2, 3, 6, x, 2x, 3x, 6x, x 2, 2x 2, 3x 2 and 6x 2. 4xy are 1, 2, 4, x, y, 2x, 2y, 4x, 4y, xy, 2xy and 4xy. The highest common factor (HCF) of 4xy and 6x 2 is 2x. The product of a(b + c) is ab + ac and the product of (a + b)(c + d) is ac + ad + bc + bd, we say, these two results are written in expanded form. When they are written in the form a(b + c) and (a + b)(c + d), we say they are in factored form. Factorisation is the process whereby an expression in expanded form, such as ab + ac, is changed to factored form, such as a(b + c).

3
3 Factorising by Grouping ax+ bx + ay +by Algebraic expressions with 4 terms do not have a factor common to every term. Such expressions can often be factorised by first grouping the terms in pairs. = x(a + b)+ y(a+ b) Examples: Fully factorise the following a) a(b - 3) + 6(b - 3) = a(b - 3) + 6(b - 3) = (b - 3)(a + 6) b) ac + ad + bc + bd = ac + ad + bc + bd = a(c + d)+ b(c + d) = (c + d)(a + b) c) 4x 2 - 4xy + x - y = 4x 2 - 4xy + x - y = 4x(x - y)+ 1(x - y) = (x - y) (4x + 1) (x + y) = (a + b)

4
4 More Examples: [negative signs as a factor] d) mn + 4n - m - 4 = mn + 4n - (m + 4) = n(m + 4) - 1(m + 4) = (m + 4) (n - 1) e) 6x 2 - 4x - 3xy + 2y = 2 x 3x x 2x - 3xy + 2y = 2x(3x - 2) - y(3x - 2) = (3x - 2)(2x - y) Exercise: Fully factorise the following a) ab + 5a + 2b + 10b) xy - 3x + 2y - 6c) 3c 2 - 2cd + 3cd 2 - 2d 3 d) ab + 2b - 3a - 6e) 2x 2 - 6x - xy + 3yf) 4c c - cd - 3d = (a + 2)(b + 5)= (x + 2)(y - 3)= (c + d 2 )(3c - 2d) = (a + 2)(b - 3)= (2x - y)(x - 3)= (4c - d)(c + 3)

5
5 Factorising the difference of 2 squares: a 2 - b 2 = (a - b)(a + b) Examples: Factorise. a) n 2 - p 2 = (n - p) (n + p) b) x = (x) 2 - (3) 2 = (x - 3) (x + 3) c) 9x = (3x) 2 - (5) 2 = (3x - 5) (3x + 5) d) 4x 2 - y 2 = (2x) 2 - (y) 2 = (2x - y) (2x + y) e) 5x = 5(x 2 - 9) = 5[(x) 2 - (3) 2 ] = 5(x - 3) (x + 3) f) 32x y 2 = 2(16x 2 - 9y 2 ) = 2[(4x) 2 - (3y) 2 ] = 2(4x - 3y) (4x + 3y)

6
6 Exercise: Factorise a) x b) 4x c) 9x y 2 d) 8x e) 18x y 2 = (x - 6)(x + 6)= (2x - 7)(2x + 7)= (3x - 5y)(3x + 5y) = 2(2x - 5)(2x + 5)= 18(x - 2y)(x + 2y) Factorisation of CUBICS: a 3 - b 3 and a 3 + b 3 a 3 - b 3 =(a - b)(a 2 + ab + b 2 )a 3 + b 3 =(a + b)(a 2 - ab + b 2 ) Must know for HSC = x = (x - 2)(x 2 + 2x ) = (x - 2)(x 2 + 2x + 4) Examples: Factorise a) x 3 - 8b) 8x = (2x) 3 + (3) 3 = (2x + 3)(4x 2 - 6x ) = (2x + 3)(4x 2 - 6x + 9)

7
7 Exercise: Factorise a) x b) 8x 3 + y 3 c) 9x y 3 = (x - 5)(x 2 + 5x + 25)= (2x + y)(4x 2 - 2xy + y 2 )= 9(x - 2y)(x 2 + 2xy + 2y 2 ) Factoring quadratic trinomials: ax 2 ± bx ± c A trinomial has 3 terms. A quadratic trinomial is an expression with 3 terms where the highest power of a term is 2. Sum and Product Method. To factorise a trinomial of the form x 2 ± bx ± c (where b and c are non-zero integers) look for two numbers P and Q such that: * Their sum is b : (P + Q = b) * Their product is c : (PQ = c) [ The first step to start may be to write out the factors of c in a systematic process.]

8
8 Examples: Factorise the following: a) x 2 + 7x +12 Sum = +7 Product = +12 Factors of = 1 x 12 = 2 x 6 = 3 x 4 Why not the (-) factors? Sum and Product are both (+) Factor sums = = = +7 So, x 2 + 7x +12 = (x + 3)(x + 4) b) x 2 - 8x +12 Sum = -8 Product = +12 Factors of = -1 x -12 = -2 x -6 = -3 x -4 Why not the (+) factors? Sum (-) and Product (+) both are (-) Factor sums = = = -7 So, x 2 - 8x +12 = (x - 2)(x - 6) c) x 2 + 2x - 24 Sum = +2 Product = -24 Factors of = -1 x 24 = -2 x 12 = -3 x 8 = -4 x 6 Why do the factors have different signs? Sum (+) and Product (-), largest number is (+) Factor sums = = = = +2 So, x 2 + 2x - 24 = (x - 4)(x + 6) d) x 2 - 3x - 40 Sum = -3 Product = -40 Factors of = 1 x -40 = 2 x -20 = 4 x -10 = 5 x -8 Why do the factors have different signs? Sum (-) and Product (-), largest number is (-) Factor sums = = = = -3 So, x 2 - 3x - 40 = (x + 5)(x - 8)

9
9 Exercise: Factorise a) x 2 - 9x + 20b) x x + 24c) x 2 - 8x - 48 d) x 2 + 2x - 48e) x x + 54e) x x - 64 = (x - 4)(x - 5)= (x + 4)(x + 6)= (x + 4)(x - 12) = (x + 8)(x - 6)= (x - 9)(x - 6)= (x + 16)(x - 4) = 2(x 2 - 5x - 14) Sum = -5 Product = -14 Factors of = 1 x -14 = 2 x -7 Factor sums = = -5 So, 2x x - 28 = 2(x + 2)(x - 7) = 4(x 2 - 5x + 6) Sum = -5 Product = +6 Factors of = -1 x -6 = -2 x -3 Factor sums = = -5 So, 4x x - 28 = 4(x - 2)(x - 3) Examples: Factorise a) 2x x - 28b) 4x x + 24

10
10 Exercise: Factorise a) 7x x + 140b) 12x x - 72c) 9x x + 27 = 7(x - 4)(x - 5)= 12(x + 2)(x - 3)= 9(x + 1)(x + 3) 3x 2 - 2x - 8 [ Variation of sum + product method] Step 1: multipy first and last term. 3 x -8 = -24 Step 2: Look for two numbers that have a sum of -2 and product of -24. (I.e. -6 and +4) Step 3: Split the middle term using the number found in step 2 3x 2 - 6x + 4x - 8 Step 4: Factorise. = 3x 2 - 6x + 4x - 8 = 3x(x - 2) + 4(x - 2) = (x - 2) (3x + 4) 6x x - 5 Step 1: multipy first and last term. 6 x -5 = -30 Step 2: Look for two numbers that have a sum of +13 and product of -30. (I.e. -2 and +15) Step 3: Split the middle term using the number found in step 2 6x 2 - 2x + 15x - 5 Step 4: Factorise. = 6x 2 - 2x + 15x - 5 = 2x(3x - 1) + 5(3x - 1) = (3x - 1) (2x + 5) Harder factorisation of quadratic trinomials: ax 2 + bx + c Examples: Factorise a) 3x 2 - 2x - 8b) 6x x - 5 If you can use the CROSS method successfully, KEEP USING IT!

11
11 Exercise: Factorise a) 2x 2 + 5x + 2b) 5x x - 12c) 6x 2 - 7x + 1 d) 4x 2 + 7x - 2e) 6x 2 - 7x - 20f) x - 3x 2 (2x + 1)(x + 2)(5x + 3)(x - 4)(6x - 1)(x - 1) (4x - 1)(x + 2)(2x - 5)(3x + 4)(2 + 3x)(4 - x)

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google