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10. FACTORISATION

 

 1.        Multiply the following and express them in their simplest form:

(i) (a + b) (a – b)                                  (ii) (2x + 3y) (2x – 3y)

            (iii) (a + b)2                                           (iv) (m – n)2     

            (v) (a + b) (a2 – ab + b2)                       (vi) (g – h)  (g2 + gh + h2)         

(vii) (3j – 2k) ( 9j2 + 6jk + 4k2)            (viii) (x – 3) (x + 2)

(ix) (L + 5) (L + 7)                               (x) (2d – 4) (3d – 5)

 

2.                  Factorise the following:

(i) p2 – q2         (ii) r2 – 9           (iii) w2 – t2        (iv) s2 – 36       (v) 25 – k2

(vi)16 – 4g2      (vii) 9m2 – 36n2                        (viii) 49p2 – 4q2 

(ix) 8b2 – 18c2              (x) 50y2 – 32z2

 

3.                  Factorise the following:

(i) a3 – b3          (ii) m3 + n3        (iii) g3 + 8         (iv) w3 – 27      (v) 64 – p3

(vi) 8y3 – z3      (vii) 125g3 + 8 (viii) 27p3 – 8q3

(ix) 64r3 + 125s3                       (x) 54m3 + 16n3

 

4.                  Factorise the following:

(i) x2 + x -12    (ii) x2 + 7x + 10            (iii) x2 – 6x – 16            (iv) x2 + 12x + 32

(v) x2 – 10x + 21          (vi) x2 – 10x + 16         (vii) x2 –x – 56

(viii) x2 – 9x + 18          (ix) x2 + 11x + 24         (x) x2 – x - 42

(xi) x2 + 5x – 24           (xii) x2 –x - 90

 

  1. Factorise the following:

(i) 3x2 + 11x + 10         (ii) 4x2 + 5x – 6            (iii) 5x2 + 33x + 18

(iv) 5x2 – 11x – 12       (v) 7x2 – 30x +8           (vi) 5x2 + 17x – 12

(vii) 2x2 + 5x – 42        (viii) 4x2 – 13x - 35      (ix) 7x2 –x -6

(x) 5x2 + 31x – 28        (xi) 6x2 + 5x + 1           (xi) 12x2 + 5x - 2

 

ANSWERS.

 

1.         (i) a2 – b2   (ii) 4x2 – 9y2  (iii) a2 + 2ab + b2   (iv) m2 – 2mn + n2

            (v) a3 + b3  (vi) g3 – h3     (vii) 27j3 – 8k3  (viii) x2 –x – 6  (ix) L2 + 12L + 35

            (x) 6d2 – 22d + 20

2.         (i) (p + q)( p – q)   (ii) (r + 3)( r – 3)   (iii) (w + t)( w – t)   (iv) (s + 6)( s – 6)   (v) (5 + k)( 5 – k)   (vi) (4 + 2g)( 4 – 2g)   (vii) (3m + 6n)( 3m – 6n)  

(viii) (7p + 2q)(7 p – 2q)   (ix) 2(2b + 3)( 2b – 3c)   (x) 2(5y + 4z)( 5y – 4z)  

3.         (i) (a – b)( a2 + ab + b2)   (ii) (m + n)( m2 – mn + n2)  (iii) (g + 2)( g2 – 2g + 4)

            (iv) (w – 3)(w2 + 3w + 9) (v) (4 – p)(16 + 4p +p2)  (vi) (2y – z)(4y2 + 2yz + z2)

            (vii) (5g + 2)(25g2 – 10g +4)      (viii) (3p – 2q)( 9p2 +6pq + 4q2)  

            (ix) (4r + 5s)( 16r2 – 20rs + 25s2)   (x) 2(3m + 2n)( 9m2 – 6mn + 4n2)

4.         (i) (x-3)(x+4)   (ii) (x + 2)( x + 5)          (iii) (x – 8)( x + 2)  (iv) ((x +4)( x + 8)

            (v) (x – 7)( x – 3)  (vi) (x – 8)( x – 2)   (vii) (x + 7)( x – 8)  (viii) (x – 3)( x – 6)

            (ix) (x + 8) (x + 3)  (x) (x + 6) (x – 7)  (xi) (x + 8)( x – 3) (xii) (x + 9)(x – 10)

5.         (i) (x + 2)( 3x + 5)   (ii) (x + 2)( 4x – 3)   (iii) (x + 6)( 5x + 3)

(iv) (5x + 4)( x – 3) (v) (7x – 2)( x – 4) (vi) (5x – 3)(x + 4)

(vii) (2x – 7)( x + 6) (viii) (4x + 7)( x – 5) (ix) (7x + 6)(x – 1)

(x) (5x – 4)( x + 7)   (xi) (2x + 1)(3x + 1)  (xii) (3x + 2)( 4x – 1)

FACTORISING TRINOMIALS

 

 Trinomials are expressions involving three terms. Those of the form ax2 + bx + c, are called “Quadratic trinomials”.

The easiest trinomials to factorise are those where the coefficient of x is 1,

i.e. x2 + bx + c

 

Consider the trinomial x2 + 5x + 6

We need two numbers that multiply to give 6 and add to give 5.

You will recognise these numbers as 2 & 3.

Hence the factors of  x2 + 5x + 6 are (x + 2)(x + 3)

 

Consider the trinomial x2 - 5x + 6

We need two numbers that multiply to give 6 and add to give -5.

You will remember that two negative numbers multiply to give a positive.

So the numbers we want are -2 & -3.

Hence the factors of  x2 - 5x + 6 are (x - 2)(x - 3)

 

Consider the trinomial x2 + x - 6

We need two numbers that multiply to give -6 and add to give +1.

You will remember that a positive and a negative multiply to give a negative.

You will recognise these numbers as -2 & 3.

Hence the factors of  x2 + x - 6 are (x - 2)(x + 3)

 

Consider the trinomial x2 - x - 6

We need two numbers that multiply to give -6 and add to give -1.

You will recognise these numbers as 2 & -3.

Hence the factors of  x2 - x - 6 are (x + 2)(x - 3)

 

So to factorise a trinomial of the form x2 + bx + c

Look at the sign of the c. If it is positive then both factors have the same sign. If it is negative then the factors have different signs.

Look at the sign of the b. If the factors have the same sign, then the sign of the b is the sign of the factors. If the factors have different signs, then the sign of the b is the sign of the larger factor.

Apply these rules to the four examples above.

 

Exercise: Factorise the following:

(i) x2 + 7x + 12 (ii) x2 - 7x + 12        (iii) x2 - x – 12           (iv) x2 + x – 12

(v) x2 + 3x – 4              (vi) x2 - 3x – 4           (vii) x2 - 7x + 10    (viii) x2 + 3x – 10

(ix) x2 - 6x + 8              (x) x2 - 5x – 14         (xi) x2 + 6x + 8        (xii) x2 - x – 20

(xiii) x2 + 4x + 21         (xiv) x2 + x – 30       (xv) x2 + 8x + 12    (xvi) x2 - 2x - 35

(xvii) x2 - 2x - 15          (xiv) x2 - 6x – 16      (xv) x2 - 9x + 20     (xvi) x2 + 11x + 28

 

Answers:

(i) (x + 4)(x + 3)           (ii) (x - 4)(x - 3)            (iii) (x - 4)(x + 3)     (iv) (x + 4)(x - 3)

(v) (x + 4)(x - 1)           (vi) (x - 4)(x + 1)          (vii) (x - 2)(x - 5)     (viii) (x - 2)(x + 5)

(ix) (x - 4)(x - 2)           (x) (x - 7)(x + 2)           (xi) (x + 4)(x + 2)     (xii) (x + 4)(x - 5)

(xiii) (x - 3)(x + 7)        (xiv) (x - 5)(x + 6)        (xv) (x + 6)(x + 2)     (xvi) (x + 5)(x - 7)

(xvii) (x - 5)(x + 3)       (xviii) (x - 8)(x + 2)       (xix) (x - 4)(x - 5)     (xx) (x + 4)(x + 7)

FACTORISATION: COMMON FACTORS

 

The simplest method of factorisation is to take out a common factor

e.g. 2x + 4 = 2(x + 2)

Note that when we multiply everything inside the bracket by the factor we have taken out, we again get the original expression.

 

Examples:

(i) 5x2 + 10x + 20 = 5(x2 + 2x + 4)

(ii) 10x2 + 5x = 5x(2x + 1)

 

Exercise 1: Factorise the following by taking out a common factor.

(i) 5x + 10        (ii) 6g + 3         (iii) x2 + 2x       (iv) 21p2 + 7p + 14      (v) 2k2 + 4k

(vi) 8y2 + 12y + 8         (vii) 5z – 15z2               (viii) 21f 2 – 14 (ix) mn + 5m

(x) m2n – 5m                (xi) m2n + mn2              (xii) p3k2 – pk2             

 

Answers:

Exercise 1: (i) 5(x + 2)     (ii) 3(g + 1)     (iii) x(x + 2)     (iv) 7(3p2 + p + 2)

(v) 2k(k + 2)     (vi) 4(2y2 + 3y + 2)     (vii) 5z(1 – 3z)     (viii) 7(3f 2 – 2)    

(ix) m(n + 5)     (x) m(mn – 5)     (xi) mn(m + n)     (xii) pk2 (p2 – 1)

 

GROUPING in PAIRS

Where there are 4 terms it is often possible to group in pairs so that there is a common factor.

Examples:

(i) ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)

(ii) ax – ay – bx + by = a(x + y) – b(x + y) = (a – b)(x + y)

(iii) 2xp + 3p + 8x + 12 = 2p(x + 3) + 4(x + 3) = (2p + 4)(x + 3)

 

Exercise 2: Factorise the following expressions by grouping in pairs:

(i) mp + np + mq + nq     (ii) 2gh + 8g + hk + 4k          (iii) x3 – 4x2 + 2x – 8

(iv) pq + q2 – p – q            (v) 2pr – 2qr + ps – qs        (vi) 2g2 – 4gh – 3g + 6h

(vii) 2p2 – 6pq – qp + 3q2     (viii) a3 – 3a2 + 6a – 18     (ix) x3 + x2 + x + 1

(x) 15x2 + 3xyz – 10xyz + 2y2z2     (xi) 3pq – 2pr – 4r + 6q     (xii) 3pq – qr – 3p + r

(xiii) m2n2 + 3m2n + 4n + 12    (xiv) 2rs – 8pr – qs + 4pq   (xv) 5f 2g2 + 20fg2 – 3f - 12

(xvi) xyz – yz2 – 4x + 4z      (xvii) 9 – 15z + 27z2 – 45z3     (xviii) x5 – x4 + x3 – x2

(xix) a3b3 + a3b2 – ab3c – ab2c              (xx) g3h – 2g3 – 4g2h – 8g3

 

Answers:

Exercise 2: (i) (m + n)(p + q)     (ii) (2g + k)(h + 4)     (iii) (x2 + 2)(x – 4)

(iv) (p + q)(q – 1)     (v) (2r + s)(p – q)     (vi) (2g – 3)(g – 2h)     (vii) (2p – q)(p – 3q)

(viii) (a2 + 6)(a – 3)     (ix) (x2 + 1)(x + 1)     (x) (5x + yz)(3x – 2yz)

(xi) (p + 2)(3q – 2r)      (xii) (q – 1)(3p – r)     (xiii) (m2n + 4)(n + 3)    

(xiv) (2r – q)(s – 4p)    (xv) (5fg2 – 3)(f + 4)     (xvi) (yz – 4)( x-z)

(xvii) 3(1 + 3z2)(3 – 5z)     (xviii) x2(x2 + 1)(x – 1)     (xix) b2(a3 – ac)(b + 1)

(xx) g(g2 – 4g)(h + 2)

 

 

 

 

FACTORISING TRINOMIALS

 The Cross Method:

This method involves writing the factors of the first term of the trinomial on the left hand side of the cross and the factors of the last term of the trinomial on the right hand side of the cross. The factors are multiplied along the diagonals of the cross and their sum written down. A sum equal to the middle term of the trinomial indicates the correct factors.

Since there are several combinations, it is often a case of trial and error.

Consider the trinomial expression 4x2 + 8x + 3

The possible combinations are:

                       

The third combination gives the correct sum for the middle term.

Collect the terms at the top and the terms at the bottom of the third combination.

                                   

Hence the factors of 4x2 + 8x + 3 are (2x + 1)(2x + 3)

 

Product & Factor Method:

Multiply the first and the last term: 4x2 . 3 = 12x2

Determine the factors of 12x2 that add to 8x.

Possible factors:            12x, x   sum = 13x

                                    6x, 2x   sum = 8x

                                    3x, 4x   sum = 7x

As can be seen from above, the correct factors are 6x and 2x. Insert these instead of 8x.

4x2 + 6x + 2x +3 Factorise by grouping in pairs

2x(2x + 3) + 1(2x + 3)

(2x + 1)(2x + 3)

 

Exercise:

 Factorise the following trinomials by each of the above methods:

(i) 2x2 + 5x + 3     (ii) 2x2 + 7x + 3     (iii) 4x2 + 7x + 3     (iv) 4x2 + 12x + 5

(v) 6x2 + 11x + 6  (vi) 6x2 + 17x + 5  (vii) 6x2 + 31x + 5  (viii) 6x2 + 25x + 25

(ix) 2x2 + 9x + 7   (x) 4x2 + 16x + 7   (xi) 4x2 + 29x + 7    (xii) 8x2 + 34x + 21

 

Answers:

(i) (2x + 3)(x + 1)     (ii) (2x + 1)(x + 3)     (iii) (4x + 3)(x + 1)   (iv) (2x + 5)(2x + 1)

(v) (3x + 2)(2x + 3)  (vi) (2x + 5)(3x + 1)   (vii) (x + 5)(6x + 1) (viii) (2x + 5)(3x + 5)    

(ix) (2x + 7)(x + 1)   (x) (2x + 1)(2x + 7)    (xi) (x + 7)(4x + 1)   (xii) (4x + 3)(2x + 7)    

 FACTORISING TRINOMIALS (negatives)

 The Cross Method:

A cross is drawn and the factors of the first and last terms are written on opposite sides of the cross, being careful to insert correct signs.

There are several combinations, so it is a case of trial and error.

Consider the trinomial expression 6x2 + 5x - 6

The possible combinations are:

                       

                       

The last combination gives the correct sum for the middle term.

Collect the terms at the top and the terms at the bottom of the last combination.

                                   

Hence the factors of 6x2 + 5x - 6 are (3x - 2)(2x + 3)

 

Product & Factor Method:

Multiply the first and the last term: 6x2 .(-6)  = -36x2

Determine the factors of -36x2 that add to 5x.

Possible factors:            (-36x)(x)          sum = -35x

                                    (36x)(-x)          sum = 35x

                                    (-6x)(6x)          sum = 0

(12x)(-3x)        sum = 9x

(-12x)(3x)        sum = -9x

(-9x)(4x)          sum = -5x

(9x)(-4x)          sum = 5x

                                   

As can be seen from above, the correct factors are 9x and -4x. Insert these instead of 5x.

6x2 + 9x - 4x - 6 Factorise by grouping in pairs

3x(2x + 3) -2 (2x + 3)

(3x - 2)(2x + 3)

 

Exercise:

 Factorise the following trinomials by each of the above methods:

(i) 2x2 + 5x - 3     (ii) 2x2 - 9x - 5     (iii) 12x2 - 16x - 3     (iv) 4x2 - 5x - 6

(v) 6x2 - 17x - 3  (vi) 4x2 - 10x + 4  (vii) 15x2 + 17x - 4  (viii) 15x2 + 2x - 8

(ix) 20x2 + x - 12   (x) 6x2 – 17x + 12   (xi)14x2 + 13x - 12    (xii) 21x2 + 2x - 8

 

Answers:

(i) (2x - 1)(x + 3)     (ii) (2x + 1)(x - 5)     (iii) (2x - 3)(6x + 1)   (iv) (4x + 3)(x - 2)

(v) (6x + 1)(x - 3)  (vi) 2(2x - 1)(x - 2)   (vii) (3x + 4)(5x - 1) (viii) (5x + 4)(3x - 2)    

(ix) (5x + 4)(4x - 3)   (x) (3x - 4)(2x - 3)    (xi) (2x + 3)(7x - 4)   (xii) (3x + 2)(7x - 4)    

DIFFERENCE OF SQUARES

 

Multiply the factors (a + b)(a – b)

You should have obtained the result (a + b)(a – b) = a2 + ab – ab – b2 = a2 – b2

This leads to the important result in factorisation that is known as “The difference of squares” i.e. a2 – b2  = (a + b)(a – b)

Example: Factorise 4x2 – 9y2

Answer: 4x2 – 9y2 = (2x)2 – (3y)2 = (2x + 3y)(2x – 3y)

 

Exercise: Factorise the following:

(i) m2 – n2     (ii) p2 – q2     (iii) 9f 2 – g2     (iv) 16k2 – 9h2     (v) 4x2 – 25y2

(vi) 121v2 – 49w2     (vii) 81g2 – 169f 2     (viii) 18x2 – 32y2     (ix) 27p2 – 12q2

(x) 8z2 – 98w2     (xi) 16 k2 – 36 h2     (xii) 100n2 – 198m2

 

Answers:

(i) (m + n)(m – n)     (ii) (p + q)(p – q)     (iii) (3f + g)(3f – g)    (iv) (4k + 3h)(4k – 3h)    

(v) (2x + 5y)(2x – 5y)     (vi) (11v + 7w)( 11v – 7w)     (vii) (9g + 13f)(9g – 13f)

(viii) 2(3x + 4y)( 3x – 4y)     (ix) 3(3p + 2q)(3p – 2q)     (x) 2(2z + 7w)(2z – 7w)

(xi) 4(2k + 3h)(2k – 3h)     (xii) 4(5n + 7m)(5n – 7m)

 

SUM & DIFFERENCE OF CUBES

 

Multiply the factors (a + b)( a2 – ab + b2)

You should get a3 – a2b + ab2 + a2b – ab2 + b3 = a3 + b3

This leads to the important result regarding factorising the sum of two cubes:

 a3 + b3 = (a + b)(a2 – ab + b2)

  

 Multiply the factors (a - b)( a2 + ab + b2)

You should get a3 + a2b - ab2 + a2b – ab2 + b3 = a3 - b3

This leads to the important result regarding factorising the difference of two cubes:

 a3 - b3 = (a - b)(a2 + ab + b2)

 

Example: Factorise 8x3 – y3

Answer: 8x3 – y3 = (2x)3 – y3 = (2x – y)( 4x2 + 2xy + y2)]

 

Exercise:

(i) m3 – n3       (ii) p3 + r3      (iii) a3 + 27      (iv) 27k3 – h3     (v) 8a3 + 27b3   

(vi) 64g3 –125     (vii) 8p3 – 27q3     (viii) 125r3 – 27s3     (ix) 2m3 + 16n3

(x) 81a3 – 3b3   (xi) 40k3 – 5h3 (xii) 24x3 + 81y3     (xiii) 135c3 + 40d3

(xiv) 2ab3c3 + 2ad3       (xv) 3wx3y3 – 24wz3

 

Answers:

(i) (m – n)(m2 + mn + n2)    (ii) (p + r)(p2 – pr + r2)     (iii) (a + 3)( a2 – 3a + 9)

(iv) (3k – h)( 9k2 + 3kh + h2)                 (v) (2a + 3b)(4a2 – 6ab + 9b2)  

(vi) (4g – 5)( 16g2 + 20g + 25) (vii) (2p – 3q)(4p2 + 6pq + q2)

(viii) (5r – 3s)(25r2 + 15rs + 9s2)          (ix) 2(m + 2n)(m2 – 2mn + n2)

(x) 3(3a – b)(9a2 + 3ab + b2)                (xi) 5(2k – h)( 4k2 + 2kh + h2)

(xii) 3(2x + 3y)(4x2 – 6xy + 9y2)           (xiii) 5(3c + 2d)(9c3 – 6cd + 4d2)

(xiv) 2a(bc + d)(b2c2 – bcd + d2)          (xv) 3w(xy – 2z)(x2y2 + 2xyz + z2)