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Find the factors of 3 x 2 + 2 x - 1 .

  1. The quadratic is in the required form.

  2. ( x ) ( x )

    Write down a set of factors for a and c . The possible factors for a are: (1,3). The possible factors for c are: (-1,1) or (1,-1).

    Write down a set of options for the possible factors of the quadratic using the factors of a and c . Therefore, there are two possible options.

    Option 1 Option 2
    ( x - 1 ) ( 3 x + 1 ) ( x + 1 ) ( 3 x - 1 )
    3 x 2 - 2 x - 1 3 x 2 + 2 x - 1
  3. ( x + 1 ) ( 3 x - 1 ) = x ( 3 x - 1 ) + 1 ( 3 x - 1 ) = ( x ) ( 3 x ) + ( x ) ( - 1 ) + ( 1 ) ( 3 x ) + ( 1 ) ( - 1 ) = 3 x 2 - x + 3 x - 1 = x 2 + 2 x - 1 .
  4. The factors of 3 x 2 + 2 x - 1 are ( x + 1 ) and ( 3 x - 1 ) .

Factorising a trinomial

  1. Factorise the following:
    (a) x 2 + 8 x + 15 (b) x 2 + 10 x + 24 (c) x 2 + 9 x + 8
    (d) x 2 + 9 x + 14 (e) x 2 + 15 x + 36 (f) x 2 + 12 x + 36
  2. Factorise the following:
    1. x 2 - 2 x - 15
    2. x 2 + 2 x - 3
    3. x 2 + 2 x - 8
    4. x 2 + x - 20
    5. x 2 - x - 20

  3. Find the factors for the following trinomial expressions:
    1. 2 x 2 + 11 x + 5
    2. 3 x 2 + 19 x + 6
    3. 6 x 2 + 7 x + 2
    4. 12 x 2 + 8 x + 1
    5. 8 x 2 + 6 x + 1

  4. Find the factors for the following trinomials:
    1. 3 x 2 + 17 x - 6
    2. 7 x 2 - 6 x - 1
    3. 8 x 2 - 6 x + 1
    4. 2 x 2 - 5 x - 3

Factorisation by grouping

One other method of factorisation involves the use of common factors. We know that the factors of 3 x + 3 are 3 and ( x + 1 ) . Similarly, the factors of 2 x 2 + 2 x are 2 x and ( x + 1 ) . Therefore, if we have an expression:

2 x 2 + 2 x + 3 x + 3

then we can factorise as:

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

You can see that there is another common factor: x + 1 . Therefore, we can now write:

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

We get this by taking out the x + 1 and seeing what is left over. We have a + 2 x from the first term and a + 3 from the second term. This is called factorisation by grouping .

Find the factors of 7 x + 14 y + b x + 2 b y by grouping

  1. There are no factors that are common to all terms.

  2. 7 is a common factor of the first two terms and b is a common factor of the second two terms.

  3. 7 x + 14 y + b x + 2 b y = 7 ( x + 2 y ) + b ( x + 2 y )
  4. x + 2 y is a common factor.

  5. 7 ( x + 2 y ) + b ( x + 2 y ) = ( x + 2 y ) ( 7 + b )
  6. The factors of 7 x + 14 y + b x + 2 b y are ( 7 + b ) and ( x + 2 y ) .

Khan academy video on factorising a trinomial by grouping.

Factorisation by grouping

  1. Factorise by grouping: 6 x + a + 2 a x + 3
  2. Factorise by grouping: x 2 - 6 x + 5 x - 30
  3. Factorise by grouping: 5 x + 10 y - a x - 2 a y
  4. Factorise by grouping: a 2 - 2 a - a x + 2 x
  5. Factorise by grouping: 5 x y - 3 y + 10 x - 6

Simplification of fractions

In some cases of simplifying an algebraic expression, the expression will be a fraction. For example,

x 2 + 3 x x + 3

has a quadratic in the numerator and a binomial in the denominator. You can apply the different factorisation methods to simplify the expression.

x 2 + 3 x x + 3 = x ( x + 3 ) x + 3 = x provided x - 3

If x were 3 then the denominator, x - 3 , would be 0 and the fraction undefined.

Simplify: 2 x - b + x - a b a x 2 - a b x

  1. Use grouping for numerator and common factor for denominator in this example.

    = ( a x - a b ) + ( x - b ) a x 2 - a b x = a ( x - b ) + ( x - b ) a x ( x - b ) = ( x - b ) ( a + 1 ) a x ( x - b )
  2. The simplified answer is:

    = a + 1 a x

Simplify: x 2 - x - 2 x 2 - 4 ÷ x 2 + x x 2 + 2 x

  1. = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) ÷ x ( x + 1 ) x ( x + 2 )
  2. = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) × x ( x + 2 ) x ( x + 1 )
  3. The simplified answer is

    = 1

Simplification of fractions

  1. Simplify:
    (a) 3 a 15 (b) 2 a + 10 4
    (c) 5 a + 20 a + 4 (d) a 2 - 4 a a - 4
    (e) 3 a 2 - 9 a 2 a - 6 (f) 9 a + 27 9 a + 18
    (g) 6 a b + 2 a 2 b (h) 16 x 2 y - 8 x y 12 x - 6
    (i) 4 x y p - 8 x p 12 x y (j) 3 a + 9 14 ÷ 7 a + 21 a + 3
    (k) a 2 - 5 a 2 a + 10 ÷ 3 a + 15 4 a (l) 3 x p + 4 p 8 p ÷ 12 p 2 3 x + 4
    (m) 16 2 x p + 4 x ÷ 6 x 2 + 8 x 12 (n) 24 a - 8 12 ÷ 9 a - 3 6
    (o) a 2 + 2 a 5 ÷ 2 a + 4 20 (p) p 2 + p q 7 p ÷ 8 p + 8 q 21 q
    (q) 5 a b - 15 b 4 a - 12 ÷ 6 b 2 a + b (r) f 2 a - f a 2 f - a
  2. Simplify: x 2 - 1 3 × 1 x - 1 - 1 2

Summary

  • Product of two binomials
  • Factorising
  • Distributive law
  • Sum and difference of cubes
  • Factorising quadratics
  • Factorising by grouping
  • Simplifying fractions

End of chapter exercises

  1. Factorise:
    (a) a 2 - 9 (b) m 2 - 36 (c) 9 b 2 - 81
    (d) 16 b 6 - 25 a 2 (e) m 2 - ( 1 / 9 ) (f) 5 - 5 a 2 b 6
    (g) 16 b a 4 - 81 b (h) a 2 - 10 a + 25 (i) 16 b 2 + 56 b + 49
    (j) 2 a 2 - 12 a b + 18 b 2 (k) - 4 b 2 - 144 b 8 + 48 b 5

  2. Show that ( 2 x - 1 ) 2 - ( x - 3 ) 2 can be simplified to ( x + 2 ) ( 3 x - 4 )

  3. What must be added to x 2 - x + 4 to make it equal to ( x + 2 ) 2

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Source:  OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2
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