Factorise polynomials

Key Notes:

Definition of Factorisation

  • Factorisation is the process of expressing a polynomial as a product of its factors.

Common Factors Method

  • Identify and take out the greatest common factor (GCF) from all terms.
  • Example: 6x2 + 9x = 3x(2x + 3).

Grouping Method

  • Group terms to find common factors within pairs.
  • Example: ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y).

Factorising Quadratic Polynomials

  • For polynomials in the form ax2 + bx + c, find two numbers that multiply to acacac and add to bbb.
  • Example: x2 + 5x + 6 = (x + 2)(x + 3).

Difference of Squares

  • Use the identity a2 − b2 = (a − b)(a + b).
  • Example: x2 − 9 = (x − 3)(x + 3).

Perfect Square Trinomials

  • Recognise and factorise expressions like a2 + 2ab + b2 = (a + b)2.
  • Example: x2 + 6x + 9 = (x + 3)2.

Cubic Polynomials (Factor Theorem)

  • If f(a) = 0, then (x − a) is a factor of f(x).
  • Use synthetic or polynomial division for further factorisation.

Trial and Error Method

  • Check for integer values of xxx that make the polynomial zero and use them to factorise.

Using Algebraic Identities

Apply formulas like:

  • (a + b)2 = a2 + 2ab + b2
  • (a − b)2 = a2 − 2ab + b2
  • (a + b)(a − b) = a2 − b2

Checking the Factors

  • Multiply the factors to verify if they give the original polynomial.

Learn with an example

🎻 Factorise.

16c2–8c+1=——

Notice that 16c2–8c+1 is a perfect square trinomial because it can be written in the form a2–2ab+b2, where a is 4c and b is 1.

a2–2ab+b2

(4c)2–2 . 4c . 1+12

16c2–8c+1

Now use the formula for factorising perfect square trinomials.

a2–2ab+b2=(a–b)2

(4c)2–2 . 4c . 1+12=(4c–1)2

16c2–8c+1=(4c–1)2

The factorised form of 16c2–8c+1 is (4c–1)2.

Finally, check your work.

(4c–1)2

(4c–1)(4c–1)Expand

16c2–4c–4c+1Apply the distributive property (FOIL)

16c2–8c+1

Yes, 16c2–8c+1=(4c–1)2.

🎻 Factorise.

3w2+8w+5=——

Look at the given quadratic:

3w2+8w+5

The product ac is 15, so you need to find a pair of factors with a product of 15. The b term is 8, so you need to find a pair of factors with a sum of 8. Since the product is positive (15) and the sum is positive (8), you need both factors to be positive.

Make a list of the possible factor pairs with a product of 15, and then find the one with a sum of 8.

Factor pairs of ac=15Sum of factor pairs
1 . 15=151+15=16
3 . 5=153+5=8

The factors 3 and 5 have a sum of 8. So, replace the quadratic’s 8w term with 3w and 5w, and then factor by grouping.

3w2+8w+5

3w2+3w+5w+5

3w(w+1)+5(w+1)Factor by grouping; the expressions in brackets should match

(3w+5)(w+1)

Finally, check your work.

(3w+5)(w+1)

3w2+5w+3w+5Apply the distributive property (FOIL)

3w2+8w+5

Yes, 3w2+8w+5=(3w+5)(w+1).

🎻 Factor.

12fg+6f+10g+5=—–

Factor by grouping.

12fg+6f+10g+5

6f(2g+1)+5(2g+1) Factor by grouping; the expressions in brackets should match

(6f+5)(2g+1) Apply the distributive property

let’s practice!