BSB Year 9 is working well at factorisation (we called it factorising in lessons; American and Canadian spelling is factorizing) and this page is written for them.

Theory: Using the first four greek letters, α,β,γ,δ;

(αx+β)(γx+δ) = αγx2 + (αδ+βγ)x +βδ  so for any quadratic expression ax2 + bx + c then ac=αβγδ that is a and c multiply to the number αβγδ which has alternative factors αδ and βγ. this leads to the understanding that the class has: if the x2 coefficient is not one then multiply the two end numbers and look for factors of that number [αβγδ] that add to make the middle (x) coefficient.

Factorise these expressions: Note that some questions are connected. Assume that if a homework is set form this page then it is intended that you do questions 1-20 OR 21-40 OR 41-50.

1     x² + 8x + 12                            21        x² + 8x + 12                                41        x² + 8x + 12

2     x² + 8x + 7                              22        x² + 4x - 12                                 42        x² + 8x + 12

3     x² + 8x +15                             23        x² + 13x + 12                              43        x² + 8x + 12

4    3x² + 16x + 5                           24        x² - x -  12                                   44        x² + 8x + 12

5    5x²  - 8x  + 3                            25        2x² + x - 6                                   45        x² + 8x + 12

6    2x²+ 16x + 24                         26        x² + 8x + 24                                46        x² + 8x + 12

7    2x² + 5x - 12                           27        2x² + 8x + 12                              47        x² + 8x + 12

8    2x² - 10x  - 12                         28        3x² + 8x + 8                                48        x² + 8x + 12

9    12x² - 10x + 2                         29        x² + 8x + 12                                49        x² + 8x + 12

10   12x² - 5x  - 2                          30        x² + 8x + 12                                50        x² + 8x + 12

Be careful to factorise a question completely.  If a term has a common factor left in it, e.g. (2x+8) then you have not quite finished, as 2x+8 = 2(x+4). Good students spot a common factor in all three terms before they start writing a pair of brackets. The best students check both before and afterwards - but the issue is, did you succeed in factorising completely?.

Solving quadratics starts off with factorisation. When the factorisation is finished there is a result of the form (αx+β)(γx+δ) = 0. This is two terms (or numbers, for a value of x) which multiply to make zero. So one of them is zero. Some people read this but do not ‘get’ it; make sure you are not among those folk.
As x changes values, there will be two solutions, one for each term. Either αx+β = 0 [which means x = -β/α]  or γx+δ = 0 [which would mean that x = -δ/γ]. Do not confuse the process of finding the value of x for which a term is zero with the factorisation process. Solving produces roots of the equation; factorising only finds factors.

Solve these.

11     x² + 8x + 12 = 0                                                                              (both roots are negative)

12     x² - 8x + 15 = 0                                                                                (both roots are positive)

13     x² - 18x + 45 = 0

14    3x² + 16x + 5 = 0

15    5x²  - 8x  + 3 = 0

....now write out all the factors of 144 (yes, really) and then attempt these harder questions:

16      x² + 15x + 144 = 0

17    2x² + 18x   - 72 = 0

18    4x²  - 10x   - 36 = 0

19    12x²  - 51x  + 12 = 0

20    24x²  - 7x   - 6 = 0

Again, I may well extend these by another ten or twenty problems.

DJS 20130415