Simplifying formulas » Factorising formulas
Contents
1. Why?2. Terms and factors
3. How do you factorise
4. Special: the trinomial (with a = 1)
5. For the advanced
1. Why?
Factorising formulas is especially important when solving (quadratic) equations.
When reducing formulas you normally have to remove all the brackets, but in special cases, for example with fractional formulas, sometimes you can use factorisation to shorten a formula.
2. Terms and factors
Term: something you add or subtract (subtracting is adding a negative number).
Factor: something you multiply.
Sum = term + term
Product = factor × factor
Example
p = 5(3q – 7)
The 5 and 3q – 7 in the formula are factors. These are multiplied.
In factor 3q – 7 are 3q and –7 terms. Those two are added.
In the term 3q you have the factors 3 and q.
3. How do you factorise?
You try to find the greatest common multiple in all terms.
This is put in front of the brackets. Your factorised formula is always of the form:
y = …(… + …) or y = …(… – …).
Example 1
y = 6x2 + 15x
y = 3x(2x + 5)
In both terms of formula y = 6x2 + 15x is a factor 3x.
Because you can write the formula as y = 2 × 3 × x × x + 3 × 5 × x.
Example 2
y = 8x2 + 4x – 20
y = 4(2x2 + x – 5)
In all terms of formula y = 8x2 + 4x – 20 is a factor 4.
Because you can write the formula as y = 2 × 4 × x × x + 4 × x – 4 × 5.
Example 3
y = 5x2 + 5x
y = 5x(x + 1)
In both terms of formula y = 5x2 + 5x is a factor 5x.
Because you can write the formula as y = 5 × x × x + 5 × x × 1.
4. Special: the trinomial (with a = 1)
A trinomial is a quadratic formula in the form of y = ax2 + bx + c.
When there is no number in front of x2 and therefore a = 1, you can use the sum-product-method to factorise the trinomial.
A sum is an answer to an addition and a product is the answer to a multiplication.
Your factorised formula is always of the form y = (x + …)(x + …)
Behind x on the open spaces will be two numbers.
These two numbers should be added b and multiplied c (from y = ax2 + bx + c).
All the work is in finding these two numbers.
Factorise y = x2 + 12x + 35.
Because b = 12 and c = 35 we have to find to numbers added 12 and multiplied 35.
It has no use starting at 12. There is an infinite number of possibilities to have to numbers added 12. For that reason, you start with 35. There are only 4 pairs of numbers multiplied 35:
1 × 35, –1 × –35, 5 × 7 and –5 × –7.
Of these four only 5 and 7 are added 12.
So these two numbers are the ones we need to fill in:
y = x2 + 12x + 35 = (x + 5)(x + 7)
| y = x2 – 10x + 24 y = (x – 4)(x – 6) |
y = x2 + 3x – 18 y = (x + 6)(x – 3) |
y = x2 – x – 12 y = (x + 3)(x – 4) |
You still find it difficult to find the two numbers?
Make a table in which you write all the possibilities.
Example
Factorise y = x2 + 7x – 60.
We have to find two numbers multiplied –60.
Make a table and start with factor 1, that is always possible.
Then you try factor 2, et cetera.
Make the addition column and find the correct pair.
| Multiplied –60 | Added? |
|
–1 × 60 1 × –60 –2 × 30 2 × –30 –3 × 20 3 × –20 –4 × 15 4 × –15 –5 × 12 5 × –12 –6 × 10 6 × –10 |
59 –59 28 –28 17 –17 11 –11 7 –7 4 –4 |
The correct pair is –5 and 12, as these have a sum of 7.
The answer is therefore y = (x – 5)(x + 12).
5. For the advanced
|
y = x3 – x2 – 6x y = x(x2 – x – 6) y = x(x + 2)(x – 3) |
b = 27a6 – 18a4 – 36a2 b = 9a2 · 3a4 – 9a2 · 2a2 – 9a2 · 4 b = 9a2(3a4 – 2a2 – 4) |
|
q = a2(2b – 1) + 10b – 5 q = a2(2b – 1) + 5(2b – 1) q = (a2 + 5)(2b – 1) |
y = x8 – x6 y = x6(x2 – 1) y = x6(x – 1)(x + 1) (see strange products) |