If you are factorising expressions you need to look for the highest common factor (hcf);
for example
15y + 10 = 5(3y+5)
hcf = 5
(5 x 3y = 15y) + (5 x 2 = 10)
(5 x 3y) + (5 x2)
Put the hcf at the beginning of the equation!
5(3y+5)
However if you are factorising quadratic expressions the method is similar but slightly similair. You need to find the factors of the last number that will make the sum of the middle number. I know your thinking it's difficult but once you know how to do it, it's easy!
p2 - 8p + 15 = (p-3)(p-5)
1. Find the factors of 15
15 x 1
3 x 5
2. Find the two which will sum up to 8
3 x 5
3. Use the squared number/letter in the expression
(p-3)(p-5)
It can get harder be aware, so if you need more help I suggest youu search the internet.
Answer B
Just to correct that the above answer is wrong for factorising 15y + 10.
The answer is 5(3y + 2). You have added 5 and 5 instead of timesing 5 and 2 which should give 10.
Also the example that you gave for p2 - 8p + 15 = (p-3)(p-5) wasn't correct but the answer was.
1. Firstly we find the factors of 15 such as:
5x3
15x1
-5x-3
these all equal to 15.
2. Then we find the factors of 15 that add up to -8
which are -5 and -3
3. Now we put it all together. P² will set up as (p )(p ) this means p² in expanding brackets.
Now it becomes: (p-5)(p-3) if we expand this answer which is doing the opposite of factorising, the answer will become back to p2 - 8p + 15.
In the same way that you would factorise any other expressions that do not contain an equality sign depending on the expressions of which none have been given.
For example: 9x2-25 = (3x-5)(3x+5) but this only applies to negative expressions it wouldn't apply to 9x2+25
Factorise fully is when brackets are involved in the equation
m(g-a)
5(3-2x)
In the same way that you would factorise any other expressions that do not contain an equality sign depending on the expressions of which none have been given.
For each of a list of algebraic expressions, find one or more common factors and factorise the expression.
Placing a question mark at the end of a list of expressions or numbers does not make it a sensible question. What do you want? To evaluate it (impossible), factorise it, something else?
For example: 9x2-25 = (3x-5)(3x+5) but this only applies to negative expressions it wouldn't apply to 9x2+25
Factorise fully is when brackets are involved in the equation
a²-a = a(a-1)
To factorise is to find the numbers that divide into the original number by only using prime numbers. For example factorise 20 = 2 times 2 times 5
to put into brackets
you do (245x)
The answer will depend on where the brackets are. In general the solution would be to expand all the brackets, combine like terms and then factorise.
6(t2s)
m(g-a)