A linear expression can only have a numeric factor that can be "taken out".
The expression will be of the form ax + b where a and b are numbers that have k as their highest common factor (HCF).
That is, a = k*c
and b = k*d
Than being the case, ax + b = kcx + kd = k*(cx + d)
4
To factorise the expression (x^2 - x - 6), we need to find two numbers that multiply to (-6) (the constant term) and add to (-1) (the coefficient of the linear term). The numbers (-3) and (2) satisfy these conditions. Therefore, we can write the factorised form as ((x - 3)(x + 2)).
a²-a = a(a-1)
you do (245x)
It is the opposite of Expanding The Brackets
4
To factorise the expression (x^2 - x - 6), we need to find two numbers that multiply to (-6) (the constant term) and add to (-1) (the coefficient of the linear term). The numbers (-3) and (2) satisfy these conditions. Therefore, we can write the factorised form as ((x - 3)(x + 2)).
If you know one linear factor, then divide the polynomial by that factor. The quotient will then be a polynomial whose order (or degree) is one fewer than that of the one that you stared with. The smaller order may make it easier to factorise.
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
you do (245x)
to put into brackets
-180
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)
It is the opposite of Expanding The Brackets