how would you factorise this?

(* represents multiplication)
3x^2+27x-30
first we use normal factorization
= 3x^2+3*9-3*10 ( 3 is common so we take it out)
3(x^2+9-10) now we use the third identity (x+a) (x+b) = x^2+(a+b)x+axb
or (x-a)(x-b) = x^2=(-a-b)x-a*-b
to solve this the same numbers must be multiplied to form 10 which are added or subtracted to form 9 and then we put them in the form of (x+a)(x+b) or (x-a)(x-b)
so
x^2 +(10-1)x+10*-1
Answer=(x+10)(x-1)

HOPE THIS HELPS.

Factorise:

3x^2 + 27x - 30

Factor out the Greatest Common Factor (GCF) = 3

3(x^2 + 9x - 10)

Factor a trinomial

3(x + 10)(x - 1)

3x^2 + 27x - 30
or,3x^2+30x-3x- 30
or,3x(x+10) -3(x+10)
or,(x+10)(3x-3)
or,3(x-1)(x+10).
this is the answer.

3x^2 + 27x - 30
= 3(x^2 + 9x - 10)
= 3(x^2 + 10x - x - 10)
= 3[(x^2 + 10x) - (x + 10)]
= 3[x(x + 10) - 1(x + 10)]
= 3(x + 10)(x - 1)

split 27 as 30-3 and solve

take out common factor... this one has a common factor of 3.

so..

3(x^2+9x-10)=0 ---------- = new eqn right??

and just solve for x.
as in.. factorise the expression inside brackets.

3(x+10)(x-1)=0

and.. just expand each bracket..

so its..

3x+10 = 0 3x-3=0
x=-10/3 or x=+1

3x^2+27x-30
=3(x^2+9x-10) [take 3 common factor outside]
=3(x+10)(x-1) {factorise the term using V method}

3(x^2+9x-10)


3(x+10)(x-1)

3x^2+27x-30 =3 (x-1) (x+10)

=3(x^2+9x-10)=3(x+10)(x-1)