factorize 30x - 5x^2 fully?

(5x + 5x)(3x -x)=12X^2 not 30x

try this:
x [ 5 ( 6-x) ]

30x - 5x² = 5x(6 - x)

30x-5x²

5x.(6-x)



kisses!1`am brazil, i not speak english very good.

x*(30-5x)= 5x*(6-x) since you dont have a constant its not a true quadratic

The above equates to 2q(p-q) which would be whether p and q are integers before everything. you will do (p-q) first, then multiply this by utilising q. Then this is not appropriate no count if q(p-q) is ordinary or maybe because of the fact any integer prolonged by utilising 2 is even. in case you have been questioning what would happen if p=q or p and q are 0, 0 is an integer and this is even. Fretty, you have further q^2 on relatively of taking it away.

30x - 5x²

well you would only have one bracket because there are only two terms.
find the highest common term in the equation - in this case its 5x - and put this outside the bracket
then work out what you need multiply 5x by to get your two terms - in this case you times 6 by 5x to get 30x, and times x by 5x to get 5x²
so it would look like:
5x(6-x)

or if you wanted to stick to the normal pattern and have two brackets,
(5x+0)(6-x)

30x - 5x^2
= 5(6x - x^2)
= 5x(6 - x)
= -5x(-6 + x)
= -5x(x - 6)

5x (6 - x)

Factorize:

30x - 5x^2

Factor out the Greatest Common Factor (GCF) = 5x

5x(6 - 1x)

The answer is:

5x(6 - 1x)

5x(6-x)