Factor 3p^2 - 3 p?

Answer 1

3p(p-1)

Answer 2

First evaluate x^3 - y^3 This components into (x - y)(x^2 + xy + y^2) putting p + q = x and p - q = y, we've (p + q - (p - q))((p + q)^2 + (p + q)(p - q) + (p - q)^2) (p + q - p + q)(p^2 + 2pq + q^2 + p^2 - q^2 + p^2 - 2pq + q^2) (2q)(3p^2 + q^2) You provide up right here in case you're factoring over the integers or the reals. in case you're factoring over the complexes, considering 3 = (sqrt(3))^2, you have (2q)(sqrt(3)p + qi)(sqrt(3)p - qi) the place i = sqrt(-a million). in spite of this, you are able to multiply out first: (p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3 (p - q)^3 = p^3 - 3p^2q + 3pq^2 - q^3 so (p + q)^3 - (p - q)^3 = 6p^2q + q^3 = (2q)(3p^2 + q^2), and so on.

Answer 3

Here's your original expression:

3p² - 3p

Now factor out "3p":

3p(p - 1)

Answer 4

take a 3p out
=
3p(p-1)

Answer 5

3 and p go into both
So take them outside the bracket (factorise)
3p(p - 1)

Answer 6

=3p(p-1)

Answer 7

3p(p-1)

Answer 8

-b-4ac
(must equal perfect square)

3-4(3)(0)

3-0

=3

Not a perfect square therefore not factorable.