please help me?

Question

the book says factor each p;oynomial completely. to begin, state which method should be applid as the 1st step, then continue.

2p-6q+pq-3q^2

and

m^4-9n^4

Answer 1

2p - 6q + pq - 3q²

2(P - 3q) + q(p - 3q)

(2 + q)(p - 3q)

- - - - - - - - - -

m⁴- 9n²

m⁴- 3m²n + 3m²n - 9n²

m²(m² - 3n) + 3n(m ² - 3n)

(m² + 3)(m² - 3n)

- - - - - - - - -s-

Answer 2

Given expression
=2p+pq-6q-3q^2
=p(2+q)-3q(2+q)
=(2+q)(p-3q)
=)q+2)(p-3q)

m^4-9n^4
=(m^2)^2-(3n^2)^2
=(m^2+3n^2)(m^2-3n*2) [a^2-b^2=(a+b)(a-b)]

Answer 3

you can write the first as

2(p-3q)+q(p-3q) = (p-3q)*(q+2)

m^4-9n^4= (m^2+3n^2)*(m^2-3n^2)
the second factor can still be factored as (m-sqrt3*n)*(m+sqrt3*n)
So you can put all factors together

Answer 4

1)First step_2(p-3q)+q(p-3q)
=(2+q) (p-3q)

2)First step_(m^2)^2 – (3n^2)^2

=(m^2+3n^2) (m^2-3n^2)

Answer 5

2p-6q+pq-3q^2
=p(2+q)-3q(2+q)
=(2+q)(p-3q)

m^4-9n^4
=(m²-3n²)(m²+3n²)