Question

factor:
9w - w^3 = 3w(3-w^2)
m^4 - n^4 = (m-n)^2(m-n)^2
are these correct?
the next few I wasn't sure how to work...
factor:
2h^2 - h - 3 = 0

x^2 - 36 = 0

x^3 = 4x

Answer 1

Your answers aren't right. You can check factorizations pretty easily by just multiplying.

(3w)(3 - w^2) = (3w)(3) - (3w)(w^2)
=9w - 3w^2

((m-n)^2)((m-n)^2) = (m-n)^4 =/= m^4 - n^4.

Some hints:
9w - w^3 : Factor out the w first. This leaves a difference of two squares. Remember that (x^2 - y^2) = (x+y)(x-y).

m^4 - n^4 : Difference of two squares. One of the factors is also a difference of two squares.

2h^2 - h - 3 = 0 : Quadratic, so factor the left side into (2h +/- a)(h +/- b).

x^2 - 36 = 0 : Left side is a difference of two squares.

x^3 = 4x : Move everything to the left side. Factor out the x. You'll be left with a difference of two squares.

Answer 2

9w - w^3
Most of your questions can be reduced to the difference of two squares.
a^2 - b^2 = (a+b)(a-b)
for example:
9w-w^3
first factor out the common variable w
w (9-w^2)
NOW we have w times (square of 3 - square of w
w (3+w)(3-w)

Answer 3

Bow look at the second one
m^4 - n^4
that is (m^2)^2 - (n^2)^2
which we know is
(m^2+n^2)(m^2-n^2)
now that second factor is also the difference of two squares so
(m^2+n^2)(m+n)(m-n)

Answer 4

2h^2 - h - 3 = 0
Now try to factor that
(2 h ? 3)(h ? 1)
or
(2 h ? 1)(h ? 3)
well
(2 h -3)(h+1) = 0 works
the left is zero if either factor is zero so two solutions
h = 3/2
or
h = -1

Answer 5

x^2 - 36 = 0
back to difference of two squares
(x+6)(x-6) = 0
then do like the last one

x^3 = 4x
well that is
x (x^2-4) = 0
or
x(x+2)(x-2) = 0
then proceed as before x = 0, x = 2, x = -2