# algebra

posted by .

The first three equations I have solved, I would just appreciate someone checking them over to make sure I'm doing them right.
Find the GCF:
16x^2z ,40xz^2 , 72z^3
= 3^3z

Factor our GCF:
a(a+1) - 3(a+1)
= a+1

Factor Polynomial:
9a^2 - 64b^2
= (3a+8b)(3a-8b)

9w - w^3
= w(3) (?really unsure if correct?)

I'm not sure how to do the following, or where to even begin. Some help getting started would be very appreciated! :)

Factor Polynomials:
x^3y + 2x^2y^2 + xy^3

x^3 + ax + 3a + 3x^2

Factor:
18z + 45 +z^2

• algebra -

16x^2z ,40xz^2 , 72z^3
the GCF I see is 4xz, I can't see how you got your answer.

a(a+1) - 3(a+1)
= (a+1)(a-3) fully factored

9a^2 - 64b^2
= (3a+8b)(3a-8b) that is ok

9w - w^3 , I see a common factor of 3w
= 3w(3 - w^2)

x^3y + 2x^2y^2 + xy^3 first go for a common factor
= xy(x^2 + 2xy + y^2)
= xy(x+y)(x+y)
= xy(x+y)^2

x^3 + ax + 3a + 3x^2
this is a "grouping" type of factoring
= x^3 + 3x^2 + ax + 3a
= x^2(x+3) + a(x+3)
= (x+3)(a+3)

18z + 45 +z^2
= z^2 + 18z + 45
= (z+15)(z+3)

• algebra -

in the first one: 16x^2z ,40xz^2 , 72z^3
the only question I have is, how can it be 4xz when they don't all have the variable 'x'? the way I factored it was like this:
16x^2z
16 = 8 & 2 x^2z
4&2 (x)(x)(z)
2&2
= 2^4

40xz^2
40 = 10 & 4 xz^2
5&2 2&2 (x)(z)(z)
= 5(2^3)

72z^3
72 = 36 & 2 z^3
18&2 (z)(z)(z)
9&2
3&3
= 3^2(2^3)

with the answer being 2^3z, which is what they all have in common

• algebra -

3w-3