Posted by kay bri on Friday, January 23, 2009 at 2:17pm.
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)(3a8b)
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  Reiny, Friday, January 23, 2009 at 2:27pm
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)(a3) fully factored
9a^2  64b^2
= (3a+8b)(3a8b) 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  kay bri, Friday, January 23, 2009 at 3:16pm
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  Anonymous, Monday, October 3, 2011 at 7:47pm
3w3
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