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March 31, 2015

March 31, 2015

Posted by **kay bri** on Friday, January 23, 2009 at 2:17pm.

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 -
**Reiny**, Friday, January 23, 2009 at 2:27pm16x^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 -
**kay bri**, Friday, January 23, 2009 at 3:16pmin 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 7:47pm3w-3

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