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# [(3x^2)(y^3)]^2 * (6x)^-3 divided by (9xy^3)^2 (X+ )(X+ )

### Resolved Question:

[(3x^2)(y^3)]^2 * (6x)^-3 divided by (9xy^3)^2

(X+?)(X+?)= 23x^2 + 12X + 32
Submitted: 6 years ago.
Category: Math Homework
Expert:  abozer replied 6 years ago.

abozer :

Hi,

abozer :

= [(3x^2)(y^3)]^2 * (6x)^-3 divided by (9xy^3)^2

Customer :

hi

Customer :

oooo

abozer :

= (3x^2)^2(y^3)^2 * (6x)^-3 divided by (81x^2y^6)

Customer :

two different problems

abozer :

= 9x^4y^6 * 6^-3x^-3 divided by (81x^2y^6)

Customer :

ok ic

abozer :

= 9xy^6 divided by (81x^2y^6)6^3

abozer :

= 1 / 1944x

abozer :

Second question.

abozer :

3x^2 + 12X + 32

abozer :

This is not factorable.

abozer :

oh sorry,

Customer :

ooo i knew it was wierd

abozer :

23x^2 + 12X + 32

abozer :

no, it is still not factorable.

abozer :

hold on.

abozer :

is it x^2 + 12X + 32 ?

abozer :

abozer :

=(x + 4)(x + 8)

abozer :

23 is the question number I guess.

Customer :

oo my bad 23 was problem number

abozer :