• Dec 9th 2007, 03:42 PM
MathMack
here they are. THANK YOU! YOUR HELP IS WELL NEEDED!

1.) 3x^2(2x^3-5x+3)

2.) (3z+4) (3z-4)

3.) (2x-8) ^2

4.) 3y^4 - 5y^3 + 9y^2 all divided by 3y^2

5.) x^3 + 4x^2 + 7x + 28 all divided by X + 4
• Dec 9th 2007, 05:37 PM
colby2152
Quote:

Originally Posted by MathMack
here they are. THANK YOU! YOUR HELP IS WELL NEEDED!

1.) 3x^2(2x^3-5x+3)

2.) (3z+4) (3z-4)

3.) (2x-8) ^2

4.) 3y^4 - 5y^3 + 9y^2 all divided by 3y^2

5.) x^3 + 4x^2 + 7x + 28 all divided by X + 4

1) $6x^5 - 15x^3 + 9x^2$
2) $9z^2-16$
3) $4x^2-32x+64$
4) $y^2 - \frac{5y}{3} + 3$
• Dec 9th 2007, 06:07 PM
MathMack
Thanks Colby, can you show me how you got them answers if you don't mind. Thanks again for your help, I appreciate it
• Dec 9th 2007, 06:14 PM
MathMack
just 3&4, I worked out the other two. Thanks again
• Dec 9th 2007, 06:29 PM
MathMack
nevermind on 3 & 4 I worked them out! Thank you again!
• Dec 9th 2007, 06:29 PM
colby2152
Quote:

Originally Posted by MathMack
just 3&4, I worked out the other two. Thanks again

$\frac{3y^4 - 5y^3 + 9y^2}{3y^2} = \frac{3y^4}{3y^2} - \frac{5y^3}{3y^2} + \frac{9y^2}{3y^2}$
$= y^2 - 1.67y + 3$
• Dec 10th 2007, 02:31 AM
Quote:

5.) x^3 + 4x^2 + 7x + 28 all divided by X + 4
$\frac {x^3+4x^2+7x+28}{x+4}$

= $\frac {x^3+4x^2}{x+4} + \frac {7x+28}{x+4}$

= $\frac{x^2(x+4)}{x+4} + \frac {7(x+4)}{x+4}$

= $x^2+7.$

The difficult part here is explaining how I knew to separate the original numerator into pairs like that. Really, its just practice and checking each possible combination of pairs for two that can cancel with the denominator. There is a way of doing this kind of problem that does not require trial and error and will always work, but it is probably not necessary for you to learn at this stage. If you are interesting in learning it you can google "polynomial long division"
• Dec 10th 2007, 07:08 AM
topsquark
Quote:

Originally Posted by colby2152
$= y^2 - 1.67y + 3$

In general, you should leave the coefficients as fractions. The only time I would say that decimal coefficients are okay is if the problem is in either Engineering or one of the Sciences.

-Dan