Cause I'm not sure of my answers :(

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- Dec 8th 2009, 04:47 AMiraisavampireBasic College Algebra (Factoring + Polynomial Operations)
Cause I'm not sure of my answers :(

http://img268.imageshack.us/img268/5342/ps45.jpg - Dec 8th 2009, 04:59 AMDefunkt
Post your answers and we'll tell you if you're right/wrong and where.

- Dec 8th 2009, 05:05 AMiraisavampire
(#s 5,8,9 are factoring)

5. i'm stuck at completing the square, dunno what to add to 20x2y2 to make it a perfect square trinomial and then a difference of two squares.

8. totally lost on this one.

9. does this end up being [(xy -3x -y -3) (x2y2 -6x2y +9x2 +2xy2-18x +y2 +6y +9)]? - Dec 8th 2009, 05:05 AMI-Think
For number 5, see that

$\displaystyle (3x^2+4y^2)^2=9x^2+24x^2y^2+16y^2

$

Hence

$\displaystyle 9x^4+20x^2y^2+16y^4=(3x^2+4y^2)^2-4x^2y^2$ - Dec 8th 2009, 05:23 AMiraisavampire
Thanks!

- Dec 9th 2009, 06:00 AMmr fantastic
Please don't post more than two questions in a thread. Otherwise the thread can get convoluted and difficult to follow. Start new threads as necessary for remaining questions. eg. If you have five questions, post two of them in two threads and start a new thread for the remaining one etc.

And if the question has more than two parts to it, it is best to post only that question and its parts in the thread and start a new thread for other questions.

I will do #8 and #9 before closing the thread:

8) The first three terms have a common factor and then can be factorised as a perfect square. Therefore you have $\displaystyle 4(a + b)^2 + 8(a + b) + 3$.

Now let $\displaystyle x = (a + b)$, factorise $\displaystyle 4x^2 + 8x + 3$ and then substitute back that $\displaystyle x = (a + b)$.

9) You have $\displaystyle x^3 (y^3 - 27) - (y^3 - 27) = (x^3 - 1)(y^3 - 27)$. Now factorise each of these (they are differences of perfect cubes).

Thread closed (pm me if you need more help on 8) and 9). Be sure to say exactly where you're stuck).