# Thread: Factoring a polynomial/simplifying

1. ## Factoring a polynomial/simplifying

I have an test tomorrow. I studied everything..but i know my teacher will give us hard questions he did not even solve in the class..

how to factor this polynomial:
4y^2 - 81

i know that a^2 - b^2 = (a-b) (a+b)

Simplify:
(m^2 n ^3)^-2 (m^-3 n^2)^-3

we learned that if we have (m^2 m^3)^2 then the answer is m^2 n^3 x m^2 n^3 but how about if we have a -ve square or -3?

√8 + √18

according to my teacher the solution is..
= √(2x2x2) + √(3x3x3)
=2√2 + 3√2
=5√3

how is that possible?

2. Originally Posted by Cherry

√8 + √18

according to my teacher the solution is..
= √(2x2x2) + √(3x3x3)
=2√2 + 3√2
=5√3

how is that possible?
the solution is
$\displaystyle \sqrt{8}+\sqrt{18}$
$\displaystyle =\sqrt{2*2*2}+\sqrt{3*3*2}$
$\displaystyle =2\sqrt{2}+3\sqrt{2}$
$\displaystyle =5\sqrt{2}$

Keep Smiling
Malay

3. Originally Posted by Cherry;28279
how to factor this polynomial:
[B
4y^2 - 81[/B]

i know that a^2 - b^2 = (a-b) (a+b)

2y=a
9=b

Keep Smiling
Malay

4. Originally Posted by Cherry

Simplify:
(m^2 n ^3)^-2 (m^-3 n^2)^-3

we learned that if we have (m^2 m^3)^2 then the answer is m^2 n^3 x m^2 n^3 but how about if we have a -ve square or -3?
do you mean
$\displaystyle \frac{1}{({m^2n^3})^2}$*$\displaystyle \frac{1}{({m^{-3}}{n^2})^3}$
it is the same with negative powers
$\displaystyle (m^{-2})^2=m^{-2}*m^{-2}=m^{(-2)+(-2)}=m^{-4}$
hence your expression simplifies to
$\displaystyle \frac{1}{{m^4n^6}}$*$\displaystyle \frac{1}{{m^{-9}}{n^6}}$
$\displaystyle =\frac{1}{{m^4m^{-9}}}$*$\displaystyle \frac{1}{{n^6n^6}}$
$\displaystyle =\frac{1}{{m^{4+(-9)}}}$*$\displaystyle \frac{1}{{n^{6+6}}}$
$\displaystyle =\frac{1}{{m^{-5}}}$*$\displaystyle \frac{1}{{n^{12}}}$
$\displaystyle =\frac{m^{5}}{n^{12}}$
Keep Smiling
Malay

5. Originally Posted by malaygoel
do you mean
$\displaystyle \frac{1}{({m^2n^3})^2}$*$\displaystyle \frac{1}{({m^{-3}}{n^2})^3}$

Keep Smiling
Malay
Nope.

6. Originally Posted by malaygoel
do you mean
$\displaystyle \frac{1}{({m^2n^3})^2}$*$\displaystyle \frac{1}{({m^{-3}}{n^2})^3}$
it is the same with negative powers
$\displaystyle (m^{-2})^2=m^{-2}*m^{-2}=m^{(-2)+(-2)}=m^{-4}$
hence your expression simplifies to
$\displaystyle \frac{1}{{m^4n^6}}$*$\displaystyle \frac{1}{{m^{-9}}{n^6}}$
$\displaystyle =\frac{1}{{m^4m^{-9}}}$*$\displaystyle \frac{1}{{n^6n^6}}$
$\displaystyle =\frac{1}{{m^{4+(-9)}}}$*$\displaystyle \frac{1}{{n^{6+6}}}$
$\displaystyle =\frac{1}{{m^{-5}}}$*$\displaystyle \frac{1}{{n^{12}}}$
$\displaystyle =\frac{m^{5}}{n^{12}}$
Keep Smiling
Malay
Ahhh I see

Thanks for helping me. I really appreciate it!