# Factoring a polynomial/simplifying

• Nov 21st 2006, 03:15 AM
Cherry
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? :confused:

• Nov 21st 2006, 03:18 AM
malaygoel
Quote:

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? :confused:

the solution is
$\sqrt{8}+\sqrt{18}$
$=\sqrt{2*2*2}+\sqrt{3*3*2}$
$=2\sqrt{2}+3\sqrt{2}$
$=5\sqrt{2}$

Keep Smiling
Malay
• Nov 21st 2006, 03:21 AM
malaygoel
Quote:

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
• Nov 21st 2006, 03:26 AM
malaygoel
Quote:

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
$\frac{1}{({m^2n^3})^2}$* $\frac{1}{({m^{-3}}{n^2})^3}$
it is the same with negative powers
$(m^{-2})^2=m^{-2}*m^{-2}=m^{(-2)+(-2)}=m^{-4}$
$\frac{1}{{m^4n^6}}$* $\frac{1}{{m^{-9}}{n^6}}$
$=\frac{1}{{m^4m^{-9}}}$* $\frac{1}{{n^6n^6}}$
$=\frac{1}{{m^{4+(-9)}}}$* $\frac{1}{{n^{6+6}}}$
$=\frac{1}{{m^{-5}}}$* $\frac{1}{{n^{12}}}$
$=\frac{m^{5}}{n^{12}}$
Keep Smiling
Malay
• Nov 21st 2006, 03:39 AM
Cherry
Quote:

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

Keep Smiling
Malay

Nope.
• Nov 21st 2006, 04:47 AM
Cherry
Quote:

Originally Posted by malaygoel
do you mean
$\frac{1}{({m^2n^3})^2}$* $\frac{1}{({m^{-3}}{n^2})^3}$
it is the same with negative powers
$(m^{-2})^2=m^{-2}*m^{-2}=m^{(-2)+(-2)}=m^{-4}$
$\frac{1}{{m^4n^6}}$* $\frac{1}{{m^{-9}}{n^6}}$
$=\frac{1}{{m^4m^{-9}}}$* $\frac{1}{{n^6n^6}}$
$=\frac{1}{{m^{4+(-9)}}}$* $\frac{1}{{n^{6+6}}}$
$=\frac{1}{{m^{-5}}}$* $\frac{1}{{n^{12}}}$
$=\frac{m^{5}}{n^{12}}$