Algebra I factoring a^2b^2x^2-18a^2b^2x+81a^2b^2

COCOA1203

New member
Joined
Jun 22, 2007
Messages
7
Please help factoring this trinomial. Instructions say that if the coefficient of the first term is negative, begin by factoring out -1

a^2 b^2 x^2 - 18 a^2 b^2 x + 81 a^2 b^2

Any help would be greatly appreciated!
 
Factor out \(\displaystyle a^{2}b^{2}\)

\(\displaystyle \L\\a^{2}b^{2}(x^{2}-18x+81)\)

Now factor what's in the parentheses and finish up.
 
Thanks. Is it necessary to reposition the order of the numbers inside the parenthesis before factoring that?
 
Yep. Except write it as \(\displaystyle \L\\a^{2}b^{2}(x-9)^{2}\)
 
You could tackle it in the same fashion you factor others with the same variables.

What two numbers when added equal 1 and when multiplied equal -72.

9 and -8?.

\(\displaystyle \L\\12m^{2}+9mn-8mn-6n^{2}\)

Now factor out common factors and finish up.
 
Same principle:

12 * -6 = -72 = 9 * -8

9 + -8 = 1

12m^2 +mn - 6n^2

= 12m^2 + 9mn - 8mn - 6n^2

= 3m(4m + 3n) - 2n(4m + 3n) ....now factor out (4m + 3n)

= (4m + 3n)(3m - 2n)
 
I understand all the steps except for the last two. Can you explain to me what happened to the 3m and -2n that were outside the parenthesis in the next to the last step. Thanks
 
I explained a bit more in my previous post - is it a little clearer?
 
Top