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

COCOA1203

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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!
 

galactus

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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.
 

COCOA1203

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Thanks. Is it necessary to reposition the order of the numbers inside the parenthesis before factoring that?
 

galactus

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No. As a matter of fact, it's an easy factor.
 

COCOA1203

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Would it be

a^2b^2(x-9)(x-9)
 

galactus

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Yep. Except write it as \(\displaystyle \L\\a^{2}b^{2}(x-9)^{2}\)
 

galactus

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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.
 

Subhotosh Khan

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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)
 

COCOA1203

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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
 

Subhotosh Khan

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I explained a bit more in my previous post - is it a little clearer?
 

COCOA1203

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Ok, that makes sense. Thanks for your help.
 
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