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Factor the expression using the two different techniques listed for Parts 1(a) and 1(b).

SamJones  Feb 24, 2018

Best Answer 

 #1
avatar+2190 
+1

a) In order to factor using this method, let's try and identify the GCF first. 9 is the greatest common factor between 36 and 81. a^4 is the greatest common factor between the a's, and b^10 is the factor for the b's. Let's factor it out!
 

\(36a^4b^{10}-81a^{16}b^{20}\) Factor out the GCF, \(9a^4b^{10}\), like I described earlier. 
\(9a^4b^{10}\left(4-9a^{12}b^{10}\right)\) Don't stop here, though! Notice that the resulting binomial is a difference of squares. 
\(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\)  
   

 

b) The beginning binomial is a difference of squares to begin with, so it is possible to start with this first! 

 

\(36a^4b^{10}-81a^{16}b^{20}\) Let's do this approach this time!
\(\left(6a^2b^5+9a^8b^{10}\right)\left(6a^2b^5-9a^8b^{10}\right)\) Don't stop yet! Both binomials have their own GCF's!
\(3a^2b^5\left(2+3a^6b^5\right)*3a^2b^5\left(2-3a^6b^5\right)\) Combine the multiplication.
\(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\)  
   

 

Well, these are the two techniques. 

TheXSquaredFactor  Feb 24, 2018
 #1
avatar+2190 
+1
Best Answer

a) In order to factor using this method, let's try and identify the GCF first. 9 is the greatest common factor between 36 and 81. a^4 is the greatest common factor between the a's, and b^10 is the factor for the b's. Let's factor it out!
 

\(36a^4b^{10}-81a^{16}b^{20}\) Factor out the GCF, \(9a^4b^{10}\), like I described earlier. 
\(9a^4b^{10}\left(4-9a^{12}b^{10}\right)\) Don't stop here, though! Notice that the resulting binomial is a difference of squares. 
\(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\)  
   

 

b) The beginning binomial is a difference of squares to begin with, so it is possible to start with this first! 

 

\(36a^4b^{10}-81a^{16}b^{20}\) Let's do this approach this time!
\(\left(6a^2b^5+9a^8b^{10}\right)\left(6a^2b^5-9a^8b^{10}\right)\) Don't stop yet! Both binomials have their own GCF's!
\(3a^2b^5\left(2+3a^6b^5\right)*3a^2b^5\left(2-3a^6b^5\right)\) Combine the multiplication.
\(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\)  
   

 

Well, these are the two techniques. 

TheXSquaredFactor  Feb 24, 2018

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