+0

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

0
200
1
+558

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

SamJones  Feb 24, 2018

#1
+2295
+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
+2295
+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