We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

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

0
322
1

Factor the expression using the two different techniques listed for Parts 1(a) and 1(b). Feb 24, 2018

### Best Answer

#1
+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.

Feb 24, 2018

### 1+0 Answers

#1
+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