How do you simplify #( 5a^2 + 20a) /( a^3-2a^2) * ( a^2-a-12) / (a^2-16)#?

2 Answers
Jul 26, 2015

Answer:

#a(5a+20)/a^2(a-2)# . #(a-4)(a+3)/(a-4)^2#

Explanation:

simplyfing the first equation:

having a common factor "a"
a(5a+20)

simplifying The denominator:
having a common factor " #a^2# "
#a^2# (a-2)

Moving to the second equation:

The numerator:
#a^2#-a- 12
This equation cannot be solved by the common factor method, because -12 has no "a".
However, It can be solved by another method:
opening 2 different parenthesis
(a-4).(a+3)

The dominator:
having the power common factor
#(a-4)^2#

Jul 26, 2015

Answer:

By factoring each expression in the numerator(top) and the denominator(bottom) and then cancelling out the commons.

Explanation:

There are #4# expressions. First, each expression must be factored.

Here's how we do it:

#color(red)((1)) 5a^2+20a=a(5a+20)=5a(a+4)#

#color(red)((2)) a^3-2a^2=a^2(a-2)#

#color(red)((3)) a^2-a-12=a^2-4a+3a-12=a(a-4)+3(a-4)=(a+3)(a-4)#

#color(red)((4)) a^2-16=a^2-4^2#

This is an expression of the form : #(A+B)(A-B)=A^2-B^2#

Hence,#color(red)((4)) a^2-16=(a-4)(a+4)#

#=>(5a^2+20a)/(a^3-2a^2)*(a^2-a-20)/(a^2-16)" "# becomes

#(5acolor(red)cancel(color(black)((a+4))))/(a^2(a-2))*(color(green)cancel(color(black)((a-4)))(a+3))/(color(green)cancel(color(black)((a-4))) color(red)cancel(color(black)((a+4))))=(5a(a+3))/(a^2(a-2))=color(blue)((5(a+3))/(a(a-2)))#