How do you simplify #(25-x^2)/12 *(6x^2)/(5-x)#?

1 Answer
Sheffy
Apr 27, 2016

#=(x^2(5+x))/2#

Explanation:

#(25-x^2)/12xx(6x^2)/(5-x)#

Here, #25-x^2# can be written as #5^2-x^2# .

This is of the form #a^2-b^2#

And we know that #a^2-b^2=(a-b)(a+b)#

Therefore, we have #5^2-x^2=(5-x)(5+x)#

Back to the given expression:

#(25-x^2)/12xx(6x^2)/(5-x) #

#= ((5-x)(5+x))/12xx(6x^2)/(5-x)#

#(5-x)# is common to the numerator and the denominator, and is cancelled out.

#=(5+x)/12xx6x^2#

Next, we know that #2xx 6 = 12#.

#=(5+x)/(2xx6)xx6x^2#

Cancel 6 from the numerator and denominator.

#=(5+x)/2xxx^2#

#=(x^2(5+x))/2#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1043 views around the world
You can reuse this answer
Creative Commons License