How do you simplify #(3z-6)/(3z^2-12)#?

1 Answer
Sep 18, 2016

#1/(z+2)#

Explanation:

The first step in simplifying is to factorise the numerator/denominator.

Numerator:

3z - 6 has a #color(blue)"common factor"# 0f 3.

#rArr3z-6=3(z-2)larr" factorised form"#

Denominator:

#3z^2-12" also has a common factor of 3"#

#rArr3z^2-12=3(z^2-4)#

The factor #z^2-4" is a difference of squares"# and, in general is factorised as follows.

#color(red)(bar(ul(|color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))#

now #(z)^2=z^2" and " (2)^2=4rArra=z" and " b=4#

Thus #z^2-4=(z-2)(z+2)#

#rArr3z^2-12=3(z-2)(z+2)larr" in factorised form"#

Transferring these results into the original expression.

#rArr(3(z-2))/(3(z-2)(z+2))#

cancelling common factors on numerator/denominator gives.

#(cancel(3)^1cancel((z-2)))/(cancel(3)^1cancel((z-2))(z+2))=1/(z+2)#