How do you simplify #((j^2-16)/(j^2+10j+16))/(15/(j+8))#?

1 Answer
Dec 18, 2016

Answer:

#=(1/15)(j-2)(j+4)#

#=1/15j^2+2/15j-8/15#

Explanation:

#frac{(frac{j^2-16}{j^2+10j+16})}{(frac{15}{j+8})}#

Rather than dividing two fractions, use the fact that dividing is the same as multiplying by the reciprocal.
#=(frac{j^2-16}{j^2+10j+16})*(frac{j+8}{15})#

#=frac{(j^2-16)(j+8)}{(j^2+10j+16)(15)}#

Now factor the terms on the numerator and denominator:
#=frac{(j-4)(j+4)(j+8)}{(j+8)(j+2)(15)}#

#=frac{(j-2)(j+2)(j+4)(j+8)}{(j+8)(j+2)(15)}#

Cancel terms that occur on both the numerator and denominator:
#=frac{(j-2)cancel((j+2))(j+4)cancel((j+8))}{cancel((j+8))cancel((j+2))(15)}#

#=frac{(j-2)(j+4)}{15}#
#=(1/15)(j-2)(j+4)#
#=(1/15)(j^2+2j-8)#