Question 201121

{{{((k^2+10k+24)/(k^2+11k+30))((k^2+5k)/(k^2-3k-28))}}} Start with the given expression.



{{{(((k+6)*(k+4))/(k^2+11k+30))((k^2+5k)/(k^2-3k-28))}}} Factor {{{k^2+10k+24}}} to get {{{(k+6)*(k+4)}}}.



{{{(((k+6)*(k+4))/((k+6)*(k+5)))((k^2+5k)/(k^2-3k-28))}}} Factor {{{k^2+11k+30}}} to get {{{(k+6)*(k+5)}}}.



{{{(((k+6)*(k+4))/((k+6)*(k+5)))((k*(k+5))/(k^2-3k-28))}}} Factor {{{k^2+5k}}} to get {{{k*(k+5)}}}.



{{{(((k+6)*(k+4))/((k+6)*(k+5)))((k*(k+5))/((k+4)*(k-7)))}}} Factor {{{k^2-3k-28}}} to get {{{(k+4)*(k-7)}}}.



{{{(k*(k+6)*(k+4)(k+5))/((k+6)*(k+5)(k+4)*(k-7))}}} Combine the fractions. 



{{{(k*highlight((k+6))highlight((k+4))highlight((k+5)))/(highlight((k+6))highlight((k+5))highlight((k+4))(k-7))}}} Highlight the common terms. 



{{{(k*cross((k+6))cross((k+4))cross((k+5)))/(cross((k+6))cross((k+5))cross((k+4))(k-7))}}} Cancel out the common terms. 



{{{k/(k-7)}}} Simplify. 



So {{{((k^2+10k+24)/(k^2+11k+30))((k^2+5k)/(k^2-3k-28))}}} simplifies to {{{k/(k-7)}}}.



In other words, {{{((k^2+10k+24)/(k^2+11k+30))((k^2+5k)/(k^2-3k-28))=k/(k-7)}}}