Answer:
[tex]\dfrac{5}{k+7}[/tex]
Step-by-step explanation:
Given multiplication:
[tex]\dfrac{k^2+7k+12}{k^2+10k+21}\cdot \dfrac{15}{3k+12},\quad k\neq -3,-4,-7[/tex]
Begin by factoring the numerator and denominator of the first fraction:
[tex]\textsf{Numerator:}\\\\k^2+7k+12\\\\k^2+4k+3k+12\\\\k(k+4)+3(k+4)\\\\(k+3)(k+4)[/tex]
[tex]\textsf{Denominator:}\\\\k^2+10k+21\\\\k^2+7k+3k+21\\\\k(k+7)+3(k+7)\\\\(k+3)(k+7)[/tex]
Therefore, the multiplication can be rewritten as:
[tex]\dfrac{(k+3)(k+4)}{(k+3)(k+7)}\cdot \dfrac{15}{3k+12}[/tex]
Now, factor the denominator of the second fraction:
[tex]\dfrac{(k+3)(k+4)}{(k+3)(k+7)}\cdot \dfrac{15}{3(k+4)}[/tex]
Divide 15 by 3:
[tex]\dfrac{(k+3)(k+4)}{(k+3)(k+7)}\cdot \dfrac{5}{(k+4)}[/tex]
Multiply the numerators and denominators:
[tex]\dfrac{5(k+3)(k+4)}{(k+3)(k+7)(k+4)}[/tex]
Cancel the common factors (k + 3) and (k + 4):
[tex]\dfrac{5}{k+7}[/tex]
Therefore, the solution to the multiplication is:
[tex]\Large\boxed{\boxed{\dfrac{5}{k+7}}}[/tex]