## Algebra 1

$32j^{35}k^{11}$
Given expression : $4j^{2}k^{6}(2j^{11})^{3}k^{5}$ By law of exponents, $(ab)^{n} = a^{n}b^{n}$ Hence, $(2j^{11})^{3} = 2^{3}$ X$( j^{11})^{3} = 8$ X $j^{33}$ Therefore, the given expression = $4$ X $j^{2}$ X $k^{6}$ X $8$ X $j^{33}$ X $k^{5}$ This becomes : $32$ X $j^{2+33}$ X $k^{6+5}$ (since the exponents are added while multipliying with the same base) This becomes : $32$ X $j^{35}$ X $k^{11}$ Final result : $32j^{35}k^{11}$