## Algebra 1

$120\sqrt 2x^{3}$
$10 \times \sqrt 4 \times \sqrt (3x^{3}) \times 2 \times \sqrt (6x^{3})$ *** $\sqrt 4$ = 2 because $2 \times 2$ = 4 $10 \times 2 \times 2 \times \sqrt (3x^{3} \times 6x^{3})$ We multiply the constants together and the numbers in the square root rogether $40 \sqrt 18x^{6}$ The factors of $18x^{6}$ are $9 \times 2x^{3}$ $40 \sqrt 9 \sqrt 2x^{3}$ 9 is a perfect square because 3 x 3 = 9 $40 \times 9 \sqrt 2x^{3}$ $120\sqrt 2x^{3}$