## Precalculus: Mathematics for Calculus, 7th Edition

The dimensions of the box: - Height: $6$ ft - Length: $15$ ft - Width: $2$ ft
$Volume = 180$, $x>0$ $x(x-4)(x+9)=180$ $x(x^{2}+9x-4x-36)=180$ $x(x^{2}+5x-36)=180$ $x^{3}+5x^{2}-36x-180=0$ $x^{2}(x+5)-36(x+5)=0$ $(x+5)(x^{2}-36)=0$ $x+5=0$ or $x^{2}-36=0$ With $x+5=0$, so $x=-5$ With $x^{2}-36=0$, so $x=6$ or $x=-6$ Because $x>0$, so $x=6$ Therefore, the dimensions of the box: - $Height=x=6$ ft - $Length=x+9=6+9=15$ ft - $Width=x-4=6-4=2$ ft