## College Algebra (10th Edition)

$(5$m$)\times(8$m$).$
If the sides are x and y, then the area is $xy=40.$ We can express y as $y=\displaystyle \frac{40}{x}$, because x is not 0. The perimeter is $\ \ 2x+2y=26$, into which we substitute $y=\displaystyle \frac{40}{x}$, $2x+2\displaystyle \cdot\frac{40}{x}=26$ and multiply both sides with x. $2x^{2}+80=26x$ Divide with 2 and write in standard form $x^{2}+40=13x$ $x^{2}-13x+40=0$ We can factor the LHS, as -8 and -5 are factors of +40 whose sum is -13.... $(x-5)(x-8)=0$ So, x can be 5 or 8. If x is 5, then $y=\displaystyle \frac{40}{5}=8$, and if x =8, then $y=\displaystyle \frac{40}{8}=5.$ Either way, the dimensions of the rectangle are $(5$m$)\times(8$m$).$