Answer: Perimeter [tex]4x^{2} -6[/tex] or [tex]2(2x^{2} -3)[/tex] units
Area [tex]-12x^{3} - 5x^{2} + 42x -20[/tex] or [tex]2(-6x^{3} -5x^{2} + 21x -20)[/tex] square units
Step-by-step explanation:
A. Perimeter is Twice the length plus twice the length. P =2(l + w)
or P = 2L + 2W . Substitute the values given into the formula and Calculate.
P = 2[(2x + 5)(x-1) + 2(2-3x)] . becomes [tex]2(2x^{2} +3x -5) + 2(2-3x)[/tex],
then [tex]4x^{2} + 6x - 10 - 6x +4[/tex] add like terms and simplify [tex]4x^{2} -6[/tex]
B. Area is Length times Width. Substitute the given values and calculate.
(2x + 5)(x-1)×(2-3x) (2x + 5)(x-1)(-3x+2)
[tex]-12x^{3} - 5x^{2} + 42x -20[/tex]
It is difficult to imagine an actual rectangle with such dimensions. It's math theory.
