Given that the length of a parallelogram is $(x^{2} - 5 x)$ $(x^2-5x)$ and the width is $(2 x^{2} + x + 3)$ $(2x^2+x+3)$ , how do you find the area of the figure in terms of x?

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Answer 1 · 2016-11-23T14:15:46

$(x^{2} - 5 x) (2 x^{2} + x + 3)$ $(x^2-5x)(2x^2+x+3)$

Area = length X width

$= (x^{2} - 5 x) (2 x^{2} + x + 3)$ $=(x^2-5x)(2x^2+x+3)$

I understand width as height.

Given that the length of a parallelogram is (x^2-5x)(x2−5x)(x^2-5x) and the width is (2x^2+x+3)(2x2+x+3)(2x^2+x+3), how do you find the area of the figure in terms of x?