Given:
Consider the dimensions of the rectangle are [tex](x^2+7x-9)[/tex] and [tex](3x^2-2x)[/tex].
To find:
The perimeter in terms of x, of the rectangle.
Solution:
Let the length of the rectangle be [tex](x^2+7x-9)[/tex] and the width of the rectangle is [tex](3x^2-2x)[/tex] units.
The perimeter of a rectangle is:
[tex]P=2(l+w)[/tex]
Where, l is the length and w is the width of the rectangle.
Substituting [tex]l=(x^2+7x-9)[/tex] and [tex]w=(3x^2-2x)[/tex] in the above formula, we get
[tex]P=2((x^2+7x-9)+(3x^2-2x))[/tex]
[tex]P=2(4x^2+5x-9)[/tex]
[tex]P=2(4x^2)+2(5x)+2(-9)[/tex]
[tex]P=8x^2+10x-18[/tex]
Therefore, the perimeter of the rectangle is [tex]8x^2+10x-18[/tex] units.
