The length and width of the rectangle that has an area of [tex]55x^6 + 22x^4[/tex] square meters are [tex]11 x^4[/tex] and [tex](5 x^2 + 2)[/tex] respectively.
Area of the rectangle [tex]= 55x^6 + 22x^4[/tex] square meters.
The width of the rectangle (in meters) is equal to the greatest common monomial factor of [tex]55x^6[/tex] and [tex]22x^4[/tex].
Take out the greatest common monomial factor from [tex]55x^6[/tex] and [tex]22x^4[/tex].
[tex]55x^6 = 5 \times 11 \times x \times x \times x \times x \times x \times x[/tex]
[tex]22x^4 = 2 \times 11 \times x \times x \times x \times x[/tex]
Of the two terms, the common factors are: [tex]11 \times x \times x \times x \times x[/tex].
Area of the rectangle [tex]= 55x^6 + 22x^4[/tex]
Area of the rectangle [tex]= (5 \times 11 \times x \times x \times x \times x \times x \times x) + (2 \times 11 \times x \times x \times x \times x)[/tex]
Area of the rectangle [tex]= 11 \times x \times x \times x \times x (5 \times x \times x + 2)[/tex]
Area of the rectangle [tex]= 11 x^4 (5 x^2 + 2)[/tex]
Area of a rectangle [tex]=[/tex] width [tex]\times[/tex] length
Now, the width of the rectangle (in meters) is equal to the greatest common monomial factor of [tex]55x^6[/tex] and [tex]22x^4[/tex].
So, width [tex]= 11 x^4[/tex] and length [tex]= (5 x^2 + 2)[/tex].
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