The rectangle below has an area of 12x^4+6x^3+15x^212x 4 + 6x 3 + 15x 2 12, x, start superscript, 4, end superscript, plus, 6, x, start superscript, 3, end superscript, plus, 15, x, start superscript, 2, end superscript square meters. The width of the rectangle (in meters) is equal to the greatest common monomial factor of 12x^4, 6x^3,12x 4,6x 3,12, x, start superscript, 4, end superscript, comma, 6, x, start superscript, 3, end superscript, comma and 15x^215x 2 15, x, start superscript, 2, end superscript. What is the length and width of the rectangle?

Question

Allison Deleon

Mathematics

25 October, 09:39

The rectangle below has an area of 12x^4+6x^3+15x^212x 4 + 6x 3 + 15x 2 12, x, start superscript, 4, end superscript, plus, 6, x, start superscript, 3, end superscript, plus, 15, x, start superscript, 2, end superscript square meters. The width of the rectangle (in meters) is equal to the greatest common monomial factor of 12x^4, 6x^3,12x 4,6x 3,12, x, start superscript, 4, end superscript, comma, 6, x, start superscript, 3, end superscript, comma and 15x^215x 2 15, x, start superscript, 2, end superscript. What is the length and width of the rectangle?

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Answers (1)

Know the Answer?

Larry Mckay · Answer 1

Width = (3x^2) [m]

Length = (4x^2 + 2x + 5) [m]

Step-by-step explanation:

Area of the rectangle

Area = width * length

Area = 12x^4+6x^3+15x^2 [m^2]

The width of the rectangle (in meters) is equal to the greatest common monomial factor of the area.

If we factor the equation, we get

12x^4+6x^3+15x^2 = (3x^2) * (4x^2 + 2x + 5)

This means that,

Width = (3x^2) [m]

Length = (4x^2 + 2x + 5) [m]