Question

The rectangle below has an area of 8x^5+12x^3+20x^28x 5 +12x 3 +20x 2 8, x, start superscript, 5, end superscript, plus, 12, x, cubed, plus, 20, x, squared. The width of the rectangle is equal to the greatest common monomial factor of 8x^58x 5 8, x, start superscript, 5, end superscript, 12x^312x 3 12, x, cubed, and 20x^220x 2 20, x, squared.

Answer 1

The question is incomplete. Here is the complete question.

The rectangle bleow has an area of
`8x^(5)+12x^(3)+20x^(2)`. The width of the rectangle is equal to the greatest common monomial factor of
`8x^(5),12x^(3),20x^(2)`. What is the length and width of the rectangle?

Answer: width =
`4x^(2)`

length =
`2x^(3)+3x+5`

Explanation: Greatest common factor is the largest number that will divide into that number without rest, i.e., it's a number that will result in an exact division. The same can be applied to a polynomial.

To find the greatest common factor:

1) Write each in prime factored form:

2.2.2.x.x.x.x.x + 2.2.3.x.x.x + 2.2.5.x.x

2) Identify the common factor among the terms:

For this polynomial, the repetitive factor is
`4x^(2)`

Therefore, the width of the rectangle is:

w =
`4x^(2)`

Area of a rectangle is the multiplication of width and length, so:

`A=w*l\\l=(A)/(w)`

To calculate length, we will have to divide polynomials:

`l=(8x^(5)+12x^(3)+20x^(2))/(4x^(2))`

`l = 2x^(3)+3x+5`

Width and length of the rectangle are 4x² and
`2x^(3)+3x+5`, respectively.