Question

A rectangle has an area of 14y^3-35y.

Answer 1

A) The expressions for the length and the width where one dimension is the GCF is:

$Length = 7y\\\\Width = 2y^2 - 5\\\\Or\\\\Length = 2y^2 - 5\\\\Width = 7y$

B) Yes, the rectangle exist with given dimension if y = 2

Solution:

Given that,

A rectangle has an area of:

$Area = 14y^3-35y$

Factor out the greatest common factor

Find GCF of 14 and 35

The factors of 14 are: 1, 2, 7, 14

The factors of 35 are: 1, 5, 7, 35

Then the greatest common factor is 7

Find the GCF of variables

$GCF\ of\ y^3\ and\ y = y$

Therefore,

Factor out 7y

$Area = 7y(2y^2 - 5)$ --------- eqn 1

The area of rectangle is given as:

$Area = length \times width$

On comparing eqn 1 with above,

$Length = 7y\\\\Width = 2y^2 - 5\\\\Or\\\\Length = 2y^2 - 5\\\\Width = 7y$

If, y = 2 we have:

$Area = 7y(2y^2 - 5)$

$Area = 7 \times 2(2(2)^2 - 5)\\\\Area = 7 \times 2(3)\\\\Area = 42$

Thus the the rectangle exist with the given dimension if y=2