Question

The area of a rectangle is (12x⁴-8x³-4x²) cm². Its width is
(3x+1) cm. What is the length of the rectangle?​

Answer 1

Answer:

area=(12x^4-8x^3-4x^2)cm^2

=4x^2(3x^2-2x-1) cm^2

=4x^2(3x^2-3x+x-1) cm^2

=4x^2{3x(x-1)+1(x-1)}cm^2

=4x^2(x-1)(3x+1) cm^2

width=(3x+1) cm

length=4x^2(x-1) cm

length of the rectangle is 4x^2(x-1) cm

Answer 2

Answer:

Length of rectangle is $4x^{2} (x-1)$ cm.  

Step-by-step explanation:

Given : Area of a rectangle = $(12x^{4}-8x^{3} -4x^{2} )$ cm²

           Width of rectangle = $(3x+1)$ cm

To find : What is the length of the rectangle?​

Solution :

• It is given that the area of a rectangle = $(12x^{4}-8x^{3} -4x^{2} )$ cm²

           Width of rectangle = $(3x+1)$ cm

• We have to find the area of a rectangle.
• Rectangle has a four sides perpendicular to each other and its opposite sides are equal.
• The formula for area of a rectangle is given by,

                  Area of rectangle = Length*Width

                            $(12x^{4}-8x^{3} -4x^{2} )$  = Length*$(3x+1)$

                              ∴ Length = $\frac{12x^{4}-8x^{3} -4x^{2} }{3x+1}$

                                            = $\frac{4x^{2} (3x^{2} -2x+1)}{3x+1}$

                                            = $\frac{4x^{2} (3x^{2} -3x+x-1)}{3x+1}$

                                            = $\frac{4x^{2} [3x(x-1)+1(x-1)]}{3x+1}$

                                            = $\frac{4x^{2} (3x+1)(x-1)}{3x+1}$

                                            = $4x^{2} (x-1)$

• ∴ Length of rectangle is $4x^{2} (x-1)$ cm.