Length of a rectangle is 4x²+12x and the area of the rectangle is 24x⁴+72x³, what is the width of the rectangle?

NEED ANSWER FAST!!!!

Answer :

The width of rectangle is [tex]6x^2[/tex]

Solution:

Given that,

[tex]\text{Length of rectangle } = 4x^2+12x[/tex]

[tex]\text{Area of rectangle } = 24x^4+72x^3[/tex]

To find: width of rectangle

The area of rectangle is given by formula:

[tex]\text{Area of rectangle } = length \times width[/tex]

Therefore, width is given as:

[tex]width = \frac{\text{Area of rectangle}}{length}[/tex]

Substituting the given values we get,

[tex]width = \frac{24x^4+72x^3}{4x^2+12x}[/tex]

Factor out 24 and [tex]x^3[/tex] from numerator

[tex]width = \frac{24x^3(x+3)}{4x^2+12x}[/tex]

Factor out 4 and x from denominator

[tex]width = \frac{24x^3(x+3)}{4x(x+3)}[/tex]

Cancel the common terms in numerator and denominator

[tex]width = 6x^2[/tex]

Thus the width of rectangle is [tex]6x^2[/tex]

Other Questions