Question

the area of a rectangle is given by the expression: $21x^{2}-19xy - 12y^{2}$ and the width is $7x + 3y$. find the length of the rectangle: ?

Explanation:

Step1: Recall area formula

The area of a rectangle is $A = l\times w$, where $l$ is length and $w$ is width. So, $l=\frac{A}{w}$. Here, $A = 21x^{2}-19xy - 12y^{2}$ and $w=7x + 3y$.

Step2: Factor the area expression

Factor $21x^{2}-19xy - 12y^{2}$. We need to find two numbers that multiply to $21\times(- 12)=-252$ and add up to $-19$. The numbers are $-28$ and $9$.
So, $21x^{2}-19xy - 12y^{2}=21x^{2}-28xy+9xy - 12y^{2}=7x(3x - 4y)+3y(3x - 4y)=(3x - 4y)(7x + 3y)$.

Step3: Calculate the length

Since $l=\frac{A}{w}$ and $A=(3x - 4y)(7x + 3y)$, $w = 7x+3y$, then $l=\frac{(3x - 4y)(7x + 3y)}{7x + 3y}=3x - 4y$.

Answer:

$3x - 4y$