Question

If the area of the rectangle is 35x^2-53x-18 and its length is 7x+2 inches, find its width

Answer 1

Answer:

5x - 9

Step-by-step explanation:

The area of the rectangle is:

$A=35x^2 -53x-18$ (1)

The area of the rectangle is the product of length (L) and width (W):

$A=LW$

where $L=7x+2$, and the width must be in the form $W=ax+b$.

We need to find the values of a and b. If we calculate the product of L and W, we get:

$A=LW=(7x+2)(ax+b)=7ax^2 + 7bx+2ax+2b = 7ax^2 +(7b+2a)x+2b$

We know that this equation must be equivalent to (1), so we immediately see that:

- the coefficient of the second order term, $x^2$, must be 35, so

$7a=35\\a=\frac{35}{7}=5$

- the zero-order term must be equal to -18, so we have

$2b=-18\\b=-\frac{18}{2}=-9$

- We can verify that using a=5 and b=-9, the coefficient of the first-order term corresponds to -53:

$7b+2a=7(-9)+2\cdot 5=-63+10=-53$

So, the width is

W = 5x - 9