Question

determine the expression for the width of a rectangle
the area, a, of a rectangle is 120x² + 78x − 90, and the length, l, of the rectangle is 12x + 15. which of the following gives the width, w, of the rectangle?
options:
10x − 19
9x + 4
8x − 6
10x − 6

Explanation:

Step1: Recall area formula for rectangle

The area \( A \) of a rectangle is given by \( A = l \times w \), where \( l \) is the length and \( w \) is the width. So, to find the width, we can use \( w=\frac{A}{l} \).

Step2: Substitute given \( A \) and \( l \)

We have \( A = 120x^{2}+78x - 90 \) and \( l=12x + 15 \). So we need to perform polynomial division or factor both numerator and denominator to simplify \( \frac{120x^{2}+78x - 90}{12x + 15} \).

First, let's factor the numerator and the denominator:

Factor out a common factor from the numerator: \( 120x^{2}+78x - 90 = 6(20x^{2}+13x - 15) \)

Factor the denominator: \( 12x + 15=3(4x + 5) \)

Now, factor the quadratic \( 20x^{2}+13x - 15 \). We need two numbers \( a \) and \( b \) such that \( a\times b=20\times(- 15)=-300 \) and \( a + b = 13 \). The numbers are \( 25 \) and \( - 12 \).

So, \( 20x^{2}+25x-12x - 15=5x(4x + 5)-3(4x + 5)=(5x - 3)(4x + 5) \)

So the numerator \( 6(20x^{2}+13x - 15)=6(5x - 3)(4x + 5) \)

The denominator is \( 3(4x + 5) \)

Now, divide: \( \frac{6(5x - 3)(4x + 5)}{3(4x + 5)} \)

Cancel out the common factors \( 3 \) and \( (4x + 5) \) (assuming \( 4x+5
eq0 \)):

\( \frac{6}{3}\times(5x - 3)=2(5x - 3)=10x-6 \)

Answer:

\( 10x - 6 \) (the option with \( 10x - 6 \))