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a rectangle has an area of $45x^{2}-42x - 48$ and a width of $5x - 8$. …

Question

a rectangle has an area of $45x^{2}-42x - 48$ and a width of $5x - 8$. what is the length of the rectangle?
to find the length of the rectangle, use $45x^{2}-42x - 48$ as the dropdown and $5x - 8$ as the dropdown.
dropdown options: divisor, dividend, quotient

Explanation:

Step1: Recall area formula

The area of a rectangle is $A = l\times w$, where $A$ is area, $l$ is length and $w$ is width. So, $l=\frac{A}{w}$. Here, $A = 45x^{2}-42x - 48$ (dividend) and $w=5x - 8$ (divisor).

Step2: Perform polynomial long - division

We divide $45x^{2}-42x - 48$ by $5x - 8$.
First, divide the leading term of the dividend $45x^{2}$ by the leading term of the divisor $5x$: $\frac{45x^{2}}{5x}=9x$.
Multiply $5x - 8$ by $9x$: $9x(5x - 8)=45x^{2}-72x$.
Subtract this from the dividend: $(45x^{2}-42x - 48)-(45x^{2}-72x)=45x^{2}-42x - 48 - 45x^{2}+72x = 30x-48$.
Then divide the leading term of the new dividend $30x$ by the leading term of the divisor $5x$: $\frac{30x}{5x}=6$.
Multiply $5x - 8$ by $6$: $6(5x - 8)=30x-48$.
Subtract this from the new dividend: $(30x - 48)-(30x - 48)=0$.
So, the quotient (length) is $9x + 6$.

Answer:

$9x + 6$