Question

the area of a rectangle is $(x^{3}-5x^{2}+3x - 15)$, and the width of the rectangle is $(x^{2}+3)$. if area = length × width, what is the length of the rectangle?
$x + 5$
$x-15$
$x + 15$
$x - 5$

Explanation:

Step1: Recall the formula for length.

Since area = length × width, then length = $\frac{\text{area}}{\text{width}}$. Given area = $x^{3}-5x^{2}+3x - 15$ and width = $x^{2}+3$, so length = $\frac{x^{3}-5x^{2}+3x - 15}{x^{2}+3}$.

Step2: Factor the numerator by grouping.

Group the terms of the numerator: $(x^{3}-5x^{2})+(3x - 15)=x^{2}(x - 5)+3(x - 5)=(x^{2}+3)(x - 5)$.

Step3: Simplify the fraction.

Substitute the factored - form of the numerator into the length formula: $\frac{(x^{2}+3)(x - 5)}{x^{2}+3}$. Cancel out the common factor $(x^{2}+3)$ in the numerator and the denominator. The result is $x - 5$.

Answer:

D. $x - 5$