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: Factor the area expression

Factor \(x^{3}-5x^{2}+3x - 15\) by grouping.
Group as \((x^{3}-5x^{2})+(3x - 15)\).
Factor out \(x^{2}\) from the first group and \(3\) from the second group: \(x^{2}(x - 5)+3(x - 5)=(x^{2}+3)(x - 5)\).

Step2: Use the formula for the area of a rectangle

Since \(area=\text{length}\times\text{width}\), and \(area=(x^{2}+3)(x - 5)\), \(width=(x^{2}+3)\).
Then \(\text{length}=\frac{\text{area}}{\text{width}}=\frac{(x^{2}+3)(x - 5)}{x^{2}+3}\).
Cancel out the common factor \((x^{2}+3)\) (assuming \(x^{2}+3
eq0\), which is true for all real \(x\) as \(x^{2}\geq0\) so \(x^{2}+3\geq3\)).

Answer:

\(x - 5\) (the fourth option)