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

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?\n$x + 5$\n$x-15$\n$x + 15$\n$x-5$

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?\n$x + 5$\n$x-15$\n$x + 15$\n$x-5$

Answer

Explanation:

Step1: Recall the area formula

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

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

Step3: Simplify the fraction

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

Answer:

D. $x - 5$