the volume of a rectangular prism is $(x^{3}-3x^{2}+5x - 3)$, and the area of its base is $(x^{2}-2)$. if…

Answer:

A. $x - 3+\frac{7x - 9}{x^{2}-2}$

Explanation:

Step1: Recall volume - base - height formula

$V=A\times h$, where $V$ is volume, $A$ is base - area and $h$ is height. So, $h=\frac{V}{A}$.

Step2: Set up the division

We have $V = x^{3}-3x^{2}+5x - 3$ and $A=x^{2}-2$. So, $h=\frac{x^{3}-3x^{2}+5x - 3}{x^{2}-2}$.

Step3: Perform polynomial long - division

Divide $x^{3}-3x^{2}+5x - 3$ by $x^{2}-2$. First, divide the leading term of the numerator $x^{3}$ by the leading term of the denominator $x^{2}$: $\frac{x^{3}}{x^{2}}=x$. Multiply $x^{2}-2$ by $x$: $x(x^{2}-2)=x^{3}-2x$. Subtract: $(x^{3}-3x^{2}+5x - 3)-(x^{3}-2x)=-3x^{2}+7x - 3$. Next, divide the leading term of $-3x^{2}+7x - 3$ (which is $-3x^{2}$) by the leading term of $x^{2}-2$ (which is $x^{2}$): $\frac{-3x^{2}}{x^{2}}=-3$. Multiply $x^{2}-2$ by $-3$: $-3(x^{2}-2)=-3x^{2}+6$. Subtract: $(-3x^{2}+7x - 3)-(-3x^{2}+6)=7x - 9$. The quotient is $x - 3$ and the remainder is $7x - 9$. So, $h=x - 3+\frac{7x - 9}{x^{2}-2}$.