Question

For functions $f\left(x\right)=x^3+2x^2-19x+12$and $g\left(x\right)=x-3$.

Find (a)$\left(\frac{f}{g}\right)\left(x\right)$and (b)$\left(\frac{f}{g}\right)\left(0\right)$.

Short answer

Part (a). The solution is $x^2+5x-4$.

Part (b). The solution is$-4$.

Step-by-step solution

Part (a). Step 1. Given information.

The functions$f\left(x\right)=x^3+2x^2-19x+12$and$g\left(x\right)=x-3$.

Part (b). Step 2. To divide the polynomial function by substituting the fx and gx in fgx.

$$\begin{gathered} \left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)} \\ \Rightarrow \frac{x^3+2x^2-19x+12}{x-3} \end{gathered}$$

Part (a). Step 3. Divide the polynomials.

$$\begin{gathered} x-3\begin{array}{ccc}{x}^2& +5x& -4\end{array}\begin{array}{cccc}{x}^3& +2x^2& -19x& +12\\ x^3& -3x^2& & \end{array} \\ \overline{5x^2-19x} \\ 5x^2-15x \\ \overline{-4x+12} \\ -4x+12 \\ \overline{0} \end{gathered}$$

Part (a). Step (4). Simplified answer.

Hence, the obtained result is$x^2+5x-4$.

Part (b). Step 1. Given information.

The functions$f\left(x\right)=x^3+2x^2-19x+12$and$g\left(x\right)=x-3$.

Part (b). Step 2. To find the value of fg0 by substituting the value of x=0 in fgx=x2+5x-4.

Solve it as follows:

$$\begin{gathered} \left(\frac{f}{g}\right)\left(0\right)=0^2+5\left(0\right)-4 \\ \Rightarrow 0+0-4 \\ \Rightarrow -4 \end{gathered}$$

Part (b). Step 3. Simplified answer.

Hence, the simplified expression is$-4$.